Solutions for Complex Calculus: Mathematical Methods for Physics and Engineering - Volume 1S 9781793016805, 1793016801

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Mathematical Methods for Physics and Engineering Volume 1S

Solutions for Complex Calculus

Solutions for Complex Calculus

Jorge L. deLyra

c 2018–2019 Jorge L. deLyra Copyright

Feedback to the author via [email protected]

Translated from the Portuguese by M. A. S. deLyra Translation revised by the author

ISBN: 978-1793016805

Version 1.1 (15Feb19)

QFTL JLDL

Dedicated to Cida, for all her support, all her patience and her constant company through all these years.

Preface This book contains detailed solutions of all the problems proposed at the end of the chapters of the corresponding volume of the series of textbooks “Mathematical Methods for Physics and Engineering”. The solution were elaborated by the author himself, they are detailed and commented, and many times constitute extensions or variations of the material covered in the text. The set of problems proposed in the text should be considered as an integral part of the text, since the best use of the text presupposes that the reader does work at the problems. In this volume the problems are presented and solved in the order in which they appear in the corresponding volume of the text, and are organized sequentially in chapters, according to the chapters of the text. The original propositions of the problems are repeated in this volume, thus making it reasonably (but not completely) self-contained, and the problems are numbered exactly as in the text. The problems are of many different types, with varying levels of difficulty and complexity. They include many things, from simple examples and calculations used as reinforcement exercises, going through problems that complete and extend the material presented in the text, and proceeding as far as to include theorem proofs not given in the text. The most difficult problems, some of which extend significantly the content of the text, as marked as “challenge problems”. In some cases the results of the problems are used in subsequent parts of the text, and such problems are marked as “reference problems”. In this way, this companion volume to the corresponding volume of the original text can serve not only as a support asset to the teachers that use it in their courses, but also as extended material for students who are interested in extending and deepening their knowledge of the subject. This is Volume 1S, which is dedicated to the complex calculus. It includes 117 problems with complete solutions, organized according to the 16 chapters of the corresponding volume of the text [1]. vii

viii

Acknowledgements The author would like to thank the contributions of the several teaching assistants that acted in the courses on Mathematical Physics under his supervision, as well as to the many students that took the courses, throughout the years of development of this text, and that in a way played the role of guinea pigs during that development. Without such able and willing test subjects, it would not have been possible to develop the text to the considerable extent which was achieved. Many students and teaching assistants contributed by finding errors in the text and in the equations, both in the text itself and in the solutions to the problems, but the main contribution of the students is always to read and use the text with a critical mind-set, as well as to ask intelligent and challenging questions, of which there were many. This certainly contributed to add both quality and scope to the text.

Contents 1 Number, the Language of Science

1

2 The Simplicity of Complex Numbers

15

3 Elementary Functions, but Not Quite

35

4 Even Less Elementary Functions

47

5 Geometrical Aspects of the Functions

63

6 Border Effects in Capacitors

75

7 Complex Calculus I: Differentiation

95

8 Complex Calculus II: Integration

109

9 Complex Derivatives and Integrals

137

10 Complex Inequalities and Series

151

11 Series, Limits and Convergence

169

12 Representation of Functions by Series

189

13 Convergence Criteria and Proofs

213

14 Laurent Series and Residues

223

15 Calculation of Integrals by Residues

231

16 Residues on Riemann Surfaces

281

ix

x

Solutions 1

Number, the Language of Science We present here complete and commented solutions to all problems proposed in Chapter 1 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. Consider the set I of the integers. Show that it does not satisfy some of the 11 properties of a field of numbers, among those that do not relate only to the operation of addition. Complete Solution: Since I is contained in R (I ⊂ R), both share the same arithmetic rules, and R satisfies all the 11 properties, it follows that I will satisfy all those involving only the integers. Thus we see that only the questions related to the closure need to be analyzed in detail. Since the sum and the product of integers produce other integers, the two operations are closed in I. It remains to be seen whether the identity elements and the inverse elements are in I. Since 0 and 1 are integers, the identity elements are in I. Since I includes the negative integers, the inverse of the sum presents no problems. Therefore, the only property that fails is the existence of the inverse for the product. Problem 2. Consider the set Q of the rational numbers. Show that it satisfies all the 11 properties of a field of numbers. Complete Solution: 1

2

SOLUTIONS 1

Since Q is contained in R (Q ⊂ R), both share the same arithmetic rules, and R satisfies all the 11 properties, it follows that Q will satisfy all those that involve only the rational numbers. Thus we see that only the questions related to the closure need to be analyzed in detail. Since I ⊂ Q and we have already seen in a previous problem that for I the only property that does not hold is the existence of the inverse of the product, it is enough to look at this property. Since by definition Q is the set of numbers of the form p/q with p and q integers (p ∈ I and q ∈ I), and with q 6= 0, we see that it was built in such a way that any non-zero element has a multiplicative inverse. If we have the element p/q with q 6= 0 by definition, and p 6= 0 so that zero is excluded, then the inverse element is given by q/p with p 6= 0, which by definition is in Q. Thus we see that Q, just as R does, satisfies all the 11 properties. Problem 3. Show that Q is a dense subset of R, considering the integers, the arithmetic averages of pairs of these numbers, the arithmetic averages of pairs of the resulting numbers, and so on ad infinitum. Complete Solution: In order to show that Q is dense in R, we will show that there is no real interval (x−δ/2, x+δ/2) around a real number x, of non-zero length δ, within which there is no rational number. Let us do this by reductio ad absurdum, imagining for a moment that there is such an interval, with δ > 0. If it exists, then it certainly does not contain any integer numbers, since the integers are rational. Let us consider the integer q0− which is immediately to the left of the interval, and the integer q0+ which is immediately to the right of the interval. These two numbers form another interval which contains the interval
(x − − + δ/2, x + δ/2). We have, of course, for the size of the interval q , q +
0 0 , that q − q − = 1. Let us now consider the number q = q + + q − /2, which is 0 0 0 0 rational, since q0+ and q0− are rational and the rationals form a field. This
number is the midpoint of the interval q0− , q0+ , and it is therefore contained therein. This number cannot fall within the interval (x−δ/2, x+δ/2), because this would violate the hypothesis on this interval. Therefore it must fall to the left of the interval, between q0− and x− δ/2, or to the right of the interval, between x + δ/2 and q0+ . One and only one of these two cases can be realized. In the first case, let us define the new points q1+ = q0+ and q1− = q, so that q1+ is to the right on the interval (x − δ/2, x + δ/2), while
q1− is to the left, and so these two new points form a new interval q1− , q1+ containing the interval (x − δ/2, x + δ/2). Similarly, in the second case let us define the

NUMBER, THE LANGUAGE OF SCIENCE

3

new points q1+ = q and q1− = q0− , so that q1+ is to the right of the interval (x − δ/2, x + δ/2),
while q1− is to the left, and so these two new points form an interval q1− , q1+ containing the interval (x − δ/2, x + δ/2).
Whatever the case may be, these two points form a new interval q1− , q1+ that contains (x − δ/2, x + δ/2) and whose length is 1/2, that is, half the length of the previous interval. If we repeat this procedure indefinitely, we generate a sequence of intervals − (qn , qn+ ), for n ∈ {0, 1, 2, 3, . . . , ∞}, each nested into the previous one, whose lengths are given by 1/2n , and all of which contain the interval (x − δ/2, x + δ/2), which has a length δ > 0. However, in the limit n → ∞ we have that 1/2n → 0, so that there certainly is some value of n for which the interval (x − δ/2, x + δ/2) does not fit within the interval (qn− , qn+ ). This is therefore absurd, because by construction all the intervals in this sequence contain (x − δ/2, x + δ/2). It follows that there can be no real interval (x − δ/2, x + δ/2), with δ > 0, containing no rational number. We also see that, given a real number x and a distance ǫ > 0, there is always a rational number at a distance less than ǫ = δ/2 from x. Problem 4. Consider a regular polygon with N sides, in which the distance from the center to a vertex is 1. (a) Calculate the perimeter of the polygon as a function of N . (b) Show that, in the limit N → ∞, the perimeter approaches the value 2π. Complete Solution: Consider the unit circle with an inscribed regular polygon, as illustrated in the diagram of Figure 1.1. The complete angle 2π is divided into N equal parts, so that we have θ = 2π/N . (a) The perimeter of P the polygon is given by P = N ℓ, where we have, by elementary trigonometry, ℓ 2 P

  θ = sin ⇒ 2 π . = 2N sin N

4

SOLUTIONS 1 θ = 2π N

1 θ/2

ℓ/2

ℓ

Figure 1.1: The unit circle with the inscribed regular N -polygon. (b) In the limit N → ∞ the argument of the sine approaches zero. Using the fact that sin(α) = 1, α→0 α lim

we have that π N = 1 ⇒ sin N →∞ π N P lim = 1. N →∞ 2π lim

We have therefore that, in the N → ∞ limit, P = 2π. Problem 5. Consider the complex numbers defined as ordered pairs of real numbers, and the definitions of the operations of addition and multiplication on them. We call this structure C. (a) Show that (0, 0) is the identity element of the addition and that (1, 0) is the identity element of the multiplication. (b) Verify that the complex numbers and the two operations satisfy all the 11 properties of a field of numbers. (c) Verify that C reduces to the field R of the real numbers for the subset of the complex numbers for which the second element of the ordered pair is zero.

NUMBER, THE LANGUAGE OF SCIENCE

5

(d) Consider the complex number (0, 1), which is a very special case. Show that (0, 1)2 = (−1, 0). Complete Solution: Since each complex number z is an ordered pair of real numbers (x, y), all proofs are made by reduction to the corresponding properties of real numbers. It is thus necessary to use the definitions of the two operations for complex numbers. If we have that z1 = (x1 , y1 ) and z2 = (x2 , y2 ) then the sum is given by z1 + z2 = (x1 + x2 , y1 + y2 ), and the product is given by z1 z2 = (x1 x2 − y1 y2 , x1 y2 + x2 y1 ). (a) In the case of the sum, making x1 = 0 and y1 = 0, we have that z1 + z2 = (0 + x2 , 0 + y2 ), and therefore, using the properties of real numbers, we conclude that z1 + z2 = (x2 , y2 ) = z2 . It follows that (0, 0) is the identity element of the sum. In the case of the product, making x1 = 1 and y1 = 0, we have that z1 z2 = (x2 − 0 y2 , y2 + 0 x2 ), and therefore, using the properties of real numbers, we conclude that z1 z2 = (x2 , y2 ) = z2 . It follows that (1, 0) is the identity element of the product. (b) Let us consider each of the properties in turn. Operation of Addition: Closure: by construction, the sum of two complex numbers is a complex number, that is, an ordered pair of real numbers. Existence of the identity element: as shown above, there is a complex number (0, 0) which is the identity element of the sum. Existence of the inverse: given an arbitrary complex number (x, y), we can construct the complex number (−x, −y), since the real numbers are a field. Using the definition of the sum, we have that (x, y) + (−x, −y) = (0, 0), thus showing that (−x, −y) is the additive inverse of (x, y). Commutativity: it suffices to reduce the problem to the use of the commutativity of the sum for the real numbers, z1 + z2 = (x1 + x2 , y1 + y2 ) = (x2 + x1 , y2 + y1 ) = z2 + z1 .

6

SOLUTIONS 1 Associativity: it suffices to reduce the problem to the use of the associativity of the sum for the real numbers, z1 + (z2 + z3 ) = z1 + (x2 + x3 , y2 + y3 ) = (x1 + [x2 + x3 ], y1 + [y2 + y3 ]) = ([x1 + x2 ] + x3 , [y1 + y2 ] + y3 ) = (x1 + x2 , y1 + y2 ) + z3 = (z1 + z2 ) + z3 . Operation of Multiplication: Closure: by construction, the product of two complex numbers is a complex number, that is, an ordered pair of real numbers. Existence of the identity element: as shown above, there is a complex number (1, 0) which is the identity element of the product. Existence of the inverse: given the non-zero complex number (x, y), that is, one for which we do not have that x = y = 0, we can construct the complex number   −y x , . x2 + y 2 x2 + y 2 Using then the definition of the product, we have that   x −y (x, y) , x2 + y 2 x2 + y 2   x (−y) −y x = x 2 −y 2 ,x +y 2 x + y2 x + y 2 x2 + y 2 x + y2  2  x + y 2 −xy + yx = , x2 + y 2 x2 + y 2 = (1, 0), thus showing that the number we have constructed is the multiplicative inverse of (x, y). Commutativity: it suffices to reduce the problem to the use the commutativity of the product for the real numbers, z1 z2 = (x1 x2 − y1 y2 , x1 y2 + x2 y1 )

= (x2 x1 − y2 y1 , x2 y1 + x1 y2 ) = z2 z1 .

NUMBER, THE LANGUAGE OF SCIENCE

7

Associativity: it suffices to reduce the problem to the use of the associativity of the product for the real numbers, z1 (z2 z3 ) = z1 (x2 x3 − y2 y3 , x2 y3 + x3 y2 )

= (x1 [x2 x3 − y2 y3 ] − y1 [x2 y3 + x3 y2 ]

, x1 [x2 y3 + x3 y2 ] + y1 [x2 x3 − y2 y3 ])

= (x1 [x2 x3 ] − x1 [y2 y3 ] − y1 [x2 y3 ] − y1 [x3 y2 ]

, x1 [x2 y3 ] + x1 [x3 y2 ] + y1 [x2 x3 ] − y1 [y2 y3 ])

= (x1 [x2 x3 ] − x1 [y2 y3 ] − y1 [x2 y3 ] − y1 [y2 x3 ]

, x1 [x2 y3 ] + x1 [y2 x3 ] + y1 [x2 x3 ] − y1 [y2 y3 ])

= ([x1 x2 ]x3 − [x1 y2 ]y3 − [y1 x2 ]y3 − [y1 y2 ]x3

, [x1 x2 ]y3 + [x1 y2 ]x3 + [y1 x2 ]x3 − [y1 y2 ]y3 )

= ([x1 x2 ]x3 − [y1 y2 ]x3 − [x1 y2 ]y3 − [y1 x2 ]y3

, [x1 x2 ]y3 − [y1 y2 ]y3 + [x1 y2 ]x3 + [y1 x2 ]x3 )

= ([x1 x2 − y1 y2 ]x3 − [x1 y2 + y1 x2 ]y3

, [x1 x2 − y1 y2 ]y3 + [x1 y2 + y1 x2 ]x3 )

= (x1 x2 − y1 y2 , x1 y2 + x2 y1 )z3

= (z1 z2 )z3 .

Distributivity of the product with respect to the sum: It suffices to reduce the problem to the use of the distributivity for the real numbers, z1 (z2 + z3 ) = z1 (x2 + x3 , y2 + y3 ) = (x1 [x2 + x3 ] − y1 [y2 + y3 ]

, x1 [y2 + y3 ] + y1 [x2 + x3 ])

= (x1 x2 + x1 x3 − y1 y2 − y1 y3

, x1 y 2 + x1 y 3 + y 1 x2 + y 1 x3 )

= ([x1 x2 − y1 y2 ] + [x1 x3 − y1 y3 ]

, [x1 y2 + y1 x2 ] + [x1 y3 + y1 x3 ])

= ([x1 x2 − y1 y2 ], [x1 y2 + y1 x2 ]) + = z1 z2 + z1 z3 .

+([x1 x3 − y1 y3 ], [x1 y3 + y1 x3 ])

(c) Since we have that z = (x, y), it is clear that the subset z = (x, 0) is identical to the set of the real numbers. It only remains to show that the

8

SOLUTIONS 1 operations between these numbers are reduced to the usual operations with real numbers. If we have z1 = (x1 , y1 ) and z2 = (x2 , y2 ), making y1 = 0 and y2 = 0 we have, starting from the definition of the sum, z1 + z2 = (x1 + x2 , 0 + 0), which is a real number given by x = x1 + x2 , which is therefore the usual real operation of addition. In the case of the product we have z1 z2 = (x1 x2 − 0 × 0, x1 × 0 + x2 × 0), which is a real number given by x = x1 x2 , which is therefore, once again, the usual real operation of multiplication. (d) Directly using the definition of the product, for ı = (0, 1) multiplied by itself, we have, making x = 0 and y = 1, (0, 1)(0, 1) = (0 − 1, 0 + 0), which is the real number (−1, 0). It follows therefore that we do have that ı2 = −1.

Problem 6. Consider the complex field C. Find a representation of this field by 2 × 2 real matrices, through the steps below. (a) Find a 2 × 2 real matrix whose square is the negative of the identity 2 × 2 real matrix. (b) Identifying the complex number (1, 0) with the 2 × 2 identity matrix and the complex number (0, 1) with the matrix found above, write an arbitrary complex number z = (x, y) in terms of these matrices. (c) Write the arbitrary complex number z = (x, y) as a real 2 × 2 matrix, and verify explicitly that this new representation preserves the form of the arithmetic operations on complex numbers. Complete Solution: As a vector space, the complex numbers can be described in terms of a basis of two elements, the real unit (1, 0) and the imaginary unit (0, 1). In terms of

9

NUMBER, THE LANGUAGE OF SCIENCE

real 2 × 2 matrices, it is reasonable to expect that the real unit is represented by the identity matrix. It remains then for us to find a matrix representation for the imaginary unit. (a) One must keep in mind that the solution to this problem may very well not be unique. The condition that a generic 2 × 2 matrix has as its square the negative of the identity matrix is





a b c d



a b c d



a b c d

a2 + bc b(a + d) c(a + d) d2 + bc

2





= = =







−1 0 0 −1

−1 0 0 −1

−1 0 0 −1

a2 + bc = −1,







⇒ ⇒ ⇒

d2 + bc = −1,

b(a + d) = 0, c(a + d) = 0.

Considering the four equations which determine the coefficients, from the first two it can be concluded that a2 and d2 must have the same value. Moreover, since these quantities are positive, and the righthand sides of these equations are negative, it is necessary that neither b nor c be zero, otherwise it would not be possible to satisfy these two equations. If we now consider the last two equations, which should result in zero, since neither b nor c are zero, it is necessary that a + d be zero, that is, d = −a. With this, only an equation remains to be satisfied, a2 + bc = −1. Since we still have three unknowns, there are many ways to satisfy this equation so that we can simply choose one, which is enough for our purposes. In the interest of simplicity, and because of the fact that the positive quantity a2 cannot contribute to generate a negative number on the right-hand side, let us pick a = 0, which also implies that d = 0. With this, we see that our matrix will have zero diagonal, in a complementary way with respect to the identity matrix. Since the

10

SOLUTIONS 1 equation bc = −1 now remains, we have that c = −1/b. Again in the interest of simplicity, we will choose c = 1 and b = −1, in such a way that our matrix has for all its elements either zero or numbers with absolute value 1, just as the identity matrix. The resulting matrix is then   0 −1 , 1 0 which we will use to represent the imaginary unit. It is easy to check directly that the square of this matrix is in fact the negative of the identity matrix.

(b) Let us now consider an arbitrary complex number z = (x, y). In our usual algebraic notation it can be written as z = x + ıy. Making the identification of the real unit with the identity matrix, and of the imaginary unit with the matrix determined above, we can write that     1 0 0 −1 z=x +y , 0 1 1 0 which is a real 2 × 2 matrix representation of an arbitrary complex number. (c) Manipulating the expression above with the usual matrix rules, we can write that   x −y z= . y x Still using the usual matrix rules, we can now show that this structure gives us the correct way to add two complex numbers z1 and z2 , resulting in a complex number z = (x, y), z1 + z2 = = = x =



x1 −y1 y1 x1





x2 −y2 + y2 x2   (x1 + x2 ) −(y1 + y2 ) (y1 + y2 ) (x1 + x2 )   x −y ⇒ y x x1 + x2 ,

y = y1 + y2 .



11

NUMBER, THE LANGUAGE OF SCIENCE

The same can be verified in the case of the multiplication of z1 and z2 , resulting in a complex number z = (x, y), z1 z2 =





x1 −y1 y1 x1



x2 −y2 y2 x2



(x1 x2 − y1 y2 ) −(x1 y2 + x2 y1 ) = (x1 y2 + x2 y1 ) (x1 x2 − y1 y2 )   x −y = ⇒ y x x = x1 x2 − y 1 y 2 ,



y = x1 y 2 + x2 y 1 .

As we see, this new structure preserves the arithmetic operations on complex numbers in their original form. Problem 7. (Challenge Problem) Show that the field of rational numbers Q is countable, building explicitly a one-to-one relation between Q and N. Complete Solution: Let us start by treating separately the number 0, which becomes the first in the sequence. Also, we will limit ourselves to just enumerating the positive rational numbers, which is equivalent to doing the same for all the negative rational numbers. That is enough, because once this is done it is easy to enumerate all the positive and negative rational numbers, by simply merging (interspersing) the two sequences. We must now decide what to do with the numbers of the form p/q, with p ∈ I, q ∈ I, p 6= 0 and q 6= 0. We will organize these numbers using an infinite table as illustrated in Table 1.1, with the values of p listed along the horizontal, those of q listed along the vertical, and the numbers p/q placed in the corresponding table elements. Let us now consider going through this table along successive diagonals, each containing a finite number of elements, so that the following sequence is defined, starting with the number zero that we separated from the rest,             1 2 1 2 3 1 , , , , , , 0, 1 2 1 3 2 1         2 3 4 1 , , , ,... . 4 3 2 1

12

SOLUTIONS 1 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 ··· q=1

1/1

2/1

3/1

4/1

5/1

6/1 · · ·

q=2

1/2

2/2

3/2

4/2

5/2

6/2 · · ·

q=3

1/3

2/3

3/3

4/3

5/3

6/3 · · ·

q=4

1/4

2/4

3/4

4/4

5/4

6/4 · · ·

q=5

1/5

2/5

3/5

4/5

5/5

6/5 · · ·

q=6 .. .

1/6 .. .

2/6 .. .

3/6 .. .

4/6 .. .

5/6 .. .

6/6 · · · .. . . . .

Table 1.1: A table of the values of p and q and the corresponding ratios. With this we are effectively enumerating, that is, constructing a sequence indexed by the natural numbers, all the table elements, which includes all possible positive rational numbers. Of course, there is some repetition in the sequence, because when we simplify fractions there are several that represent the same number. For example, both the second element of the above sequence, (1/1), and the sixth element, (2/2), represent the number 1. However, it is easy to overcome this problem by simply adopting in the construction algorithm of the sequence, the procedure of skipping a number whenever it is equal to some previous number already included in the sequence. By doing this we obtain the sequence 0,

            1 2 1 3 1 1 , , , , , , 1 2 1 3 1 4           3 4 1 5 2 , , , , ,... . 3 2 1 5 1

Including now the negative numbers, we have the following sequence         1 1 1 1 0, 1, −1, ,− , 2, −2, ,− , 2 2 3 3         1 1 2 2 3, −3, ,− , ,− , 4 4 3 3

NUMBER, THE LANGUAGE OF SCIENCE

13

        3 1 1 3 ,− , 4, −4, ,− ,... , 2 2 5 5 that is sufficient to display a bijection between Q and N, thus showing that the two sets have the same number of elements. Historical note: this construction is due to the great German mathematician Georg Cantor. Problem 8. (Challenge Problem) Show that the field of real numbers R is not countable. In order to do this, by reductio ad absurdum, show that, if we assume that a sequence enumerating the real numbers is presented, then it is always possible to construct a real number that is none of the elements in that sequence. Hint: use the decimal representation of the real numbers. Complete Solution: We will limit ourselves to showing that it is not possible to enumerate the real numbers contained within the interval [0, 1), that is, greater than or equal to zero and strictly smaller than 1. This is enough because if one cannot enumerate these real numbers, then of course one cannot enumerate all the real numbers. Let us do this by reductio ad absurdum, showing that, given any enumeration that is presented, we can construct a real number that is not included in that enumeration. Any real number in [0, 1) can be represented in decimal notation as 0.d1 d2 d3 . . . , where each di is a digit, that is, an integer between 0 and 9. Of course, a change in any single digit of the number is enough to change the number. Let us assume for a moment that a sequence of these real numbers that represents an enumeration of them has been presented to us, that is, a sequence that contains all possible real numbers within [0, 1), as shown in the table below, 0.d1,1 d1,2 d1,3 d1,4 d1,5 d1,6 . . . 0.d2,1 d2,2 d2,3 d2,4 d2,5 d2,6 . . . 0.d3,1 d3,2 d3,3 d3,4 d3,5 d3,6 . . . 0.d4,1 d4,2 d4,3 d4,4 d4,5 d4,6 . . . 0.d5,1 d5,2 d5,3 d5,4 d5,5 d5,6 . . .

14

SOLUTIONS 1 0.d6,1 d6,2 d6,3 d6,4 d6,5 d6,6 . . . ... .

In order to construct a real number contained in [0, 1), described in the form 0.d1 d2 d3 . . ., which is none of the elements listed in this infinite sequence, it suffices to proceed as follows: choose for d1 any digit that is not d1,1 , for d2 any digit that is not d2,2 , for d3 any digit that is not d3,3 , and so on ad infinitum. The result is a real number contained in [0, 1) that is none of the numbers presented in the sequence. It follows that the sequence presented cannot be a complete list of all real numbers contained in [0, 1), because there is at least one of these numbers that is not included in the sequence. In fact, there are many, because in each of the choices that we made we could use any of 9 among the 10 digits available. Historical note: this construction is due to the great German mathematician Georg Cantor.

Solutions 2

The Simplicity of Complex Numbers We present here complete and commented solutions to all problems proposed in Chapter 2 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1.

Starting from the Euler formula, prove the DeMoivre theorem.

Complete Solution: The DeMoivre theorem states that for any real θ, [cos(θ) + ı sin(θ)]n = cos(nθ) + ı sin(nθ). We will prove this result by using the Euler formula, which was proved in the text, and which states that for any real θ, eıθ = cos(θ) + ı sin(θ). It suffices to write the left-hand side of the DeMoivre equation in terms of the exponential with imaginary argument,  n [cos(θ) + ı sin(θ)]n = eıθ =

eı(nθ) ,

where we used the properties of the exponential. Using again the Euler formula, this time for the case of the exponential with the imaginary argument ı(nθ), we have 15

16

SOLUTIONS 2

[cos(θ) + ı sin(θ)]n = cos(nθ) + ı sin(nθ), which establishes the DeMoivre theorem. Problem 2. Show that the constant function w(z) = 1 is analytic, that is, that it satisfies the Cauchy-Riemann conditions. Starting from this fact, prove by finite induction that the function w(z) = z n , with n a positive integer, is analytic on the entire complex plane. Complete Solution: For the function w(z) = 1 we have that u(x, y) = 1 and v(x, y) = 0. It follows therefore that all partial derivatives of u(x, y) and v(x, y) are zero, so that ∂v ∂u = 0 = , ∂x ∂y ∂u ∂v = 0 = − , ∂y ∂x that is, the Cauchy-Riemann conditions are satisfied for any values of x and of y, and thus the function is analytic in the entire complex plane. Note that we have the same results for the case of the function w(z) = 0 as well. We will now extend this result to all the powers of z, by finite induction. In order to do this, we assume the result for the case n − 1, that is, for wn−1 (z) = z n−1 , wn−1 (z) = un−1 (x, y) + ıvn−1 (x, y), ∂vn−1 ∂un−1 = , ∂x ∂y ∂vn−1 ∂un−1 = − . ∂y ∂x Let us now consider the next case, wn (z) = z n = zwn−1 (z), for which we therefore have wn (z) = un (x, y) + ıvn (x, y) = (x + ıy)[un−1 (x, y) + ıvn−1 (x, y)] = [xun−1 (x, y) − yvn−1 (x, y)] +

+ı[xvn−1 (x, y) + yun−1 (x, y)]

un (x, y) = xun−1 (x, y) − yvn−1 (x, y), vn (x, y) = xvn−1 (x, y) + yun−1 (x, y).

⇒

17

THE SIMPLICITY OF COMPLEX NUMBERS

Calculating the partial derivatives involved in the first condition, for the case n, we have ∂un ∂x ∂vn ∂y

∂un−1 ∂vn−1 −y , ∂x ∂x ∂un−1 + un−1 + y . ∂y

= un−1 + x = x

∂vn−1 ∂y

Using now the Cauchy-Riemann relations for the case n − 1 in the equation above, we get ∂un ∂x ∂vn ∂y ∂un ∂x

∂un−1 ∂vn−1 −y , ∂x ∂x ∂un−1 ∂vn−1 = un−1 + x −y ⇒ ∂x ∂x ∂vn = , ∂y = un−1 + x

thereby establishing the first condition for the case n. Doing the same for the case of the second condition we have ∂un ∂y ∂vn ∂x

∂un−1 ∂vn−1 − vn−1 − y , ∂y ∂y ∂vn−1 ∂un−1 = vn−1 + x +y . ∂x ∂x

= x

Using now the Cauchy-Riemann relations for the case n − 1 in the equation above, we get ∂un ∂y ∂vn ∂x ∂un ∂y

∂un−1 ∂vn−1 − vn−1 − y , ∂y ∂y ∂un−1 ∂vn−1 = −x + vn−1 + y ∂y ∂y ∂vn = − , ∂x = x

⇒

thus establishing the second condition for the case n. As we have shown that the case n − 1 implies the case n, and that the conditions hold for the case n = 0, it follows that the functions wn (z) = z n are analytic throughout the complex plane, for every positive value of n. Note that this proof could be greatly simplified through the use of the theorem, that we have seen previously, that tells us that the product of two analytic functions is also an analytic function within the common domain of

18

SOLUTIONS 2

analyticity of the two functions. Once we have established that the function w1 (z) = z is analytic throughout the complex plane, it follows that w2 (z) = w1 (z)w1 (z) is also analytic throughout the complex plane. Iterating this argument several times, we have that if w1 (z) = z and wn−1 (z) = z n−1 are analytic throughout the complex plane, then it follows that wn (z) = w1 (z)wn−1 (z) is also analytic throughout the complex plane. This again extends our result by finite induction, for any positive integer value of n. Problem 3. Complete the proof given in the text that the function w(z) = 1/z is analytic. From this fact prove, by finite induction, that the function w(z) = 1/z n , with n a strictly positive integer, is analytic on the whole complex plane except at the origin z = 0. Complete Solution: As noted in the text, we have for the complex function w(z) = 1/z w(z) = u(x, y) + ıv(x, y), x , u(x, y) = 2 x + y2 y . v(x, y) = − 2 x + y2 The partial derivatives involved in the first Cauchy-Riemann condition are ∂u ∂x

= =

∂v ∂y

= =

1 2x2 − x2 + y 2 (x2 + y 2 )2 −x2 + y 2 , (x2 + y 2 )2 −1 2y 2 + x2 + y 2 (x2 + y 2 )2 −x2 + y 2 , (x2 + y 2 )2

from which follows the first relation, ∂u ∂v = . ∂x ∂y The partial derivatives involved in the second Cauchy-Riemann condition are

THE SIMPLICITY OF COMPLEX NUMBERS ∂u ∂y ∂v ∂x

2xy , + y 2 )2 2xy , 2 (x + y 2 )2

= − =

19

(x2

from which follows the second relation, ∂v ∂u =− . ∂y ∂x Thus we see that the conditions are satisfied in the whole complex plane except for the point z = 0, at which none of the partial derivatives is well defined. It follows that the function w(z) = 1/z is analytic in the whole complex plane except at the point z = 0. We will now extend this result to all the negative powers of z, by finite induction. For this, we assume the result for the case n − 1, that is, for wn−1 (z) = 1/z n−1 , wn−1 (z) = un−1 (x, y) + ıvn−1 (x, y), ∂vn−1 ∂un−1 = , ∂x ∂y ∂vn−1 ∂un−1 = − . ∂y ∂x Let us now consider the next case, wn (z) = 1/z n = wn−1 (z)/z, for which we have therefore wn (z) = un (x, y) + ıvn (x, y) x − ıy = [un−1 (x, y) + ıvn−1 (x, y)] x2 + y 2 xun−1 (x, y) + yvn−1 (x, y) + = x2 + y 2 xvn−1 (x, y) − yun−1 (x, y) +ı ⇒ x2 + y 2 xun−1 (x, y) + yvn−1 (x, y) , un (x, y) = x2 + y 2 xvn−1 (x, y) − yun−1 (x, y) vn (x, y) = . x2 + y 2 Calculating the partial derivatives involved in the first condition, for the case n, we have

20

SOLUTIONS 2 ∂un ∂x

=

=

=

∂vn ∂y

=

=

=

−2x(xun−1 + yvn−1 ) un−1 + 2 + 2 2 2 (x + y ) x + y2 x y ∂un−1 ∂vn−1 + 2 + 2 2 2 x + y ∂x x + y ∂x 2 2 2 −2x un−1 − 2xyvn−1 + (x + y )un−1 + (x2 + y 2 )2 ∂un−1 ∂vn−1 y x + 2 + 2 2 2 x + y ∂x x + y ∂x 2 2 (−x + y )un−1 − 2xyvn−1 + (x2 + y 2 )2 y ∂un−1 ∂vn−1 x + 2 , + 2 2 2 x + y ∂x x + y ∂x −2y(xvn−1 − yun−1 ) un−1 − 2 + 2 2 2 (x + y ) x + y2 y ∂vn−1 ∂un−1 x − 2 + 2 x + y 2 ∂y x + y 2 ∂y −2xyvn−1 + 2y 2 un−1 − (x2 + y 2 )un−1 + (x2 + y 2 )2 ∂vn−1 ∂un−1 y x − 2 + 2 2 2 x + y ∂y x + y ∂y 2 2 (−x + y )un−1 − 2xyvn−1 + (x2 + y 2 )2 x ∂vn−1 ∂un−1 y + 2 − 2 . 2 2 x + y ∂y x + y ∂y

Using now the Cauchy-Riemann relations for the case n − 1 in the equation above, we get ∂un ∂x

=

∂vn ∂y

=

∂un ∂x

=

(−x2 + y 2 )un−1 − 2xyvn−1 + (x2 + y 2 )2 x y ∂un−1 ∂vn−1 + 2 + 2 , 2 2 x + y ∂x x + y ∂x (−x2 + y 2 )un−1 − 2xyvn−1 + (x2 + y 2 )2 ∂un−1 ∂vn−1 y x + 2 ⇒ + 2 x + y 2 ∂x x + y 2 ∂x ∂vn , ∂y

21

THE SIMPLICITY OF COMPLEX NUMBERS

thereby establishing the first condition for the case n. Doing the same for the case of the second condition we have ∂un ∂y

−2y(xun−1 + yvn−1 ) vn−1 + 2 + 2 2 2 (x + y ) x + y2 x ∂un−1 ∂vn−1 y + 2 + 2 2 2 x + y ∂y x + y ∂y 2 2 2 −2xyun−1 − 2y vn−1 + (x + y )vn−1 + (x2 + y 2 )2 ∂un−1 ∂vn−1 y x + 2 + 2 2 2 x + y ∂y x + y ∂y 2 2 (x − y )vn−1 − 2xyun−1 + (x2 + y 2 )2 y ∂un−1 ∂vn−1 x + 2 , + 2 2 2 x + y ∂y x + y ∂y −2x(xvn−1 − yun−1 ) vn−1 + 2 + 2 2 2 (x + y ) x + y2 y ∂vn−1 ∂un−1 x − 2 + 2 2 2 x + y ∂x x + y ∂x 2 2 2 −2x vn−1 + 2xyun−1 + (x + y )vn−1 + (x2 + y 2 )2 x ∂vn−1 ∂un−1 y + 2 − 2 2 2 x + y ∂x x + y ∂x 2 2 (−x + y )vn−1 + 2xyun−1 + (x2 + y 2 )2 x y ∂vn−1 ∂un−1 + 2 − 2 . 2 2 x + y ∂x x + y ∂x

=

=

=

∂vn ∂x

=

=

=

Using now the Cauchy-Riemann relation for the case n − 1 in the equation above, we get ∂un ∂y

∂vn ∂x

(x2 − y 2 )vn−1 − 2xyun−1 + (x2 + y 2 )2 y ∂un−1 ∂vn−1 x + 2 , + 2 2 2 x + y ∂y x + y ∂y (x2 − y 2 )vn−1 − 2xyun−1 = − + (x2 + y 2 )2 x ∂un−1 ∂vn−1 y − 2 − 2 2 2 x + y ∂y x + y ∂y =

⇒

22

SOLUTIONS 2 ∂un ∂y

= −

∂vn , ∂x

thus establishing the second condition for the case n. As we have shown that the case n − 1 implies the case n, and that the conditions hold for the case n = 1, it follows that the complex functions wn (z) = 1/z n are analytic in the whole complex plane except at the origin z = 0, for all strictly positive values of the integer n. Note that this proof could be greatly simplified through the use of the theorem, which we have seen previously, that tells us that the product of two analytic functions is also an analytic function within the common domain of analyticity of the two functions. Once we have established that the function w1 (z) = 1/z is analytic in the whole complex plane except for the origin, it follows that w2 (z) = w1 (z)w1 (z) is also analytic in the whole complex plane except at the origin. Iterating this argument several times, we have that if w1 (z) = 1/z and wn−1 (z) = 1/z n−1 are analytic in the whole complex plane except at the origin, then it follows that wn (z) = w1 (z)wn−1 (z) is also analytic in the whole complex plane except at the origin. This again extends our result by finite induction, for any strictly positive value of the integer n. Problem 4. Consider the movement of a solid disk which rotates around its axis with a constant angular speed ω. Consider a point of this disk at a distance ρ from the center, whose initial position forms an angle θ0 with the x axis of a system of Cartesian coordinates (x, y) on the plane of the disk. (a) Given that (x, y) are the coordinates of a point of the disk in this coordinate system, represent the position of the point as a complex number z = (x, y), giving z as a function of the time t and of the initial position θ0 of the point. (b) Give the complex velocity z˙ of the point of the disk as a function of the time t and of the initial position θ0 of the point. Also, write z˙ in terms of z. (c) What is the effect on the two-dimensional vector z = (x, y) of the operation of multiplying it by ı = (0, 1)? (d) Give the complex acceleration z¨ of the point of the disk as a function of the time t and of the initial position θ0 of the point. Also, write z¨ in terms of z.

23

THE SIMPLICITY OF COMPLEX NUMBERS y ω ρ θ 0

x

ω

Figure 2.1: The disk rotating at a constant angular velocity ω. (e) Rewrite the results for z, z˙ and z¨ using the exponential form of the complex numbers. Complete Solution: Consider the disk, its axis, a point on the disk and the coordinate system as illustrated in Figure 2.1. (a) The position vector of the point on the disk is given by ~r = (x, y). In terms of ρ and θ, where ρ is a constant and θ(t) = θ0 + ωt, we have x(t) = ρ cos(θ0 + ωt), y(t) = ρ sin(θ0 + ωt). We can represent the position by means of the complex number z(t) = ρ [cos(θ0 + ωt) + ı sin(θ0 + ωt)] . (b) Having the representation of the position in terms of z(t), it suffices to differentiate with respect to the time in order to obtain the corresponding representation of the velocity z(t), ˙ z(t) = ρ [cos(θ0 + ωt) + ı sin(θ0 + ωt)]

⇒

z(t) ˙ = ρω [− sin(θ0 + ωt) + ı cos(θ0 + ωt)] .

24

SOLUTIONS 2 Note that we can write, by putting a factor of ı in evidence, that z(t) ˙ = ıρω [cos(θ0 + ωt) + ı sin(θ0 + ωt)] = ıρωz(t).

(c) From the physics of the problem we know that the velocity is perpendicular to the position vector. This means that ız is perpendicular to z, therefore the multiplication by ı rotates the vector by an angle of π/2 in the positive direction, that is, counterclockwise. The orthogonality can be verified directly by making the scalar product (which is not the same as the complex product) of the vector ız and the vector z. (d) Having the representation of the velocity in terms of z(t), ˙ it suffices to differentiate with respect to the time in order to obtain the corresponding representation of the acceleration z¨(t), z(t) ˙ = ρω [− sin(θ0 + ωt) + ı cos(θ0 + ωt)] z¨(t) = −ρω 2 [cos(θ0 + ωt) + ı sin(θ0 + ωt)] .

⇒

Note that we can immediately write that z¨(t) = −ρω 2 z(t), which shows that the acceleration is in the direction opposite to the position vector. It is a centripetal acceleration. (e) Using the polar representation of the complex numbers and the Euler formula, we can write immediately that z(t) = ρ eı(θ0 +ωt) , z(t) ˙ = ıρω eı(θ0 +ωt) , z¨(t) = −ρω 2 eı(θ0 +ωt) . As one can see, in this way the derivatives and corresponding physical consequences are all quite immediately visible.

THE SIMPLICITY OF COMPLEX NUMBERS Problem 5. number.

25

Consider the complex number z = exp(ıθ), where θ is a real

(a) Determine the real and imaginary parts of z and of its complex conjugate z ∗ . √ (b) Calculate the absolute value of z, that is, |z| = z ∗ z. (c) Determine the complex number z 2 , the square of z, using the exponential form, and write its real and imaginary parts. (d) Calculate z 2 starting from z, without using the exponential form, that is, first writing z in terms of its real and imaginary parts, and then taking the square. (e) Comparing the results obtained in the two previous items, derive the trigonometric identities for the sine and the cosine of the double arc. Complete Solution: Consider the complex number z = exp(ıθ), which can be written in terms of its real and imaginary parts as z = u(θ) + ıv(θ). (a) Using the Euler formula, we find that the real and imaginary parts of z are u(θ) = cos(θ), v(θ) = sin(θ). Thus we see that the real part u′ (θ) and the imaginary part v ′ (θ) of z ∗ are u′ (θ) = cos(θ), v ′ (θ) = − sin(θ). (b) We can calculate the absolute value of z in two ways, q cos2 (θ) + sin2 (θ) |z| = = 1, √ |z| = eıθ e−ıθ √ = eıθ−ıθ √ e0 = = 1.

26

SOLUTIONS 2

(c) Calculating the square of z, using the exponential form, we have z2 = =



eıθ

eı2θ

2

= cos(2θ) + ı sin(2θ). (d) Calculating the square of z, using the form u + ıv, we have z 2 = [cos(θ) + ı sin(θ)]2   = cos2 (θ) − sin2 (θ) + ı [2 cos(θ) sin(θ)] .

(e) Comparing the results of the previous two items, we get the following trigonometric identities, cos(2θ) = cos2 (θ) − sin2 (θ), sin(2θ) = 2 sin(θ) cos(θ).

Problem 6. Show that the sum-function w(z) = f (z)+g(z), of two analytic functions f (z) and g(z), is also analytic in the common domain of analyticity of f (z) and g(z). Complete Solution: Consider the sum-function w(z) = f (z) + g(z), where f (z) = uf (x, y) + ıvf (x, y) and g(z) = ug (x, y) + ıvg (x, y). Within the common domain of analyticity of f (z) and g(z), we have that all partial derivatives exist, and that they satisfy ∂uf ∂x ∂uf ∂y ∂ug ∂x ∂ug ∂y

∂vf , ∂y ∂vf = − , ∂x ∂vg = , ∂y ∂vg = − . ∂x =

We therefore have for the sum-function w(z) = u(x, y) + ıv(x, y), where by definition of the sum of complex numbers u(x, y) = uf (x, y) + ug (x, y) and v(x, y) = vf (x, y) + vg (x, y), within this domain, that

THE SIMPLICITY OF COMPLEX NUMBERS ∂u ∂x

= = =

27

∂uf ∂ug + ∂x ∂x ∂vf ∂vg + ∂y ∂y ∂v , ∂y

which shows that the first Cauchy-Riemann condition is satisfied for w(z). Similarly, for the other condition, ∂u ∂y

∂uf ∂ug + ∂y ∂y ∂vf ∂vg − = − ∂x ∂x ∂v . = − ∂x =

It follows therefore that w(z) is analytic in the common domain of analyticity of f (z) and g(z). Problem 7. Show that the product-function w(z) = f (z)g(z), of two analytic functions f (z) and g(z), is also analytic in the common domain of analyticity of f (z) and g(z). Complete Solution: Consider the product-function w(z) = f (z)g(z), where f (z) = uf (x, y) + ıvf (x, y) and g(z) = ug (x, y) + ıvg (x, y). Within the common domain of analyticity of f (z) and g(z), we have that all partial derivatives exist, and that they satisfy ∂uf ∂x ∂uf ∂y ∂ug ∂x ∂ug ∂y

∂vf , ∂y ∂vf = − , ∂x ∂vg = , ∂y ∂vg = − . ∂x =

We therefore have for product-function w(z) = u(x, y) + ıv(x, y), where by definition of the product of complex numbers

28

SOLUTIONS 2 u(x, y) = uf (x, y)ug (x, y) − vf (x, y)vg (x, y), v(x, y) = uf (x, y)vg (x, y) + vf (x, y)ug (x, y),

within this domain, that ∂u ∂x

∂uf ∂vf ∂ug ∂vg + ug − vf − vg ∂x ∂x ∂x ∂x ∂vf ∂uf ∂vg ∂ug uf + ug + vf + vg ∂y ∂y ∂y ∂y     ∂uf ∂vf ∂vg ∂ug + vg + ug uf + vf ∂y ∂y ∂y ∂y ∂ ∂ (uf vg ) + (vf ug ) ∂y ∂y ∂ (uf vg + vf ug ) ∂y ∂v , ∂y

= uf = = = = =

which shows that the first Cauchy-Riemann condition is satisfied for w(z). Similarly, for the other condition, ∂u ∂y

∂uf ∂vf ∂ug ∂vg + ug − vf − vg ∂y ∂y ∂y ∂y ∂vf ∂uf ∂ug ∂vg − ug − vf − vg −uf ∂x  ∂x ∂x    ∂x ∂uf ∂vf ∂ug ∂vg + vg + ug − vf − uf ∂x ∂x ∂x ∂x ∂ ∂ − (uf vg ) − (vf ug ) ∂x ∂x ∂ − (uf vg + vf ug ) ∂x ∂v − . ∂x

= uf = = = = =

It follows therefore that w(z) is analytic in the common domain of analyticity of f (z) and g(z). Problem 8. Show that the composite function w(z) = f [g(z)] obtained by the composition of two analytic functions f (z) and g(z) is also analytic, within the domain that is given by the intersection of the image of g(z) with the analyticity domain of f (z).

THE SIMPLICITY OF COMPLEX NUMBERS

29

Hint: consider the composition of two complex functions involving three complex planes, so that one of the functions maps the first plane onto the second and the other function maps the second plane onto the third. Complete Solution: Consider the composite function w(z) = f [g(z)], where g(z) = ug (x, y) + ıvg (x, y). In the analyticity domain of g(z) we have that ∂ug ∂x ∂ug ∂y

∂vg , ∂y ∂vg = − . ∂x

=

Similarly, in the domain of analyticity of f (z) we have that f (z) = uf (x, y) + ıvf (x, y) and that ∂uf ∂x ∂uf ∂y

∂vf , ∂y ∂vf = − . ∂x =

Let us now consider the portion of the image of g(z) that falls within the domain of analyticity of f (z), that is, the intersection of the image of g(z) with the analyticity domain of f (z). Within this region we have that f [g(z)] has real and imaginary parts given by f (z) = uf (ug , vg ) + ıvf (ug , vg ). Since f (g) is analytic with respect to the complex variable g, which has real and imaginary parts given by ug and vg , we have that ∂uf ∂ug ∂uf ∂vg

∂vf , ∂vg ∂vf = − . ∂ug =

We therefore have for w(z) = u(x, y) + ıv(x, y), where by the definition of the composition of two functions u(x, y) = uf (ug , vg ) and v(x, y) = vf (ug , vg ), within this region, by the chain rule, that

30

SOLUTIONS 2 ∂u ∂x

∂uf ∂ug ∂uf ∂vg + ∂ug ∂x ∂vg ∂x ∂vf ∂vf ∂vg ∂ug + (−1) (−1) ∂vg ∂y ∂ug ∂y ∂vf ∂ug ∂vf ∂vg + ∂ug ∂y ∂vg ∂y ∂v , ∂y

= = = =

where in this last passage we use the chain rule in the reversed direction. This shows that the first Cauchy-Riemann condition is satisfied for w(z). Similarly, for the other condition, ∂u ∂y

∂uf ∂ug ∂uf ∂vg + ∂ug ∂y ∂vg ∂y ∂vf ∂ug ∂vf ∂vg − = − ∂vg ∂x ∂ug ∂x   ∂vf ∂ug ∂vf ∂vg + = − ∂ug ∂x ∂vg ∂x ∂v = − , ∂x =

where once again we use the chain rule in the reversed direction in the last passage. It follows therefore that w(z) is analytic in the intersection of the image of g(z) with the analyticity domain of f (z). Problem 9. Starting from the fact that, for any complex numbers z1 , z2 and z3 , it is true that z3 = z1 + z2

⇒

|z3 | ≤ |z1 | + |z2 |,

show that if we have a sum of n complex numbers zi , zs =

n X

zi

i=1

= z1 + z2 + z3 + . . . + zn−1 + zn , it follows that

THE SIMPLICITY OF COMPLEX NUMBERS

|zs | ≤

n X i=1

31

|zi |.

Complete Solution: Consider the basic triangle inequality |zc | ≤ |za | + |zb |, valid for any complex numbers za , zb and zc , and the sum zs =

n X

zi .

i=1

We may write this sum as

zs = z1 +

n X

zi ,

i=2

and using the basic inequality we have

n X zi . |zs | ≤ |z1 | + i=2

We may now write for the remaining sum n X

zi = z2 +

n n X X zi , zi ≤ |z2 | + i=3

i=2

and therefore that

zi ,

i=3

i=2

from which it follows that

n X

n X zi . |zs | ≤ |z1 | + |z2 | + i=3

Iterating this procedure for each possible value of i in the sum, we end up getting as the final result the general triangle inequality |zs | ≤

n X i=1

|zi |.

Note that this remains true even if the sum extends indefinitely to infinity, that is, for n → ∞, provided that the infinite sums involved converge.

32

SOLUTIONS 2

Problem 10. (Challenge Problem) Consider the complex number given by z = exp(ı2π/N ), where N is a strictly positive integer. (a) Determine the real and imaginary parts of the powers z k of z, for k ∈ {1, . . . , N }, and write z k in terms of their real and imaginary parts, leaving the result in terms of k and N . (b) Determine the sum of all the N numbers z k , that is, calculate

S=

N X

zk .

k=1

Hint: draw vectors z k in the complex plane, for a definite and not too large value of N , for example, for N = 6, N = 7 or N = 8. Complete Solution: Consider the complex number z = exp(ı2π/N ), where N > 0. (a) The N powers z k , for k ∈ {1, . . . , N }, are given by z k = exp(ı2πk/N ). Using the Euler formula, we see that their real and imaginary parts are given by h i ℜ z k = cos(2πk/N ), h i ℑ z k = sin(2πk/N ). (b) Let us now calculate the sum of all these N vectors in the complex plane,

S=

N X

zk .

k=1

Since z is a unit vector in the complex plane, which makes an angle θ = 2π/N with the real axis, each z k is also a unit vector, but pointing in a different direction. Since for k = N we have that z N = exp(ı2π) = 1, and each multiplication by z corresponds to a rotation by the angle θ, we see that these N vectors point in N directions equally spaced by

33

THE SIMPLICITY OF COMPLEX NUMBERS y 2

1

3

6 x

4

5

Figure 2.2: The unit circle with N equally spaced unit vectors, in the case N = 6. the angle θ. We can therefore draw all N vectors along the unit circle, as exemplified in the diagram of Figure 2.2. This figure suggests that the sum is zero by symmetry. In the cases where N is even, like the one shown in the figure, that is immediate, since the vectors cancel off in pairs. In the cases where N is odd, the result is still suggested by the figure, but it is not so immediate. We can see that in fact the sum is zero for any N , using the following reduction ad absurdum argument. Imagine for a moment that the sum is not zero. It follows that it is a complex number and is therefore represented by some non-zero vector in the complex plane, such as the lower vector shown in dashed line in Figure 2.2. If we now rotate all the N vectors by the angle θ, in the positive direction, we note that the figure formed by them remains invariant, and therefore has the same sum as before. However, by this same rotation the sum-vector is rotated by the angle θ, and does not remain invariant, which is absurd. It follows that the sum must be a vector which is invariant under a rotation through an angle θ 6= 0. But the only vector that has this property is the zero vector, and hence we conclude that S = 0. We can express these same ideas arithmetically. Since the rotation by θ corresponds to a multiplication by z, we have that the process described can be represented as

S =

N X k=1

zk

⇒

34

SOLUTIONS 2

zS =

N X

z k+1 ,

k=1

where z 6= 0 and z 6= 1. Now, it is easy to see that, in the lower sum, z N +1 is equal to z, because z N +1 =

eı2π(N +1)/N

=

eı2π eı2π/N

=

eı2π/N

= z, so that we have for the sum after the rotation by θ implemented by the multiplication by z, N X

z k+1 =

k=1

N −1 X

z k+1 + z N +1

k=1

= z+

N X

zk

′

k ′ =2

=

N X

zk .

k=1

This shows that the sum is invariant. It follows from the preceding equations that zS = S, where z 6= 1. The only way to satisfy this equation is to have S = 0, which thus establishes the required result. There are other ways to prove this relation for all values of N , that will be seen later on, in this series of books.

Solutions 3

Elementary Functions, but Not Quite We present here complete and commented solutions to all problems proposed in Chapter 3 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. By means of a transformation of variables in the complex plane, from the Cartesian coordinates (x, y) to the polar coordinates (ρ, θ), write the two Cauchy-Riemann conditions in terms of the variables ρ and θ, that is, in terms of derivatives with respect to these variables, where x = ρ cos(θ), y = ρ sin(θ). Complete Solution: Recalling that we have the functions u(x, y) and v(x, y), and that by means of the transformation of variables x(ρ, θ) = ρ cos(θ), y(ρ, θ) = ρ sin(θ), we also have the functions u(ρ, θ) and v(ρ, θ), we calculate the partial derivatives of u and v with respect to ρ and θ, using the chain rule. The first derivative that appears in the first Cauchy-Riemann condition is 35

36

SOLUTIONS 3 ∂u ∂ρ

= =

∂u ∂x ∂u ∂y + ∂x ∂ρ ∂y ∂ρ ∂u ∂u cos(θ) + sin(θ). ∂x ∂y

The other derivative that appears in this first condition is ∂v ∂θ

∂v ∂x ∂v ∂y + ∂x ∂θ ∂y ∂θ ∂v ∂v = − ρ sin(θ) + ρ cos(θ). ∂x ∂y =

Using now the Cauchy-Riemann conditions, in their Cartesian form, in order to write the derivatives of v in terms of derivatives of u, we can write this as ∂v ∂θ

∂u ∂u ρ sin(θ) + ρ cos(θ) ∂y ∂x   ∂u ∂u = ρ cos(θ) + sin(θ) . ∂x ∂y =

Comparing this with the previous equation we have that ∂u 1 ∂v = , ∂ρ ρ ∂θ which is the first Cauchy-Riemann condition in polar coordinates. Examining now the other relevant derivatives, we have ∂v ∂ρ

= =

∂v ∂x ∂v ∂y + ∂x ∂ρ ∂y ∂ρ ∂v ∂v cos(θ) + sin(θ), ∂x ∂y

and ∂u ∂θ

∂u ∂x ∂u ∂y + ∂x ∂θ ∂y ∂θ ∂u ∂u ρ sin(θ) + ρ cos(θ). = − ∂x ∂y

=

Using now the Cauchy-Riemann conditions, in their Cartesian form, in order to write the derivatives of u in terms of derivatives of v, we can write this as

ELEMENTARY FUNCTIONS, BUT NOT QUITE ∂u ∂θ

37

∂v ∂v ρ sin(θ) − ρ cos(θ) ∂y ∂x   ∂v ∂v = −ρ cos(θ) + sin(θ) . ∂x ∂y

= −

Comparing this with the previous equation we have that 1 ∂u ∂v =− , ρ ∂θ ∂ρ which is the second Cauchy-Riemann condition in polar coordinates. √ Problem 2. Show that the complex function w(z) = z is analytic, that is, that it satisfies the Cauchy-Riemann conditions throughout the complex plane, except for the origin z = 0. Complete Solution: Writing z in polar coordinates, z = ρ exp(ıθ), we have for the function w(z) = √ z √ ıθ/2 w(z) = ρe      θ θ √ ρ cos + ı sin . = 2 2 We can now verify the analyticity using the polar version of the CauchyRiemann conditions. For the first condition we have   θ 1 ∂u = cos , √ ∂ρ 2 ρ 2   1 ∂v θ 1 = , √ cos ρ ∂θ 2 ρ 2 so that the first condition is satisfied throughout the whole complex plane except for the point ρ = 0, which corresponds to z = 0, at which the derivatives do not exist. For the second condition we have   θ 1 ∂u 1 = − √ sin , ρ ∂θ 2 ρ 2   θ 1 ∂v = , √ sin ∂ρ 2 ρ 2 so that the second condition is also satisfied, once again throughout the whole complex plane except for the point z = 0.

38

SOLUTIONS 3

Problem 3. Given the complex number z = x + ıy, consider the complex function w(z) = u(x, y) + ıv(x, y) in each case below. In each case, determine the real part u(x, y) and the imaginary part v(x, y) of w(z), and show that the function w(z) satisfies the two Cauchy-Riemann conditions. (a) w(z) = exp(z). (b) w(z) = cos(z). (c) w(z) = sin(z). (d) w(z) = cosh(z). (e) w(z) = sinh(z). Complete Solution: In each case, we simply write explicitly the real and imaginary parts of the function w(z) = u(x, y) + ıv(x, y), and calculate the appropriate partial derivatives of u(x, y) and v(x, y). (a) w(z) = exp(z): w(z) = = ∂u ∂x ∂u ∂y

ex [cos(y) + ı sin(y)] ex cos(y) + ı ex sin(y)

⇒

∂v , ∂y ∂v = − ex sin(y) = − . ∂x

=

ex cos(y)

=

(b) w(z) = cos(z): eız + e−ız 2 e−y eıx + ey e−ıx = 2 e−y [cos(x) + ı sin(x)] + ey [cos(x) − ı sin(x)] = 2 = cos(x) cosh(y) − ı sin(x) sinh(y) ⇒

w(z) =

ELEMENTARY FUNCTIONS, BUT NOT QUITE ∂u ∂x ∂u ∂y

∂v , ∂y ∂v = − . ∂x

= − sin(x) cosh(y) = = cos(x) sinh(y)

(c) w(z) = sin(z): eız − e−ız 2ı e−y eıx − ey e−ıx = 2ı e−y [cos(x) + ı sin(x)] − ey [cos(x) − ı sin(x)] = 2ı = sin(x) cosh(y) + ı cos(x) sinh(y) ⇒

w(z) =

∂u ∂x ∂u ∂y

∂v , ∂y ∂v = − . ∂x

= cos(x) cosh(y) = = sin(x) sinh(y)

(d) w(z) = cosh(z): ez + ez 2 x e eıy + e−x e−ıy = 2 x e [cos(y) + ı sin(y)] + e−x [cos(y) − ı sin(y)] = 2 = cosh(x) cos(y) + ı sinh(x) sin(y) ⇒

w(z) =

∂u ∂x ∂u ∂y

∂v , ∂y ∂v = − cosh(x) sin(y) = − . ∂x = sinh(x) cos(y)

(e) w(z) = sinh(z): w(z) =

ez − ez 2

=

39

40

SOLUTIONS 3 ex eıy − e−x e−ıy 2 ex [cos(y) + ı sin(y)] − e−x [cos(y) − ı sin(y)] = 2 = sinh(x) cos(y) + ı cosh(x) sin(y) ⇒

=

∂u ∂x ∂u ∂y

∂v , ∂y ∂v = − sinh(x) sin(y) = − . ∂x = cosh(x) cos(y)

=

Problem 4. Show that the complex function of the variable z = x + ıy given by w(z) = x2 + ıy 2 is not analytic. Show the same for the complex function w(z) = z ∗ z. Complete Solution: In the case of the function w(z) = x2 + ıy 2 we have that u = x2 and v = y 2 , so that we have the partial derivatives ∂u ∂x ∂v ∂y ∂u ∂y ∂v ∂x

= 2x, = 2y, = 0, = 0.

Thus we see that the first Cauchy-Riemann condition is not satisfied, although the second is. In the case of the function w(z) = z ∗ z = x2 + y 2 we have that u = x2 + y 2 and v = 0, so that we have the partial derivatives ∂u ∂x ∂v ∂y ∂u ∂y ∂v ∂x

= 2x, = 0, = 2y, = 0.

ELEMENTARY FUNCTIONS, BUT NOT QUITE

41

Thus we see that in this case neither of the Cauchy-Riemann conditions is satisfied. In fact, we can see that in both cases there is a single isolated point where the two conditions are satisfied, the point z = 0. However, this is not enough for us to say that the functions are analytic at that point, because the analyticity condition at a given point is, in fact, that the Cauchy-Riemann conditions be satisfied at least in an infinitesimal neighborhood of that point, and not just at the point. Problem 5. In each case below, determine the domain of analyticity of the function, that is, determine the points of the complex plane where the function is not analytic. (a) w(z) = tan(z). (b) w(z) = cot(z). (c) w(z) = sec(z). (d) w(z) = csc(z). Complete Solution: In each case, we must determine the location of all possible zeros in denominator, as well as other sources of non-analyticity, if any. (a) w(z) = tan(z): since we have that w(z) =

sin(z) , cos(z)

where both the sin(z) and the cos(z) are analytic functions throughout the complex plane, we only need to find the set of all the zeros of the function cos(z). This function can be written as cos(z) = cos(x) cosh(y) − ı sin(x) sinh(y), where both the real part and the imaginary part should be zero. In the real part we have that cosh(y) never vanishes, so that it is necessary that cos(x) vanish. However, at the points where cos(x) is zero, the function sin(x) appearing in the imaginary part is not zero, hence it is

42

SOLUTIONS 3 necessary that sinh(y) vanish in the imaginary part. The only point at which sinh(y) vanishes is y = 0, so that the zeros of the function cos(z) are all on the real axis, being characterized by cos(x) = 0 and y = 0. In other words, the only zeros of the complex function cos(z) are the real zeros of the real function cos(x). It follows that the complex function w(z) = tan(z) is analytic in the entire complex plane except for the set of points given by y = 0 and x=

π + nπ, 2

where n is an arbitrary positive or negative integer. (b) w(z) = cot(z): since we have that w(z) =

cos(z) , sin(z)

where both the sin(z) as the cos(z) are analytic functions throughout the complex plane, we only need to find the set of all the zeros of the function sin(z). This function can be written as sin(z) = sin(x) cosh(y) + ı cos(x) sinh(y), where both the real part and the imaginary part should be zero. An analysis similar to the one that was developed in the previous item implies that, in this case, we should have sin(x) = 0 and y = 0. In other words, the only zeros of the complex function sin(z) are the real zeros of the real function sin(x). It follows that the complex function w(z) = cot(z) is analytic in the entire complex plane except for the set of points given by y = 0 and x = nπ, where n is an arbitrary positive or negative integer. (c) w(z) = sec(z): since we have that w(z) =

1 , cos(z)

where cos(z) is an analytic function in the whole complex plane, we only need to find the set of all the zeros of the function cos(z). The analysis of the problem is thus identical to that of the first item, so that

ELEMENTARY FUNCTIONS, BUT NOT QUITE

43

the complex function w(z) = sec(z) is analytic in the entire complex plane except for the set of points given by y = 0 and x=

π + nπ, 2

where n is an arbitrary positive or negative integer. (d) w(z) = csc(z): since we have that w(z) =

1 , sin(z)

where sin(z) is an analytic function in the whole complex plane, we only need to find the set of all the zeros of the function sin(z). The analysis of the problem is thus identical to that of the second item, so that the complex function w(z) = csc(z) is analytic in the entire complex plane except for the set of points given by y = 0 and x = nπ, where n is an arbitrary positive or negative integer. Problem 6. (Challenge Problem) Determine the branch points, the branch cuts and the Riemann leaves of the functions that follow. Cut and paste the leaves to produce Riemann surfaces on which the functions are completely well defined and analytic, except for the singularities at the branch points. √ (a) w(z) = z 2 − 1. √ (b) w(z) = 1/ z 2 − 1. Hint: consider the behavior of the functions along circles around the points where the square root vanishes. Complete Solution: In both cases to be examined, we can write that p p z 2 − 1 = (z − 1)(z + 1),

so that it is clear that the two special points to consider during the analysis are z = 1 and z = −1. We can see this because if we have, for example, z ≈ 1, it follows that z + 1 ≈ 2 is essentially a constant, and therefore we

44

SOLUTIONS 3 y

θ′

θ′

θ 0

x

−1

1

Figure 3.1: The singular points and the three test circles. effectively have a square root of the variable z ′ = z − 1. In order to make the analysis, we will go around these two points along the circles illustrated in the diagram of Figure 3.1. Adopting the polar representation for z, that is z = ρ exp(ıθ), we will first write the quantity z 2 − 1 in polar form, z 2 − 1 = λ exp(ıα). We thus have z 2 − 1 = λ eıα

= ρ2 e2ıθ − 1     = ρ2 cos(2θ) − 1 + ı ρ2 sin(2θ) .

We can now calculate λ, λ2 =

 2 2  2 ρ cos(2θ) − 1 + ρ2 sin(2θ)

= ρ4 + 1 − 2ρ2 cos(2θ) ⇒ p λ = ρ4 + 1 − 2ρ2 cos(2θ).

Similarly, we can now calculate exp(ıα), eıα = p

ρ2 cos(2θ) − 1

ρ4 + 1 − 2ρ2 cos(2θ)

+ıp

ρ2 sin(2θ) ρ4 + 1 − 2ρ2 cos(2θ)

.

45

ELEMENTARY FUNCTIONS, BUT NOT QUITE

√ 2 (a) In this case we have √ w(z) = z − 1, which is expressed in terms of λ and α as w(z) = λ exp(ıα/2). In order to go around the big circle in the complex plane shown in the diagram of Figure 3.1, so as to verify whether or not there is a change of leaf when we do this, we will calculate exp(ıα) in the limit ρ → ∞, lim eıα =

ρ→∞

=

= =

lim

ρ→∞

lim

ρ→∞

"

"

p

p

ρ2 cos(2θ) − 1

ρ4 + 1 − 2ρ2 cos(2θ) +ıp

+

ρ2 sin(2θ)

ρ4 + 1 − 2ρ2 cos(2θ)

cos(2θ) − 1/ρ2

1 + 1/ρ4 − 2 cos(2θ)/ρ2 +ıp

#

+

sin(2θ)

1 + 1/ρ4 − 2 cos(2θ)/ρ2  cos(2θ) sin(2θ) √ +ı √ 1 1 2ıθ e .



#

Similarly, we have in this limit λ = ρ2 . Thus we see that in this limit we simply have that w(z) = ρ exp(ıθ), that is, the function approaches the identity. Thus we see that there is no change of leaf when we vary θ from 0 to 2π with ρ → ∞, so that there is no branch cut that extends to infinity. On the other hand, if we approach the point z = 1, p we can write the function in terms of the variable z ′ =√ z − 1 as w(z) = z ′ (z ′ + 2). In the limit in which z ′ → 0 it approaches 2z ′ , which can be written in terms of a polar coordinate system (ρ′ , θ ′ ) centered at the √point z = 1, as illustrated in the diagram of Figure 3.1, since w(z) ≈ 2ρ′ exp(ıθ ′ /2). Thus we see that, if we follow an infinitesimal circle around z = 1, with ρ′ constant and θ ′ going from 0 to 2π, there is indeed a change of leaf. It follows that there is a branch cut connected to the point z = 1, which, however, does not extend to infinity. A similar analysis around the point z = 1 leads to the same conclusions for that point. It follows that there is a single branch cut of finite extent, which connects the two branch points z = 1 and z = −1. Any simple curve that connects the two points can be chosen as the branch cut, but the simplest choice is the real segment [−1, 1].

46

SOLUTIONS 3

The Riemann surface consists therefore of two copies of the complex plane, except for the branch points z = 1 and z = −1, and with a cut from one of these points to the other. Each side of this cut in one of the two planes is continuously connected to the other side of the cut in the other plane, so that by crossing the branch cut one changes from one plane to the other. √ (b) In this case we have p w(z) = 1/ z 2 − 1, which is expressed in terms of λ and α as w(z) = 1/λ exp(−ıα/2), which already shows that the analysis is not very different from that of the previous case. Furthermore, we can reduce this problem explicitly to the previous problem, because we have that w(z) = = =

1 √ 2 z −1 √ z2 − 1 z2 − 1 √ z2 − 1 , (z − 1)(z + 1)

where the function in denominator is analytic and single-valued, so that the structure of branch cuts and points is determined only by the numerator. There are, of course, additional singularities at z = 1 and z = −1, due to the zeros in denominator, but this does not change the structure of the leaves or of the Riemann surface. It follows that in this case we also have that z = 1 and z = −1 are two branch points, with a branch cut which connects the two points. The Riemann surface is the same as the one in the previous item, so that it consists of two copies of the complex plane, except for the branch points z = 1 and z = −1, and with a cut from one of these points to the other. Each side of this cut in one of the two planes is continuously connected to the other side of the cut in the other plane, so that by crossing the branch cut one changes from one plane to the other. Note that it is not really necessary to take the limits ρ → ∞ and z → 1 or z → −1 in order to do this analysis. The limits simplify the analysis, and are sufficient due to the topological nature of the problem, with continuous simple curves that connect points or that extend to infinity, but they are not really essential. For the analysis it suffices to consider closed curves that go independently around each one of the two branch points, and that go around the two points at once.

Solutions 4

Even Less Elementary Functions We present here complete and commented solutions to all problems proposed in Chapter 4 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. that follow.

Calculate, giving their real and imaginary parts, the numbers

(a) z = 1ı . (b) z = eı . (c) z = sin(ı). (d) z = cos(ı). Complete Solution: In each case, the idea is to reduce the given expressions to forms that are already known, using the elementary functions that were discussed in the text, as well as their properties. (a) z = 1ı : taking logarithms we have that ln(z) = ı ln(1), where ln(1) = 0 + 2nπı for some integer n, where n = 0 corresponds to the usual leaf of the logarithm function. It follows that we have ln(z) = 2nπ and 47

48

SOLUTIONS 4 therefore that z = exp(2nπ), where n is an arbitrary integer, and the particular value n = 0 corresponds to the particular value z = 1 for this imaginary power.

(b) z = eı : using the Euler formula we have that z = cos(1) + ı sin(1), where cos(1) and sin(1) are well-defined real values of these functions. (c) z = sin(ı): by the algorithmic definition of the function sin(z), we have that z = = =

eıı − e−ıı 2ı e−1 − e1 2ı  ı 1 e− , 2 e

which is purely imaginary and is written in terms of the well-defined real number e. Note that this result can also be written as ı sinh(1). (d) z = cos(ı): by the algorithmic definition of the function cos(z), we have that z = = =

eıı + e−ıı 2 e−1 + e1 2  1 1 e+ , 2 e

which is real and is written in terms of the well-defined real number e. Note that this result can also be written as cosh(1). Problem 2. Consider the complex function w(z) = z 1/n = is an integer.

√ n z where n > 1

(a) Show that w(z) is a multivalued function, with n different values for each z. (b) Show that these n values are evenly distributed along a circle in the complex plane.

EVEN LESS ELEMENTARY FUNCTIONS

49

(c) Determine the radius of this circle and the angles corresponding to each one of the n possible values. (d) Build a Riemann surface with n leaves to represent the domain of the function. Determine the singular point. (e) Show that the function is analytic throughout this Riemann surface. Complete Solution: In order to define the complex function w(z) = z 1/n , it is necessary to use the logarithm function, in which we must explicitly consider all the leaves of the Riemann surface. Thus, with z = ρ exp(ıθ) where θ ∈ [−π, π], we have that w(z) =

1

e n ln(z) , where

ln(z) = ln(ρ) + ıθ + ık2π, where k is an arbitrary integer, and ln(ρ) is the usual real logarithm function of the positive real quantity ρ. (a) Writing w(z) explicitly, we have w(z) = =

1

e n [ln(ρ)+ıθ+ık2π] k θ 1 e n ln(ρ) e(ı n ) e(ı n 2π)

k θ 1 = ρ n e(ı n ) e(ı n 2π) .

Since θ is between −π and π, and n > 1, θ/n never makes a complete turn around the origin, so that both the factor involving ρ and the first imaginary exponential are single-valued complex functions. On the other hand, the second exponential attributes to the function various values, as k varies along the integers. However, since both n and k are integers, a periodicity is established in this function, because its value for k = 0 coincides with its value for k = n, since in this case the argument of the exponential is 2πı. In fact, all values of k given by k = mn, where m is an arbitrary integer, result in the same value for the function, including the case of negative multiples of n. Thus, there are only n different values of the function, which can be represented, for example, by k ∈ {0, . . . , n − 1}. Thus we see that the function w(z) = z 1/n is a function of n values, which assigns n different complex values to each value of z.

50

SOLUTIONS 4

(b) As we already have the n values of the function explicitly written in polar form, k θ 1 w(z) = ρ n e[ı( n + n 2π)] , for k ∈ {0, . . . , n − 1},

we verify immediately that the radius ρ1/n does not depend on k, so that it is the same for all the n values of k, which are therefore located on the circle with this radius. Furthermore, each one of the values has an angle equal to the previous one increased by the constant quantity 2π/n, so that the n values are equally spaced along the circle. (c) As we have already seen in previous items, the radius is ρ1/n and the n angles are given by 2π θ + k , for k ∈ {0, . . . , n − 1}. n n (d) The singular point is the point z = 0, where the angles that differentiate the Riemann leaves from one another are undefined. Therefore, in order to build the Riemann surface, it is necessary to remove this point from the domain, and consider as leaves n superposed copies of the complex plane, each one with the origin removed. We assign to each of these n leaves a value of k in {0, . . . , n − 1}. Each of these leaves must be cut from the origin to infinity, along an arbitrary simple curve, the same for all leaves, for example the negative real semi-axis. Then, one side of each of these cuts is glued to the opposite side of the cut on the next leaf, thus forming a kind of spiraling surface with n turns. Finally, the remaining sides of the first and last leaves are glued together to form the Riemann surface that represents the domain of the function, and over which the image of the function can be represented completely and unambiguously. (e) Again, we start from the explicit polar form of the function, θ k 1 w(z) = ρ n e[ı( n + n 2π)]      1 θ θ k k + 2π + ı sin + 2π = ρ n cos n n n n     1 1 k k θ θ n n + 2π + ıρ sin + 2π , = ρ cos n n n n

EVEN LESS ELEMENTARY FUNCTIONS

51

and we use the polar version of the Cauchy-Riemann conditions in order to check the analyticity. We have for the relevant expressions involving the partial derivatives, ∂u ∂ρ 1 ∂v ρ ∂θ 1 ∂u ρ ∂θ ∂v − ∂ρ

= = = =

  1 1 −1 k θ n ρ + 2π , cos n n n   1 θ k 1 −1 ρn cos + 2π , n n n   1 θ k −1 (−1) n ρ sin + 2π , n n n   (−1) 1 −1 k θ ρ n sin + 2π , n n n

from which we find that the two Cauchy-Riemann conditions are satisfied for all values of ρ and θ except for ρ = 0, because in this case the derivatives do not exist, since the exponent (1/n) − 1 is negative, given that n > 1. Furthermore, we see that the two conditions are satisfied for any value of k, from which it follows that the function is analytic over the entire Riemann surface. Problem 3. For each one of the inverse functions listed below, prove that they can be written in terms of the logarithm as shown in each case.   √ (a) sin−1 (z) = −ı ln ız ± 1 − z 2 .

  √ (b) cos−1 (z) = −ı ln z ± z 2 − 1 .

  z . (c) tan−1 (z) = ı2 ln ıı + −z   √ (d) sinh−1 (z) = ln z ± z 2 + 1 .

  √ (e) cosh−1 (z) = ln z ± z 2 − 1 . (f) tanh−1 (z) =

1 2

  +z . ln 1 1−z

Complete Solution:

52

SOLUTIONS 4

In each case, we write z as the corresponding direct function of w, and isolate the exponential of w in terms of z, which always reduces to finding the roots of a quadratic equation. We use the Baskara formula for this, since it only depends on the operations of the field and on the square root function, both of these being concepts that are transparently generalized from the real context to the complex context. (a) w(z) = sin−1 (z): we have that z = sin(w), and therefore eıw − e−ıw 2ı ıw −ıw e −e − 2ız = 0 ⇒ z =

⇒

( eıw )2 − 2ız ( eıw ) − 1 = 0

⇒ p ( e ) = ız ± 1 − z 2 ⇒   p w(z) = −ı ln ız ± 1 − z 2 . ıw

(b) w(z) = cos−1 (z): we have that z = cos(w), and therefore eıw + e−ıw 2 ıw −ıw e +e − 2z = 0 ⇒ z =

( eıw )2 − 2z ( eıw ) + 1 = 0

⇒

⇒ p

z2 − 1 ⇒   p w(z) = −ı ln z ± z 2 − 1 .

(e ) = z ± ıw

(c) w(z) = tan−1 (z): we have that z = tan(w), and therefore 1 eıw − e−ıw ı eıw + e−ıw = 0 ⇒

z = eıw − e−ıw − ız eıw − ız e−ıw

(1 − ız) ( eıw )2 − (1 + ız) = 0 ⇒ r 1 + ız ıw (e ) = ⇒ 1 − ız   ı 1 + ız w(z) = − ln 2 1 − ız   ı+z ı ln = . 2 ı−z

⇒

53

EVEN LESS ELEMENTARY FUNCTIONS (d) w(z) = sinh−1 (z): we have that z = sinh(w), and therefore ew − e−w 2 ew − e−w − 2z = 0 ⇒ z =

( ew )2 − 2z ( ew ) − 1 = 0

⇒

⇒ p

( ew ) = z ± z 2 + 1 ⇒   p w(z) = ln z ± z 2 + 1 .

(e) w(z) = cosh−1 (z): we have that z = cosh(w), and therefore ew + e−w 2 ew + e−w − 2z = 0 ⇒ z =

( ew )2 − 2z ( ew ) + 1 = 0

⇒

⇒ p

( ew ) = z ± z 2 − 1 ⇒   p w(z) = ln z ± z 2 − 1 .

(f) w(z) = tanh−1 (z): we have that z = tanh(w), and therefore ew − e−w ew + e−w = 0 ⇒

z = ew − e−w − z ew − z e−w

⇒

(1 − z) ( ew )2 − (1 + z) = 0 ⇒ r 1+z w (e ) = ⇒ 1−z   1+z 1 ln . w(z) = 2 1−z

Each of these functions has multiple values, which can be determined by the analysis of the infinitely many Riemann leaves of the logarithm function, as well as of the two Riemann leaves of the square root function, which are explicitly represented in these formulas, where appropriate, by the symbol ±.

54

SOLUTIONS 4

Problem 4. Starting from the definition of the function Γ(x) in terms of a parametric integral, show that it can be written in each one of the three forms below. Z ∞ dt e−t e(x−1) ln(t) . (a) Γ(x) = 0

(b) Γ(x) =

Z

(c) Γ(x) =

Z

∞

0

dt −t x ln(t) e e . t

t=∞

d[ln(t)] e−t ex ln(t) .

t=0

Complete Solution: We start from the original parametric integral that defines Γ(x), Z ∞ Γ(x) = dt e−t tx−1 . 0

(a) First, we recall the definition of a power with an arbitrary real exponent, in terms of the logarithm function. Since the function in which we are interested is real, we take the n = 0 leaf of the logarithm. We have therefore tx−1 = Ze(x−1) ln(t) ⇒ ∞ dt e−t e(x−1) ln(t) . Γ(x) = 0

(b) Separating the argument of the second exponential in two parts in the result of the previous item, we have that e(x−1) ln(t) =

e(−1) ln(t) ex ln(t) 1 x ln(t) e ⇒ = Zt ∞ dt −t x ln(t) Γ(x) = e e . t 0

(c) Considering now that 1 d ln(t) = , dt t

55

EVEN LESS ELEMENTARY FUNCTIONS we have that dt/t = d[ln(t)], so that we obtain Γ(x) =

Z

t=∞

d[ln(t)] e−t ex ln(t) ,

t=0

where the integration interval is still written in terms of t. Note that by setting ξ = ln(t), that is, t = exp(ξ), this can also be written as Γ(x) = Problem 5. Γ(x).

Z

∞

dξ e− exp(ξ) exξ .

−∞

Calculate explicitly the following real values of the function

(a) Γ(1). (b) Γ(2). (c) Γ(1/2). Answer:

√

π.

Complete Solution: This involves the explicit calculation, in some particular cases, of the parametric integral that defines Γ(x), Z ∞ dt e−t tx−1 . Γ(x) = 0

(a) For x = 1 we have Γ(1) = =

Z

∞

Z0 ∞ 0

dt e−t t(1−1) dt e−t ∞

= − e−t

0

= −(0 − 1) = 1,

that is, Γ(1) = 1.

56

SOLUTIONS 4

(b) For x = 2 we have ∞

Z

Γ(2) =

Z0 ∞

=

dt e−t t(2−1) dt t e−t .

0

In this case we integrate by parts, ∞

Z

dt t e−t 0 ∞ Z −t = −t e +

Γ(2) =

∞

dt e−t

0

0

= −(0 − 0) − e−t = −(0 − 1)

∞ 0

= 1,

that is, Γ(2) = 1. (c) For x = 1/2 we have Γ(1/2) = = =

∞

Z

Z0 ∞

dt e−t t1/2−1 dt e−t t−1/2

Z0 ∞ 0

dt √ e−t . t

Making the transformation of variables t = ξ 2 , that is, ξ = dt = 2ξ dξ, we have Γ(1/2) = =

∞

Z

Z0 ∞ 0

= 2

Z

dt √ e−t t 2ξ dξ −ξ 2 e ξ

∞

0

2

dξ e−ξ .

√ t, with

57

EVEN LESS ELEMENTARY FUNCTIONS

Since the integrand is an even function, we can write the integral in the form Z ∞ 2 dξ e−ξ Γ(1/2) = 2 0 Z ∞ 2 dξ e−ξ = −∞ √ π, = where we used the result for the Gaussian integral over the whole real line, which is obtained in another problem of this chapter, in the case √ α = 1. In short, we have that Γ(1/2) = π. Problem 6. Show that Γ(z + 1) = zΓ(z) for any complex z except for the origin z = 0 and the negative integers. Complete Solution: We start from the original parametric integral that defines Γ(x), generalized to a complex argument z, which we can do without problems since x or z are only fixed parameters for this integration, which remains a real integration on t, Z ∞ dt e−t tz−1 . Γ(z) = 0

The only concern we must have is that the integral exist, and as we have seen in the text it exists so long as ℜ(z) > 0. We simply write Γ(z + 1), and integrate by parts, to obtain Z ∞ dt e−t tz Γ(z + 1) = 0 ∞ Z ∞ −t z dt e−t ztz−1 = −e t + 0 0 ∞ Z ∞ dt e−t tz−1 +z = − e−t ez ln(t) 0 0 ∞ = − e−t e(x+ıy) ln(t) + zΓ(z), 0

where we wrote z as x + ıy in the exponent of the second exponential in the integrated term. Joining the two exponentials in the integrated term, we have for that term

58

SOLUTIONS 4 − e−t e(x+ıy) ln(t) = − e−t+x ln(t)+ıy ln(t)

= − e−t+x ln(t) eıy ln(t) .

In the limit t → ∞ the second exponential, which has as real and imaginary parts real periodic functions, simply oscillates, remaining with absolute value limited by 1. In this same limit the argument of the first exponential goes to −∞, because the ln(t) goes to ∞ much more slowly than −t goes to −∞. Thus we see that this first exponential goes to zero, as well as the entire integrated term, when t → ∞. On the other hand, in the limit t → 0 we have that ln(t) goes to −∞, so that the second exponential still remains limited and oscillates. In this other limit the argument of the first exponential goes to −∞ since x > 0, in which case this exponential also vanishes. Thus we see that the integrated term vanishes in both limits, so long as ℜ(z) > 0. Going back to the previous equation, we therefore have that Γ(z + 1) = zΓ(z), for ℜ(z) > 0. Using the procedure described in the text, we can now use this property to extend the analytic definition of this function to successive strips of the complex plane with x < 0. Of course, in this way the property remains valid throughout the region to which it is possible to extend the function. For example, for x > −1 we can write that Γ(z) =

Γ(z + 1) , z

since the right-hand side of this equation is well defined for x > −1. Of course, due to the zero in denominator, this definition cannot be used for the case z = 0, at which point the function remains undefined, having there an isolated singularity. Continuing with the process for x > −2 we can write that Γ(z) =

Γ(z + 2) , z(z + 1)

which remains undefined at z = 0 and at z = −1. Continuing this process indefinitely, we can extend the function to the whole complex plane except for the zero and the negative integers, which is also the domain where the property Γ(z + 1) = zΓ(z) holds.

EVEN LESS ELEMENTARY FUNCTIONS Problem 7.

59

Consider the function   (t − t0 )2 , f (t) = exp − 2τ 2

where t0 is a real constant and τ 6= 0 is a strictly positive real constant, and consider also the definition of the average value of another function g(t) in the statistical distribution defined by f (t), which is given by Z ∞ g(t)f (t) dt −∞ hgi = Z ∞ . f (t) dt −∞

(a) Calculate the average value of t, that is, hti. Answer: t0 . (b) Calculate the dispersion of t, that is the quantity σt =

p

h(t − hti)2 i.

Answer: τ . Complete Solution: These are two examples of calculation of average values via Z ∞ 2 2 dt g(t) e−(t−t0 ) /(2τ ) Z ∞ . hgi = −∞ 2 2 dt e−(t−t0 ) /(2τ ) −∞

By means of a simple transformation of variables, we can see that the integral in the denominator is an example of Gaussian integral, whose value is given by the result that is obtained in another problem of this chapter. Making ξ = t − t0 we have that dt = dξ, and therefore that Z ∞ Z ∞ 2 2 2 2 dξ e−ξ /(2τ ) dt e−(t−t0 ) /(2τ ) = −∞ −∞ p = (2τ 2 )π √ 2π τ. =

60

SOLUTIONS 4

Here we used the result for the Gaussian integral that is obtained in another problem of this chapter, with α = 2τ 2 . Let us give √ the name N to this normalization denominator, that is, we have that N = 2π τ . (a) We must calculate the following integral, 1 hti = N

Z

∞

2 /(2τ 2 )

dt t e−(t−t0 )

.

−∞

Making the transformation of variables ξ = t − t0 , t = ξ + t0 , dt = dξ, we have Z 1 ∞ 2 2 hti = dξ (ξ + t0 ) e−ξ /(2τ ) N −∞ Z Z t0 ∞ 1 ∞ 2 2 2 2 dξ ξ e−ξ /(2τ ) + dξ e−ξ /(2τ ) . = N −∞ N −∞ The first of these two integrals is zero because it is the integral of an odd function on a symmetric domain. The second integral is again simply N , so that we have for the average value hti = t0 . (b) We must calculate the following integral, σt2 = = =

 (t − t0 )2 Z 1 ∞ 2 2 dt (t − t0 )2 e−(t−t0 ) /(2τ ) N −∞ Z 1 ∞ 2 dt (t − t0 )2 e−β(t−t0 ) , N −∞



2 where √ we defined the parameter β = 1/(2τ ), that is, we have that τ = 1/ 2β. We can now write this integral as a derivative with respect to β of another integral, which ends up being once again the normalization integral of value N ,

1 N 1 = − N

σt2 = −

Z ∞ ∂ 2 dt e−β(t−t0 ) ∂β −∞ ∂N . ∂β

61

EVEN LESS ELEMENTARY FUNCTIONS Writing N =

√

2π τ in terms of β and taking the derivative, we have  1 ∂ √ 2π τ N ∂β √  1 ∂ π √ = − N ∂β β √ 1 1 = π √ . 2N β 3

σt2 = −

Substituting again the value of N = σt2

√ = π

p

r

π/β in terms of β we have

β 1 1 √ π 2 β3

1 1 √ 2 β2 1 = 2β = τ 2. =

It follows therefore that we have for the dispersion around t0 σt = τ. Problem 8. (Challenge Problem) Show that, for any real, strictly positive number α,  2 Z ∞ √ t dt = απ. exp − α −∞ Hint: calculate the square of the integral, and transform to a polar coordinate system on the plane. Complete Solution: We will calculate the following integral, in which the integration variable can have any name, so that we chose the name x for it, Z ∞ 2 dx e−x /α , I= −∞

62

SOLUTIONS 4

where α > 0. Following the hint given, we will calculate I 2 , because it is easier to do this than to calculate I directly. Since we will have the product of two integrals, we need to use two different integration variables, to avoid confusion, so we chose x and y, Z ∞ Z ∞ 2 2 dx e−x /α I2 = dy e−y /α −∞ Z−∞ ∞ Z ∞ 2 2 = dx dy e−(x +y )/α . −∞

−∞

We now make a transformation in the integration variables, interpreting the pair (x, y) as Cartesian coordinates in an infinite plane, and changing to the corresponding polar coordinates r and θ, where x = r cos(θ), y = r sin(θ). The transformation of the integration element is dx dy = r dr dθ, and we also have that x2 + y 2 = r 2 , so that we can write the integral on the plane R2 in the form Z 2 2 2 dx dy e−(x +y )/α I = 2 ZR 2 r dr dθ e−r /α = R2 Z ∞ Z 2π 2 dr r e−r /α . dθ = 0

0

The integral over θ is now immediate, and the integral over r can be done by reversing the chain rule, due to the additional factor of r which appeared in the integrand. Thus we have Z ∞ Z 2π 2 2 dr r e−r /α dθ I = 0 0  (−α) −r2 /α ∞ e = 2π 2 0 = −πα(0 − 1) = απ.

We have therefore that I 2 = απ, and hence it follows that I =

√

απ.

Solutions 5

Geometrical Aspects of the Functions We present here complete and commented solutions to all problems proposed in Chapter 5 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. Let us call stationary those points where a function f (x, y) of two variables (x, y) has its two first partial derivatives with respect to these variables equal to zero. Assuming that at least one of the two second partial derivatives is non-zero, a stationary point of this type can be a point of local maximum of the function, a point of local minimum of the function, an inflection point or a saddle point. At a point of local minimum the two second partial derivatives are strictly positive, at a point of local maximum the two second partial derivatives are strictly negative, at an inflection point one of the second derivatives is zero while the other is strictly positive or strictly negative, and at a saddle point one of the second derivatives is strictly positive while the other is strictly negative. Prove that the functions u(x, y) and v(x, y) which constitute an analytic function cannot have any points of local minimum, local maximum, or inflection, but only saddle points. Complete Solution: As noted in the text, the functions u(x, y) and v(x, y) which are the real and imaginary parts of an analytic function are always harmonic functions, that is, continuous and differentiable functions that satisfy the Laplace equation in two dimensions, 63

64

SOLUTIONS 5

∂2 ∂2 f (x, y) + f (x, y) = 0, ∂x2 ∂y 2 at all points where the function is analytic, where we are using f as a generic name for either u or v. It follows that there can be no point where the two second partial derivatives are strictly positive, in which case the sum would be strictly positive, and this equation could not be satisfied, which eliminates the possibility of points of local minimum. The same can be said for the case in which the two second partial derivatives are strictly negative, which eliminates the possibility of points of local maximum. Furthermore, if one of the two second partial derivatives is zero it follows that the other must also be zero, which eliminates the possibility of any inflection points as described in the statement of the problem. Therefore, the only possibilities are that both second derivatives be zero, or that one be strictly positive while the other be strictly negative. As we assumed by definition that at least one of the two partial second derivatives is non-zero, the only remaining possibility is that there are saddle points. Problem 2. Consider an arbitrary analytic function w(z) = u(x, y) + ıv(x, y) of the complex variable z = x + ıy. (a) Write the gradient vectors in the (x, y) plane of the two harmonic functions u(x, y) and v(x, y). (b) Show that these two gradient vectors are orthogonal to each other, at any point of the (x, y) plane where w(z) is analytic. (c) Show that these two gradient vectors have the same absolute value at every point of the (x, y) plane where w(z) is analytic. Complete Solution: We have to deal here with some basic concepts of the vector calculus in two dimensions. (a) We have for the gradient vectors,  ∂u ∂u , , ∂x ∂y   ∂v ∂v ~ , . ∇v = ∂x ∂y

~ ∇u =



GEOMETRICAL ASPECTS OF THE FUNCTIONS

65

Since the function is analytic, we can use the Cauchy-Riemann conditions in order to write this only in terms of partial derivatives of u(x, y),  ∂u ∂u , , ∂x ∂y   ∂u ∂u ~ ∇v = − , . ∂y ∂x

~ ∇u =



(b) Calculating the dot product of the two gradient vectors, using the latter form for them, we have    ∂u ∂u ∂u ∂u , , · − ∂x ∂y ∂y ∂x ∂u ∂u ∂u ∂u + = − ∂x ∂y ∂y ∂x = 0.

~ · ∇v ~ ∇u =



It follows therefore that the two gradient vectors are orthogonal to one another at all points where the Cauchy-Riemann conditions hold, that is, at all points where the function w(z) is analytic. (c) Calculating the absolute value of the gradient vector in each case, using the latter form for them, we have s

 ∂u 2 + ∂y s 2  2 ∂u ∂u ~ | = |∇v − + ∂y ∂x s    ∂u 2 ∂u 2 + . = ∂x ∂y

~ | = |∇u

∂u ∂x

2



It follows therefore that the two gradients have the same absolute value, at all points where the Cauchy-Riemann conditions hold, that is, at all points where the function w(z) is analytic.

66

SOLUTIONS 5

Problem 3. Consider the analytic function w(z) = 1/z as a mapping from the complex plane z = (x, y) onto the complex plane w = (u, v). (a) Show that the unit circle is mapped onto itself, but not as the identity map. (b) Find the fixed points of the mapping, that is, points that are mapped to themselves. Answer: ±1. (c) Show that the interior of the unit circle is mapped onto its exterior, and vice-versa. Complete Solution: Writing the function w(z) = 1/z in the polar representation, where z = ρ exp(ıθ), we have w(z) = u + ıv 1 −ıθ e . = ρ (a) Making ρ = 1 we also have that 1/ρ = 1, so that we are on the unit circle in both the (x, y) plane and the (u, v) plane, which makes it clear that the unit circle of (x, y) is mapped on the unit circle of (u, v). In detail, we have the mapping z = cos(θ) + ı sin(θ) w = cos(θ) − ı sin(θ),

−→

which shows that there is a reversal of the orientation of the unit circle, since θ is replaced by −θ by the mapping, so that the mapping is not the point-to-point identity on the circle. Nevertheless, the unit circle of (x, y) is, in its entirety, mapped onto the unit circle of (u, v). (b) The fixed points of the mapping are those at which w = z, that is, u = x and v = y. Of course this is only possible if ρ = 1, because otherwise the distance from the point to the origin in the (x, y) plane would not be preserved when one maps the point onto the corresponding point in the (u, v) plane. In other words, the fixed points are necessarily on the

GEOMETRICAL ASPECTS OF THE FUNCTIONS

67

unit circle. The condition u = x is an identity, but the condition v = y implies that sin(θ) = − sin(θ), which in turn implies that sin(θ) = 0, and therefore that θ = 0 or θ = π. It follows that the two fixed points of the mapping are z = 1 and z = −1. (c) This is immediate because, if z is within the unit circle, then ρ < 1, which implies that 1/ρ > 1, so that w is outside the unit circle. In other words, the interior of the unit circle of (x, y) is mapped onto the exterior of the unit circle of (u, v). The reverse situation is also true because, if z is outside the unit circle, then ρ > 1, which implies that 1/ρ < 1, so that w is within the unit circle. In other words, the exterior of the unit circle of (x, y) is mapped onto the interior of the unit circle of (u, v). Problem 4.

Consider the complex function 1 w(z) = z + , z

where z = x + ıy and w = u + ıv, seen as a transformation that maps the complex (x, y) plane onto the complex (u, v) plane. (a) Show that this transformation maps the unit semicircle with positive imaginary part of the (x, y) plane onto the real segment (−2, 2) of the (u, v) plane. (b) Show that this transformation maps the radii of the unit disk that fall strictly within this upper unit semicircle onto curves starting in the real segment (−2, 2), extending to infinity on the (u, v) half-plane with negative imaginary part. (c) Show that the curves mentioned in the previous item intersect the segment (−2, 2) perpendicularly, on the (u, v) plane. (d) Show that the interior of the semicircle is mapped onto the whole lower half-plane in the (u, v) plane.

68

SOLUTIONS 5

Complete Solution: Writing explicitly u and v in terms of polar variables ρ and θ on the (x, y) plane we have w(z) = z +

1 z

1 = ρ eıθ + e−ıθ ρ     1 1 = ρ+ cos(θ) + ı ρ − sin(θ) ρ ρ   1 u(ρ, θ) = ρ+ cos(θ), ρ   1 sin(θ). v(ρ, θ) = ρ− ρ

⇒

(a) The upper unit semicircle is described by ρ = 1 and θ ∈ [0, π], so that we have that w is real, given by w = 2 cos(θ). In other words, we have v = 0, so that we are on the real axis of (u, v), and u ∈ [−2, 2], because the cosine varies from −1 to 1. It follows that the semicircle is mapped onto this segment. (b) Let us start with some particular cases for the radius. In the case θ = 0 and ρ ∈ [0, 1], which is one of the two limiting cases of those described in the statement of the problem, we have that the cosine is 1 and that the sine is zero, so that w is real,   1 . w = ρ+ ρ We see that v = 0, so that we are on the real axis of (u, v), and that for ρ → 1 we have that u → 2, while for ρ → 0 we have that u → ∞, so that this radius is mapped onto the real semi-axis [2, ∞) of the (u, v) plane. On the other hand, in the particular case of the vertical radius given by θ = π/2 and ρ ∈ [0, 1] we have that the cosine vanishes and that the sine is 1, so that w is purely imaginary,   1 ı. w = ρ− ρ We see that u = 0, so that we are on the imaginary axis of (u, v), and that for ρ → 1 we have that v → 0, while for ρ → 0 we have that

GEOMETRICAL ASPECTS OF THE FUNCTIONS

69

v → −∞, so that the radius is mapped onto the negative imaginary semi-axis of the (u, v) plane. Since the transformation from (x, y) to (u, v) is clearly symmetric by reflection about the axis y, whereby the imaginary part of w does not change, but the real part changes sign, it suffices to examine in detail the interval [0, π/2] of values of θ. The cases θ = 0 and θ = π/2 have been already analyzed, corresponding respectively to the part of the positive real semi-axis above u = 2 and to the negative imaginary semi-axis of the (u, v) plane. For values within the interval we have cos(θ) > 0 and sin(θ) > 0, so that  1 cos(θ), u(ρ, θ) = ρ+ ρ   1 v(ρ, θ) = ρ− sin(θ), ρ 

where u is always positive for any value of ρ ∈ [0, 1], while v is always negative. As we have seen, when ρ → 1 we have that v vanishes and u is on the real interval [−2, 2] of the (u, v) plane, showing that all the curves in the image begin at this real segment. On the other hand, when ρ → 0 we have that u → ∞ while v → −∞, so that the curves extend to infinity in the lower right quadrant of the (x, y) plane. The slope of these curves, which are described by the parameter ρ with θ constant, is given by ∂v ∂u

= = =

∂v ∂ρ ∂ρ ∂u (1 + 1/ρ2 ) sin(θ) (1 − 1/ρ2 ) cos(θ) (ρ2 + 1) sin(θ) . (ρ2 − 1) cos(θ)

Taking the limit ρ → 0, corresponding to the asymptotic limit for the curves in (u, v), we have that the slope has the limit ∂v ρ→0 ∂u lim

sin(θ) cos(θ) = − tan(θ). = −

The slopes of the curves are therefore all negative, and interpolate continuously between the positive real semi-axis for θ = 0 and the

70

SOLUTIONS 5 negative imaginary semi-axis for θ = π/2. By symmetry, a similar situation happens for negative x and u, on the other halves of the respective planes.

(c) As we saw in the previous item, the slope of the curves in (u, v) is given by (ρ2 + 1) sin(θ) ∂v = 2 , ∂u (ρ − 1) cos(θ) where both the cosine and the sine are non-zero for θ ∈ (0, π), excluding θ = 0 and θ = π, provided that we also have that θ 6= π/2. Taking the limit ρ → 1, which corresponds to the real segment (−2, 2) on the (u, v) plane, we have that lim

ρ→1

∂v = ∓∞, ∂u

where the sign depends on whether we are within the interval θ ∈ (0, π/2) (negative), or within the interval θ ∈ (π/2, π) (positive). This shows that the curves tend to be vertical as we approach the horizontal segment (−2, 2) on the (u, v) plane, and are therefore perpendicular to it. This result can be extended to the case θ = π/2, because in this case the slope is ∓∞, with a sign that depends on how we take the limit to the point π/2, whatever the value of ρ, and therefore we have the negative imaginary semi-axis, which is vertical and therefore perpendicular to the segment. Note that we can take the two limits, ρ → 1 and θ → π/2, in any order, always obtaining the same result. However, the result cannot be extended to the cases θ = 0 and θ = π, because in these cases the results depend on the order in which we take the two limits. If we take the limit ρ → 1 first, the result for the slope is ∓∞, indicating the perpendicularity, but if we take the limit θ → 0 first, the result is zero, indicating a curve that tends to become horizontal and therefore parallel to the segment. (d) As we have seen in the previous items, we can build an infinite family of curves that covers completely and continuously the domain, namely the set of all the radii of the upper unit semicircle, so that the corresponding family of curves on the image covers in a complete and continuous way the entire lower half-plane, stretching to infinity in all directions and continuously interpolating between the real positive semi-axis and

GEOMETRICAL ASPECTS OF THE FUNCTIONS

71

the real negative semi-axis. It follows that the interior of the upper semicircle of the (x, y) plane is mapped to the entire lower half-plane of the (u, v) plane. Problem 5. Consider the two oriented curves C1 and C2 in the complex (x, y) plane, which intersect at a certain point, and the transformation defined by the analytic function w(z), which maps these two curves onto two other curves C1′ and C2′ in the complex (u, v) plane, as discussed in the text. Assume that the gradients of u(x, y) and v(x, y) are not zero at the intersection point. Consider two infinitesimal variations of z at the intersection point in the (x, y) ~ 1 and dz ~ 2 , each one tangent to and plane, denoted in vector language by dz pointing in the positive direction of the corresponding curve. (a) Show that the sine of the angle θ between the two curves in the (x, y) plane is given by sin(θ) =

dx1 dy2 − dx2 dy1 , dz ~ dz ~ 1

2

~ 1 = dx1 + ıdy1 and dz ~ 2 = dx2 + ıdy2 . Show in the same way where dz ′ that the sine of the angle θ between the corresponding curves C1′ and C2′ in the complex (u, v) plane, at the intersection point, is given by
du1 dv2 − du2 dv1 , sin θ ′ = dw ~ 1 dw ~ 2

~ 1 = du1 + ıdv1 and dw ~ 2 = du2 + ıdv2 . where dw Hint: consider using vector (cross) products.

(b) Prove, using the analyticity properties of the transformation function w(z), that sin(θ ′ ) = sin(θ). Together with the result that was shown in the text, cos(θ ′ ) = cos(θ), this suffices to ensure that θ ′ = θ, thus completing the proof started in the text.

Complete Solution:

72

SOLUTIONS 5

~ 1 = dx1 + ıdy1 and (a) Considering the two infinitesimal displacements dz ~ 2 = dx2 +ıdy2 on the (x, y) plane, we can build a vector perpendicular dz ~ 1 × dz ~ 2 . On to the plane by means of the cross or vector product dz the one hand, the component of this product in the positive direction along the normal to the plane can be written as dx1 dy2 − dx2 dy1 . On the other hand, the absolute value of the product, which is equal to the ~ 1 dz ~ 2 sin(θ). absolute value of this component, can be written as dz It follows that we can write for the sin(θ), including the sign associated with the orientation of the normal to the plane, sin(θ) =

dx1 dy2 − dx2 dy1 . dz ~ dz ~ 1

2

Since the two planes have the same structure, it is clear that this same ~ 1 = du1 +ıdv1 argument, applied to the corresponding displacements dw ~ 2 = du2 + ıdv2 on the (u, v) plane, implies the corresponding and dw relation for θ ′ ,
du1 dv2 − du2 dv1 . sin θ ′ = dw ~ 1 dw ~ 2

(b) Let us now calculate sin(θ ′ ) using the formula derived in the previous item. Using the expression of the differentials of u(x, y) and v(x, y) in ~ 1,2 as functions of terms of the variations of x and y, we can write dw the corresponding (dx, dy)1,2 , as was done in the text. Omitting, for the moment, the indices that identify the curves, we have ~ = (du, dv) dw   ∂u ∂v ∂v ∂u dx + dy, dx + dy . = ∂x ∂y ∂x ∂y Using the Cauchy-Riemann conditions we can write this solely in terms of the partial derivatives of u, ~ = dw



 ∂u ∂u ∂u ∂u dx + dy, − dx + dy . ∂x ∂y ∂y ∂x

As shown in the text, we have for the absolute values in the denominator of our formulas, for each one of the two curves,

GEOMETRICAL ASPECTS OF THE FUNCTIONS

73

dw ~ = |∇u ~ . ~ | dz

We can also calculate the product in the numerator involving the vari~ 1 and dw ~ 2 . Of course, this time we need to keep the indices ations dw of the curves explicitly at every step, and we have du1 dv2 − du2 dv1    ∂u ∂u ∂u ∂u − dx1 + dy1 dx2 + dy2 + = ∂x ∂y ∂y ∂x    ∂u ∂u ∂u ∂u − dx2 + dy2 dx1 + dy1 − ∂x ∂y ∂y ∂x ∂u ∂u ∂u ∂u dx1 dx2 + dx1 dy2 + = − ∂x ∂y ∂x ∂x ∂u ∂u ∂u ∂u − dx2 dy1 + dy1 dy2 + ∂y ∂y ∂y ∂x ∂u ∂u ∂u ∂u dx1 dx2 − dx2 dy1 + + ∂x ∂y ∂x ∂x ∂u ∂u ∂u ∂u + dx1 dy2 − dy1 dy2 . ∂y ∂y ∂y ∂x Just as it was the case in the text, we find that due to the CauchyRiemann conditions several of the terms that appear cancel each other off. This time only the mixed products are left, so that we have du1 dv2 − du2 dv1

 ∂u 2 (dx1 dy2 − dx2 dy1 ) + = ∂x  2 ∂u + (dx1 dy2 − dx2 dy1 ) ∂y 

2

~ | (dx1 dy2 − dx2 dy1 ) . = |∇u We can now assemble our formula for sin(θ ′ ), sin θ ′




=

du1 dv2 − du2 dv1 dw ~ 1 dw ~ 2 2

=

~ | (dx1 dy2 − dx2 dy1 ) |∇u ~ 1 dz ~ 2 ~ |2 dz |∇u

74

SOLUTIONS 5 =

(dx1 dy2 − dx2 dy1 ) dz ~ dz ~ 1

2

= sin(θ),

~ | 6= 0. We have therefore that sin(θ ′ ) = where we are assuming that |∇u sin(θ), and as was shown in the text, we also have that cos(θ ′ ) = cos(θ). This is enough to ensure that θ ′ = θ, thus completing the demonstration that the transformations generated by analytic functions preserve the angles between curves, and are therefore conformal transformations, at ~ | 6= 0, and therefore |∇v ~ | 6= 0. points where |∇u

Solutions 6

Border Effects in Capacitors We present here complete and commented solutions to all problems proposed in Chapter 6 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. Consider the complex variables z = x + ıy and w = u + ıv, and the analytic function w(z) = z + ez , interpreted as a mapping or transformation from the complex plane z = (x, y) onto the complex plane w = (u, v). (a) Show that this transformation maps the straight line y = 0 of the (x, y) plane onto the straight line v = 0 of the (u, v) plane. (b) Show that this transformation maps the two semi-axes of the (x, y) plane given by y = ±π with x ≥ 0 onto the two semi-axes of the (u, v) plane given by v = ±π with u ≤ −1, respectively, and that it maps the two semi-axes of the (x, y) plane given by y = ±π with x ≤ 0 onto these same two semi-axes of the (u, v) plane. (c) Show that this transformation maps the strip given by the interval −π ≤ y ≤ π of the (x, y) plane onto the entire (u, v) plane. In order to do this consider the mappings of families of curves. ~ | and/or |∇v ~ |, show that this transformation is analyti(d) Calculating |∇u cally invertible within the strip of the (x, y) plane between y = −π and 75

76

SOLUTIONS 6 y = π, including the boundary of this region, except for the two points (0, π) and (0, −π) located on this boundary.

Complete Solution: Writing the analytic function w(z) = z + exp(z) explicitly in terms of real functions, with z = x + ıy, we have w(z) = u(x, y) + ıv(x, y) = x + ıy + ex [cos(y) + ı sin(y)] x

u(x, y) = x + e cos(y),

⇒

v(x, y) = y + ex sin(y). (a) Making y = 0, we immediately have that v = 0, while for u we have that u(x, 0) = x + ex . We have therefore the line v = 0 in the image plane, or a part of it. It is easy to see that if we make x → ∞, we will have u → ∞, because both the term x and the exponential exp(x) diverge to ∞. On the other hand, if we make x → −∞, the exponential vanishes quickly, and we have that the limit is determined by the term x, that is, we have that u → −∞. We see in this way that we travel along the whole straight line v = 0 in the image plane, that is, that the line y = 0 of the (x, y) plane is mapped on the line v = 0 of the (u, v) plane. (b) Making y = ±π, we have that sin(y) = 0 and that cos(y) = −1, so that we have for u and v, u(x, ±π) = x − ex , v(x, ±π) = ±π.

With this we can already see that we are on the straight lines v = ±π in the image plane. We now need to determine the maximum value of u(x), so that we compute its derivative and equate it to zero, ∂u = 1 − ex = 0 ⇒ ∂x x = 0.

77

BORDER EFFECTS IN CAPACITORS

Calculating also the second derivative, and applying at the point x = 0, we have ∂2u ∂x2

= − ex ⇒

∂2u (0) = −1 ∂x2


0 are mapped onto curves in the upper half-plane of the (u, v) plane, and those with y < 0 are mapped onto curves in the lower half-plane. Thus, it suffices to analyze in detail the case y > 0. Therefore, let us begin by examining here one more particular case, the case y = π/2. Making y = π/2, we have that sin(y) = 1 and that cos(y) = 0, so that we have for u and v, u(x, π/2) = x, π + ex . v(x, π/2) = 2 In the limit in which x → −∞, we have that v quickly approaches π/2, while u always equals x. We see therefore that in this case the mapping tends rapidly to the identity, so that in this limit the strip is

78

SOLUTIONS 6 mapped onto itself. On the other hand, in the limit in which x → ∞, we still have that u = x, but now v begins to grow exponentially with x. The curve in the (u, v) plane therefore has an exponential shape, and extends to infinity. Its slope is given by ∂v ∂u

= =

∂v ∂x ex ,

which tends to ∞ when x → ∞. It follows that in this limit the angle between the tangent to the curve and the u axis is π/2. In all other cases in which 0 < y < π/2 the corresponding curve will also extend to infinity, but will asymptotically make angles between 0 and π/2 with the axis u, with positive tangent. If we have values of y in this interval, both cos(y) and sin(y) will be positive and non-zero, and we will therefore have u(x, y) = x + ex cos(y), v(x, y) = y + ex sin(y), where both u and v tend to ∞ when x → ∞, so that again these curves extend to infinity in the (u, v) plane. The slope of the curves is given by ∂v ∂u

= =

∂v ∂x ∂x ∂u ex sin(y) . 1 + ex cos(y)

In the limit x → ∞ the slope tends to ∂v x→∞ ∂u lim

ex sin(y) x→∞ 1 + ex cos(y) sin(y) = cos(y) = tan(y), =

lim

which for the interval of values of y in consideration here is positive, indicating an angle between 0 and π/2 between the tangent to the curve

79

BORDER EFFECTS IN CAPACITORS

and the u axis. In the complementary case, if we have values of y in the interval (π/2, π), we still have sin(y) strictly positive, but in this case cos(y) will be negative and not zero, and we will therefore have u(x, y) = x + ex cos(y), v(x, y) = y + ex sin(y), where v still tends to ∞, but u tends to −∞ when x → ∞, so that again these curves extend to infinity in the (u, v) plane, but now with negative slope, contained in the second quadrant. The slope of the curves is given by the same formula as before, and in the limit x → ∞ the slope tends once again to tan(y), that for the interval of values of y in consideration here is this time negative, indicating an angle between π/2 and π between the tangent to the curve and the u axis. Thus we see that the part with negative x of the strip in the (x, y) plane is mapped essentially onto the same strip in the (u, v) plane, while the part with x and y both positive in the strip in the (x, y) plane is mapped on the part of the upper half-plane of the (u, v) plane which is outside of the strip. The curves continuously change between the semi-axis defined by v = 0 and u > 0 and the semi-axis defined by v = π and u < 0. A mapping symmetric to this one exists in the lower half-plane. Thus we see, by continuity, that the whole (u, v) plane is mapped from the strip −π ≤ y ≤ π in the (x, y) plane. (d) Calculating the gradient of u(x, y), as well as its absolute value, we have ~ ∇u(x, y) = [1 + ex cos(y)] x ˆ − [ ex sin(y)] yˆ 2 ∇u(x, ~ y) = [1 + ex cos(y)]2 + [ ex sin(y)]2

⇒

= 1 + e2x + 2 ex cos(y).

It is not necessary to independently calculate the gradient of v(x, y), since the two gradients have the same absolute value. The question now is to determine at what points this expression vanishes. The easiest way to do this is to check in what conditions the gradient is zero, because its modulus is zero if and only if it is zero itself. In this case we have the two conditions 1 + ex cos(y) = 0, ex sin(y) = 0.

80

SOLUTIONS 6 From the second condition we can conclude that sin(y) must vanish, because the exponential of a real number is always strictly positive, so that we have that y = kπ, where k is some integer. The values falling within the interval of interest, y ∈ [−π, π], are y = 0 and y = ±π. In the first case, the first condition above is reduced to 1 + exp(x) = 0, which has no solution for x, because the exponential of a real number is always positive. There remains therefore the alternative y = ±π, for which the first condition reduces to 1 − exp(x) = 0, which is only satisfied for x = 0. It follows that we have only two points belonging to the strip where the two gradients are zero, at which the transformation fails to have an analytic inverse, and therefore is not conformal: the points (0, π) and (0, −π).

Problem 2.

Show that the integral Z −1 −u0

du , 1 − ex(u)

where u(x) = x − ex , is finite for any finite value of the constant u0 in the interval [1, ∞). Complete Solution: Since we have that u(x) = x − exp(x), it follows that we also have that du = dx − ex dx

= (1 − ex ) dx.

In addition to this, the value u = −1 corresponds to x = 0, and denoting by x0 the finite value of x corresponding to the finite value u0 = x0 − exp(x0 ), we have that Z −1 Z 0 du 1 − ex dx = x(u) 1 − ex −u0 1 − e 0 Z−x x0 dx = 0

= x0 .

It follows that the integral is finite for any finite value of x0 , which is the case for any value of u0 in the given interval.

81

BORDER EFFECTS IN CAPACITORS

Problem 3. Consider the complex function w(z) = z 2 where z = x + ıy and w = u + ıv. (a) Find the equipotential curves of u and v, that is, those in which these quantities are constant. ~ u = −∇u ~ and E ~ v = −∇v. ~ (b) Calculate the electric-field vectors E ~u · E ~ v = 0 at all points, where the dot represents the (c) Show that E dot-product of vectors. ~ u = E ~ v at all points, where the absolute values shown (d) Show that E are vector magnitudes in the usual sense, that is,

~ v . and similarly for E

q E ~u · E ~u. ~ u = E

(e) Show that the equipotential curves of u and v are the integral curves ~ v and E ~ u respectively. (field lines) of E Complete Solution: Since we have that w(z) = z 2 , with z = x + ıy and w = u + ıv, it follows that w(z) = (x2 − y 2 ) + ı2xy 2

2

u(x, y) = x − y ,

⇒

v(x, y) = 2xy.

(a) The curves u = u0 are given by p x2 − y 2 = u0 . If the constant u0 is positive, then we have that x = ± u0 + y 2 , which is a pair of hyperbolas asymptotic to the straight lines x = ±y, located in the half-planes x > 0 and √ x < 0. If the constant u0 is negative, then we have that y = ± −u0 + x2 , which is a pair of hyperbolas asymptotic to the straight lines x = ±y, located in the half-planes y > 0 and y < 0. The curves v = v0 are given by y = v0 /(2x), for any sign of v0 . In the case in which v0 > 0 this defines a pair of hyperbolas asymptotic to the coordinate axes, located in the quadrants (x > 0, y > 0) and (x < 0, y < 0). In the case in which v0 < 0 this defines a pair of hyperbolas asymptotic to the coordinate axes, located in the quadrants (x > 0, y < 0) and (x < 0, y > 0).

82

SOLUTIONS 6

(b) Calculating the gradients we have ~ u = −∇u(x, ~ E y)

= −(2x, −2y)

= −2(x, −y), ~ ~ Ev = −∇v(x, y) = −(2y, 2x) = −2(y, x).

(c) Calculating directly we have ~u · E ~ v = 4 (x, −y) · (y, x) E = 4 (xy − yx) = 0. (d) Calculating directly we have E ~ = u E ~ = v E ~ = u

p 2 x2 + (−y)2 , p 2 y 2 + x2 ⇒ E ~v p = 2 x2 + y 2 .

(e) First of all we must determine the integral curves of the fields, or “lines ~ u , for example, these curves are given by relaof force”. In the case of E tions such as y = f (x) or x = f (y). More generally, we can parametrize the curves through an arc length parameter along them, so that we have for their description the pair of functions [x(t), y(t)]. The tangent to the curve is given by a derivative with respect to t, which is therefore ~ u . In short, the integral curves of E ~ u are proportional to the field E determined by ~ u = −2(x, −y) E   dx dy , , = A dt dt where A is a real number, which can depend on the position along the curve, that is, on t. Equating the components of the vectors, we therefore have the pair of differential equations

83

BORDER EFFECTS IN CAPACITORS dx , dt dy 2y = A . dt

−2x = A

Dividing the second equation by the first we eliminate all references to t, thus obtaining the differential equation dy y = ⇒ x dx dx dy = − ⇒ y x ln(y) = B − ln(x) ⇒ C y = , x −

where we have the integration constant C = exp(B). Thus we see that ~ u are in fact associated with the equipotentials these integral curves of E of v(x, y), it being enough to identify C as v0 /2, in order to complete ~ v we have the association. Repeating the argument in the case E ~ v = −2(y, x) E   dx dy , . = A dt dt Equating the components of the vectors, we have therefore the pair of differential equations dx , dt dy −2x = A . dt −2y = A

Dividing the second equation by the first we obtain the differential equation x dy = ⇒ y dx x dx = y dy ⇒ x2 = B + y 2

2

x −y

2

= B,

⇒

84

SOLUTIONS 6 where we have the integration constant B. Thus we see that these ~ v are in fact associated with the equipotentials of integral curves of E u(x, y), it being enough to identify C as u0 , in order to complete the association.

Problem 4. Consider the complex function w(z) = and w = u + ıv.

√

z where z = x + ıy

(a) Write the real part u(x, y) using polar coordinates and show that it is the potential of the electrostatic problem of a semi-infinite grounded metal plate whose cross section takes up the negative real semi-axis. Note: the conducting plate being grounded means that it is at zero electrical potential. (b) For angles θ 6= ±π determine the behavior of the potential for ρ → ∞. ~ = −∇u. ~ (c) Calculate the electric-field vector E Write the result in terms of the variables ρ and θ. Answer: sin(θ/2) ~ = − cos(θ/2) x ˆ− yˆ. E √ √ 2 ρ 2 ρ ~ for ρ constant and θ → ±π. (d) Calculate the limits of E (e) After calculating these limits, determine the limits of the results obtained when we make ρ → ∞. (f) Calculate the surface charge density σ on the plate, using the fact that En =

σ , 2ε0

where En is the component of the electric field that is normal to the plate, pointing away from it. Determine at which point σ(ρ) has a singular behavior. (g) Since we are, in fact, looking at a two-dimensional section of a threedimensional problem, consider that the negative real semi-axis is a slice of unit width of the infinite half-plane. Calculate the total electric charge within this slice between the origin ρ = 0 and a particular value ρ0 of ρ. √ Answer: −2ε0 ρ0 .

85

BORDER EFFECTS IN CAPACITORS Complete Solution:

Using the polar representation z = ρ exp(ıθ), we have for the function w(z) = √ ρ exp(ıθ/2), where we will use for θ the interval [−π, π]. (a) We have for the real part of w = u + ıv, u(ρ, θ) =

√

ρ cos(θ/2),

which of course is a solution of the Laplace equation. The condition u(ρ, θ) = 0 is satisfied in two cases, ρ = 0, which is the origin of the complex plane, and cos(θ/2) = 0, that within the chosen interval of values of θ implies that θ = ±π. It follows that this function is zero on the negative real semi-axis, including the origin. It can therefore be interpreted as the potential of a grounded electrically conducting semi-infinite plate, whose section is placed on this semi-axis. (b) If we have θ 6= ±π, then the potential u(ρ, θ) is not zero, so that we have lim u(ρ, θ) =

ρ→∞

lim

ρ→∞

√

ρ cos(θ/2)

= ∞, that is, the potential goes to infinity as a square root of the distance from the origin. A potential that tends to infinity at infinity, instead of tending to zero, is characteristic of distributions of sources (electric charges) which extend to infinity, in this case along the half-plane where the plate is. ~ = −∇u ~ in two ways, using polar coordinates or (c) We can calculate E ~ operator. Let us start making the Cartesian coordinates for the ∇ calculation in polar coordinates, which is simpler but less familiar. In ~ this case we have for the operator ∇, ~ ∇

= ρˆ

∂ ∂ρ

+

1ˆ ∂ θ ρ ∂θ

,

where ρˆ and θˆ are the two versors of the coordinate system, so that we ~ with extreme simplicity, obtain for E,

86

SOLUTIONS 6 ~ = −∇u ~ E ∂u(ρ, θ) 1 ˆ ∂u(ρ, θ) − θ = −ˆ ρ ∂ρ ρ ∂θ cos(θ/2) sin(θ/2) ˆ = − ρˆ + θ. √ √ 2 ρ 2 ρ We can also do the calculation in the Cartesian system. For this, we must write u in terms of x and y, which may be done by means of the use of the formula of the cosine of the half-arc, √

ρ cos(θ/2) r √ 1 + cos(θ) = ρ 2 r r ρ ρ+x = 2 ρ 1 √ ρ + x, = √ 2

u(x, y) =

p x2 + y 2 . We can now easily calculate the two partial where ρ = derivatives with respect to x and to y, recalling that ∂ρ ∂x ∂ρ ∂y

= =

x , ρ y , ρ

so that we have ~ = −∇u ~ E ∂u(x, y) ∂u(x, y) = −ˆ x − yˆ ∂x ∂y   y x 1 1 1 1 1+ = −ˆ x √ √ − yˆ √ √ ρ + x ρ ρ + x ρ 2 2 2 2 1 1 1 1 [1 + cos(θ)] − yˆ √ √ sin(θ) = −ˆ x√ p 8ρ 1 + cos(θ) 8ρ 1 + cos θ 1 p 1 sin(θ) = −ˆ x√ . 1 + cos(θ) − yˆ √ √ 8ρ 8ρ 1 + cos θ

BORDER EFFECTS IN CAPACITORS

87

We can write this result once again in terms of θ/2, using the trigonometric formulas of the double arc, cos(θ) = cos2 (θ/2) − sin2 (θ/2), sin(θ) = 2 sin(θ/2) cos(θ/2),

thus obtaining sin(θ/2) ~ = − cos(θ/2) x ˆ− yˆ. E √ √ 2 ρ 2 ρ This is curiously similar to the previous result, written in the basis ˆ instead of (ˆ (ˆ ρ, θ) x, yˆ). In fact, it is so similar that, since these are two very different bases, one may have the impression that one of the two results must be wrong. However, it can be shown that these two bases, in fact, satisfy the curious identity x ˆ cos(θ/2) + yˆ sin(θ/2) = ρˆ cos(θ/2) − θˆ sin(θ/2), which is somewhat unexpected, but true. Check it out! (d) For θ → ±π, we have that θ/2 → ±π/2, and therefore that cos(θ/2) → 0 and sin(θ/2) → ±1. Using each of the two ways in which we wrote ~ we therefore have E, sin(θ/2) ˆ cos(θ/2) ρˆ + θ √ √ 2 ρ 2 ρ 1 ˆ → ± √ θ, 2 ρ sin(θ/2) ~ = − cos(θ/2) E x ˆ− yˆ √ √ 2 ρ 2 ρ 1 → ∓ √ yˆ. 2 ρ

~ E

=

−

The sign difference is due to the fact that, for θ = ±π, we have that ~ is normal to the plate, and that in θˆ = −ˆ y . Thus we see that the field E either one of the two sides it points towards the plate, which indicates the existence of negative electrical charges on the plate.

88

SOLUTIONS 6

√ (e) Since the absolute value of the field is proportional to 1/ ρ, it follows that in the limit ρ → ∞ along the plate the field goes to zero. Thus, it is zero if we distance ourselves infinitely from the edge, despite being singular on the edge itself, ρ = 0. (f) The field component normal to the plate is given by the component θˆ or yˆ on the plate, with the appropriate sign, which is determined by the external normal to the material of the plate, which in this case is negative, −1 En = √ . 2 ρ Since we have that σ = 2ε0 En , it follows that −ε0 σ(ρ) = √ . ρ Just like the field, this surface charge density is singular on the edge of the plate at ρ = 0, where it diverges to infinity, and goes to zero as ρ → ∞. (g) On a unit-width strip, the total charge is given by the integral of σ(ρ) on ρ, from the edge ρ = 0 to some value ρ0 of ρ, Z

ρ0

dρ σ(ρ) Z ρ0 1 dρ √ = −ε0 ρ 0  ρ0 √ = −ε0 2 ρ √ 0 = −2ε0 ρ0 .

Q =

0

As we already expected, this total charge is negative. Thus we see that, although the charge density is singular on the edge, the total charge on a finite area that includes the edge is finite.

BORDER EFFECTS IN CAPACITORS

89

Problem 5. (Challenge Problem) Consider a capacitor formed by two semi-infinite plates. Each plate has an edge with the form of an infinite straight line and the two plates are placed forming a wedge with an angle θ0 ≪ π/4, with the two edges placed together but electrically isolated from one another. Determine the capacitance per unit area for a unit-width slice of the plates located at distances between r1 > 0 and r2 > r1 from the edges. Do this by means of a conformal transformation in the complex plane, that maps an infinite capacitor on this wedge capacitor, except for the point corresponding to the edge of the plates. Answer: ε0 ln(r2 ) − ln(r1 ) . θ0 r2 − r1 The solution to this problem was found by Prof. Henrique Fleming. It will be addressed again later in this series of books, through the use of other techniques. Complete Solution: First, let us describe the conformal map that solves the problem. Let us consider the complex planes z = x + ıy and w = u + ıv, and the mapping w(z) = exp(αz/2), where α is a strictly positive real constant. It follows that we have for u(x, y) and v(x, y), u(x, y) =

eαx/2 cos(αy/2),

v(x, y) =

eαx/2 sin(αy/2).

We will determine the images in the collection of straight lines x ∈ (−∞, ∞), y = y0 of the (x, y) plane. If we describe the (u, v) plane in terms of polar variables, p ρ = u2 + v 2 =

θ =

eαx/2 , α y, 2

we verify that these lines have as images the semi-axes that form angles θ = αy0 /2 with the axis x, on which x → −∞ corresponds for ρ → 0 and x → ∞ corresponds to ρ → ∞, that is, ρ ∈ (0, ∞), and the semi-axes extend from the origin to infinity, but do not include the origin. Thus we

90

SOLUTIONS 6

see that this transformation takes a capacitor of horizontal parallel plates in the (x, y) plane onto a capacitor with angled plates in the (u, v) plane, which do not touch at the origin due to the fact that it is outside the domain of the mapping. Of course we will have a singular behavior at the origin of the (u, v) plane. We can therefore start from a pair of plates located at y = ±1 with opposite electrical potentials, x ∈ (−∞, ∞),

y = ±1, V0 , φ(x, 1) = 2 V0 φ(x, −1) = − , 2

which represents an infinite parallel-plate capacitor, in which there is a potential difference V0 . In terms of dimensional variables X = xd/2 and Y = yd/2, such that Y = d/2 corresponds to y = 1, and where d is the distance between the capacitor plates, we can write a complex potential Φ(X, Y ) on the (x, y) plane, Φ(X, Y ) = = = =

V0 z 2 V0 (x + ıy) 2   2Y V0 2X +ı 2 d d V0 (X + ıY ). d

Since the function w(z) = z is analytic, both the real part and the imaginary part of this function are solutions of the Laplace equation in the (x, y) plane. The imaginary part of Φ(X, Y ) is the electrical potential inside the capacitor in the (x, y) plane, φ(X, Y ) =

V0 Y , d

which also satisfies the boundary conditions, because Φ(X, d/2) = V0 /2 and Φ(X, −d/2) = −V0 /2. It follows that the corresponding complex potential in the (u, v) plane is given by Φ′ (w) = Φ[z(w)], where z(w) is the inverse mapping of the mapping w(z), that is, z(w) = 2 ln(w)/α, so that we have

91

BORDER EFFECTS IN CAPACITORS V0 2 ln(w) 2 α ln(ρ) + ıθ . = V0 α

Φ′ (w) =

This is an analytic function in the (u, v) plane, so that its real and imaginary parts are solutions of the Laplace equation in that plane. We can now see that the imaginary part of this expression represents the potential within the capacitor in the (u, v) plane, because we have that θ , α

φ′ (ρ, θ) = V0

so that for θ = α/2 we have the constant potential V0 /2, and for θ = −α/2 we have the constant potential −V0 /2. Thus, the potential difference in the image capacitor is also V0 , and the angle between the plates is α, so that we have α = θ0 , and therefore we have φ′ (ρ, θ) = V0

θ . θ0

Note that the potential does not depend on ρ at all. The next step is to calculate the electric field inside the capacitor. For this, it is easier to use the expression of the gradient in polar coordinates. In this case we have for ~ the operator ∇, ~ = ρˆ ∂ + 1 θˆ ∂ , ∇ ∂ρ ρ ∂θ ~ = −∇φ ~ ′ , very simply, so that we to obtain E ~ = ρˆ 0 − 1 θˆ V0 1 . E ρ θ0 V0 ˆ = − θ. θ0 ρ Thus we see that the electric field has no radial component, but only the ˆ Furthermore, this component angular component, in the direction of −θ. does not depend on θ, but only on ρ. It follows that it is very simple to see that the component of the electric field in the direction of the external normal at the top plate is given by En =

V0 , θ0 ρ

92

SOLUTIONS 6

so that the surface charge density is given by σ(ρ) = ε0 En ε0 V0 = . θ0 ρ We must now calculate the total amount of charge contained in the top plate for values of ρ between r1 and r2 . Note that the hypothesis that θ0 be small is only necessary for us to actually cut out this part from the full capacitor, and still be able to neglect the edge effects. We therefore have for the charge, considering a unit-width slice of the capacitor in the direction perpendicular to the complex plane, Q = = = = =

Z

r2

dρ σ(ρ)

Zr1r2

ε0 V0 dρ θ0 ρ r1 Z r2 1 ε0 V0 dρ θ 0 r1 ρ  r2 ε0 V0 ln(ρ) θ0 r  1 ε0 V0 r2 ln . θ0 r1

Since the capacitance is given by C = Q/V0 and the capacitor area is proportional to r2 − r1 , we have for capacitance C per unit area C = =

ε0 ln(r2 /r1 ) θ0 r2 − r1 ε0 ln(r2 ) − ln(r1 ) . θ0 r2 − r1

Note that the result diverges if we make r1 → 0, because in this case we approach the singularity at the common edge of the two plates. We can recover the result for a parallel plate capacitor, in the approximation in which the edge effects are neglected, if we make both r1 and r2 very large, while we make θ0 ever smaller, in such a way that the distance between the plates, which is given approximately by the arc d = r1 θ0 , remains constant. Making r2 = r1 + ∆r, for a constant ∆r, we have in the limit in which r1 → ∞

BORDER EFFECTS IN CAPACITORS C = ≈ =

93

ε0 ln(1 + ∆r/r1 ) θ0 ∆r ε0 ∆r/r1 θ0 ∆r ε0 , r1 θ0

where we see appearing in the denominator the distance d between the plates. Since this is the capacitance per unit area, for a capacitor with total area A we therefore have C=

ε0 A , d

which is the usual result for this type of approximation.

94

SOLUTIONS 6

Solutions 7

Complex Calculus I: Differentiation We present here complete and commented solutions to all problems proposed in Chapter 7 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. Prove that the Leibniz formula for the derivative of a product applies to the product of two analytic functions. In other words, given two analytic functions w1 (z) and w2 (z), each one of which has, therefore, a welldefined complex derivative, show that the derivative of the product-function w(z) = w1 (z)w2 (z) is given by dw1 dw2 dw (z) = (z) w2 (z) + w1 (z) (z). dz dz dz Complete Solution: The simplest way to prove the validity of the Leibniz rule for complex derivatives is to start from the definition of the derivative of the product-function w(z) = w1 (z)w2 (z), dw(z) dz

w(z + δz) − w(z) δz w1 (z + δz)w2 (z + δz) − w1 (z)w2 (z) = lim δz→0 δz  w1 (z + δz)w2 (z + δz) − w1 (z + δz)w2 (z) + = lim δz→0 δz =

lim

δz→0

95

96

SOLUTIONS 7  w1 (z + δz)w2 (z) − w1 (z)w2 (z) + , δz

where we added and subtracted the term w1 (z + δz)w2 (z) in the numerator, in order to isolate the variations of w1 (z) and w2 (z). We therefore have, making this separation, dw(z) dz

w1 (z + δz)w2 (z + δz) − w1 (z + δz)w2 (z) + δz→0 δz w1 (z + δz)w2 (z) − w1 (z)w2 (z) + lim δz→0 δz   w2 (z + δz) − w2 (z) = lim w1 (z + δz) + δz→0 δz   w1 (z + δz) − w1 (z) + lim w2 (z) δz→0 δz   w2 (z + δz) − w2 (z) + = w1 (z) lim δz→0 δz   w1 (z + δz) − w1 (z) + lim w2 (z) δz→0 δz dw2 (z) dw1 (z) + w2 (z). = w1 (z) dz dz =

lim

It is also possible to write the complex derivative of w = u + ıv in terms of the partial derivatives of u and v, for example with respect to x, to do the same with w1 and w2 , and to manipulate these expressions in order to reduce the problem to the use of the Leibniz rule for these real functions, for which we already know that it is valid. Problem 2.

Starting from the complex derivative d −1 1 z = − 2, dz z

prove by finite induction the more general formula, for a positive integer n, n d −n z = − n+1 . dz z Complete Solution: Let us use the fact, assumed known, that

COMPLEX CALCULUS I: DIFFERENTIATION

97

dz −1 = −z −2 . dz In order to show the general result by induction, we assume its validity in the case n − 1, that is, we assume that dz −(n−1) = −(n − 1)z −n . dz We now consider the derivative of z −n , write this function as the product of z −1 with z −(n−1) , and use the Leibniz rule for this product,   d z −1 z −(n−1) dz −n = dz dz −1 dz dz −(n−1) = z −(n−1) + z −1 . dz dz Using the results that have been assumed, we have dz −n dz

= −z −2 z −(n−1) − z −1 (n − 1)z −n = −z −(n+1) − (n − 1)z −(n+1)

= −nz −(n+1) .

which proves the case n, and therefore the general case, that is, we do have, in fact, that dz −n = −nz −(n+1) , dz for all n > 0. Problem 3. Starting from the known real derivatives, derive the formulas for the derivatives of the analytic functions that follow. (a) w(z) = cosh(z). (b) w(z) = sinh(z). (c) w(z) = cos(z). (d) w(z) = sin(z).

98

SOLUTIONS 7

Complete Solution: In each case, we write the function in terms of known real functions, and choose an appropriate direction for the variation dz in the complex plane, in order to simplify the calculation, since the derivative is independent of the direction of variation. (a) w(z) = cosh(z): we make dz = dx, and get cosh(z) = = d cosh(z) dz

= = =

ez + e−z 2 ex eıy + e−x e−ıy 2 ex eıy − e−x e−ıy 2 ez − e−z 2 sinh(z).

⇒

(b) w(z) = sinh(z): we make dz = dx, and get sinh(z) = = d sinh(z) dz

= = =

ez − e−z 2 ex eıy − e−x e−ıy 2 ex eıy + e−x e−ıy 2 z −z e +e 2 cosh(z).

⇒

(c) w(z) = cos(z): we make dz = ıdy, and get cos(z) = = d cos(z) dz

= = =

eız + e−ız 2 ıx −y e e + e−ıx ey ⇒ 2 − eıx e−y + e−ıx ey ı−1 2 eız − e−ız − 2ı − sin(z).

99

COMPLEX CALCULUS I: DIFFERENTIATION (d) w(z) = sin(z): we make dz = ıdy, and get sin(z) = = d sin(z) dz

= = =

eız − e−ız 2ı eıx e−y − e−ıx ey ⇒ 2ı − eıx e−y − e−ıx ey ı−1 2ı eız + e−ız 2 cos(z).

Problem 4. Derive the second-order complex differential equations and the auxiliary conditions that are satisfied by the following analytic functions. (a) w(z) = cosh(z). (b) w(z) = sinh(z). (c) w(z) = cos(z). (d) w(z) = sin(z). Complete Solution: In each case, we start from the known derivatives of each function, and build a second-order differential equation satisfied by it. Since each complex function is reduced to the corresponding real function on the real axis, it suffices to look for the auxiliary conditions on the real axis. (a) w(z) = cosh(z): the first two derivatives are given by d cosh(z) dz d sinh(z) dz d2 cosh(z) dz 2

= sinh(z), = cosh(z)

⇒

= cosh(z).

We have therefore the differential equation and the auxiliary conditions

100

SOLUTIONS 7 d2 f (z) − f (z) = 0, dz 2 f (0) = 1, df (z) (0) = 0. dz

(b) w(z) = sinh(z): the first two derivatives are given by d sinh(z) dz d cosh(z) dz d2 sinh(z) dz 2

= cosh(z), = sinh(z)

⇒

= sinh(z).

We have therefore the differential equation and the auxiliary conditions d2 f (z) − f (z) = 0, dz 2 f (0) = 0, df (z) (0) = 1. dz (c) w(z) = cos(z): the first two derivatives are given by d cos(z) dz d sin(z) dz d2 cos(z) dz 2

= − sin(z), = cos(z)

⇒

= − cos(z).

We have therefore the differential equation and the auxiliary conditions d2 f (z) + f (z) = 0, dz 2 f (0) = 1, df (z) (0) = 0. dz

101

COMPLEX CALCULUS I: DIFFERENTIATION (d) w(z) = sin(z): the first two derivatives are given by d sin(z) dz d cos(z) dz 2 d sin(z) dz 2

= cos(z), = − sin(z)

⇒

= − sin(z).

We have therefore the differential equation and the auxiliary conditions d2 f (z) + f (z) = 0, dz 2 f (0) = 0, df (z) (0) = 1. dz Problem 5. Show that the chain rule applies to the composition of analytic functions. In other words, show that, if f (z) and g(z) are two analytic functions, and w(z) = f (g(z)) is the composite function, then   df (z) dg(z) dw(z) = . (g) dz dz dz Complete Solution: In this case the simplest thing to do is to work out the direct proof, similar to what is done in the real case. We have that w(z) = f [g(z)], so that by definition the derivative is written as dw(z) dz

w(z + δz) − w(z) δz→0 δz f [g(z + δz)] − f [g(z)] . = lim δz→0 δz =

lim

While g(z + δz) is not equal to g(z), that is, so long as g(z) is not constant in a neighborhood of the point z, we can multiply and divide by the difference of these two values, thereby obtaining   f [g(z + δz)] − f [g(z)] g(z + δz) − g(z) dw(z) = lim . δz→0 dz g(z + δz) − g(z) δz

102

SOLUTIONS 7

Note that the case in which g(z) is constant within the neighborhood must be examined separately, but it is trivial, because in this case g(z) = g0 and thus f [g(z)] = f (g0 ), which is also constant. It follows that the derivative of w(z) with respect to z is zero at this point, as is the derivative of g(z) with respect to z, so that the chain rule is satisfied, whatever the value of the derivative of f (g) with respect to g, as long as all involved functions are differentiable, which they always are, since they are analytic at the point z. Since all the derivatives exist, we can separate the two ratios in the above formula, into two different limits,    f [g(z + δz)] − f [g(z)] g(z + δz) − g(z) dw(z) = lim lim δz→0 δz→0 dz g(z + δz) − g(z) δz df (z) dg(z) = (g) , dz dz so that the chain rule is proved. Problem 6. Consider the following analytic function as a conformal transformation, that is, a transformation that preserves the angles between oriented curves that intersect each other, mapping the (x, y) plane onto the (u, v) plane, w(z) = u(x, y) + ıv(x, y) 1 = z+ . z ~ u = −∇u ~ (a) Calculate the electric field vectors associated with u and v, E ~ ~ and Ev = −∇v. ~ u = 0 at the points (−1, 0) and (1, 0) of the complex (b) Show that E ~ v = 0 at these points. plane (x, y), and that therefore we also have E (c) Show that a radius inside the semicircle of the (x, y) plane with θ very small and ρ ∈ [ρ0 , 1], where ρ0 > 0 is a small number, is mapped on a curve which is very close to the positive real semi-axis of the (u, v) plane, a curve which is, however, still perpendicular to the real segment (−2, 2) of the (u, v) plane, crossing this segment at a point near (2, 0).

(d) Show that in the θ → 0 limit the curve tends to the positive real semiaxis [2, ∞), which is obviously not perpendicular to the segment (−2, 2) of the (u, v) plane.

103

COMPLEX CALCULUS I: DIFFERENTIATION

(e) Show that at the points (−1, 0) and (1, 0), the complex derivative of the function w(z) vanishes, that is, that dw (−1, 0) = 0, dz dw (1, 0) = 0. dz Just as an informative note, let us recall that at the points where we have E ~ u = 0 the proof that the conformal transformation preserves angles does not apply, so that the preservation of angles does not hold for curves that cross at these points. Since the derivative of the function is zero at these points, it is not analytically invertible at them, which means that the inverse function of w(z) has singularities at these points. It can be said that these points are singular points of the conformal transformation, in a sense that is not the same as the notion of singularity of the complex function, which is analytic at these points. It is the inverse of this analytic function that has singularities at these points. Complete Solution: Writing explicitly u(x, y) and v(x, y) we have w(z) = z +

1 z

x − ıy = x + ıy + 2 x + y2   1 u(x, y) = x 1 + 2 , ρ   1 v(x, y) = y 1 − 2 , ρ

⇒

where, in order to simplify the differentiations, we wrote the expression in terms of ρ2 = x2 + y 2 . (a) In order to obtain the fields, in Cartesian coordinates, we calculate the relevant partial derivatives, of which two are naturally dependent on the other two, due to the Cauchy-Riemann conditions, and we thus obtain   1 2x2 ∂u = 1+ 2 − 4 ∂x ρ ρ

104

SOLUTIONS 7 = = ∂u ∂y ∂v ∂x ∂v ∂y

= = = = =

~u = E ~v = E

ρ4 + ρ2 − 2x2 ρ4 ρ4 − x 2 + y 2 , ρ4 2xy − 4 , ρ 2yx , ρ4   1 2y 2 1− 2 + 4 ρ ρ 4 2 2 ρ − ρ + 2y ρ4 ρ4 − x 2 + y 2 ⇒ ρ4 2xy ρ4 − (x2 − y 2 ) x ˆ − 4 yˆ, 4 ρ ρ 4 2 2xy ρ − (x − y 2 ) x ˆ + yˆ. ρ4 ρ4

(b) For the field vectors to be zero, all their components must be zero. From two of the components it follows that we must have xy = 0, which means that x = 0, y = 0 or both. From the other two components we have that ρ4 = x2 − y 2 . Thus we see that we cannot have x = 0, because this would reduce this equation to y 4 = −y 2 , which is impossible because y is real. Thus, it is necessary that y = 0, and it follows therefore that x4 = x2 , from which it follows that x2 = 1 and therefore that x = ±1. We have therefore that the two gradient vectors are zero at only two points, (1, 0) and (−1, 0). (c) Writing u(x, y) and v(x, y) completely in terms of the polar coordinates ρ and θ, we have  1 cos(θ), u(ρ, θ) = ρ+ ρ   1 sin(θ). v(ρ, θ) = ρ− ρ 

For θ ≪ 1 and ρ ∈ [ρ0 , 1] with ρ small and positive, we have that

COMPLEX CALCULUS I: DIFFERENTIATION

105



 1 u(ρ, θ) = ρ+ (1 − θ 2 /2), where ρ (ρ + 1/ρ) ∈ [2, U1 ],   1 θ, where v(ρ, θ) = ρ− ρ (ρ − 1/ρ) ∈ [−U2 , 0], where for ρ0 near zero the values of U1 and U2 are positive and large, but finite. Thus we see that in the θ → 0 limit the coordinate u of the image in the (u, v) plane tends to (ρ + 1/ρ), which is within the interval [2, U1 ], while the corresponding coordinate v tends to zero. Therefore, the curve on the image tends to approach the interval [2, U1 ] on the positive real semi-axis of the (u, v) plane. On the other hand, we can determine the slope of the tangent to the curve on the image, calculating the variations of u and v, when we vary ρ with a fixed value of θ,  1 du = 1 − 2 cos(θ)dρ, ρ   1 dv = 1 + 2 sin(θ)dρ ⇒ ρ 2 du ρ − 1 cos(θ) = . dv ρ2 + 1 sin(θ) 

For θ small but non-zero, the second ratio is a large but finite number, so that in the ρ → 1 limit, in which we approach the point defined by u = 2 cos(θ) and v = 0, we have for this ratio of variations du = 0, dv which means that we approach a point of the segment [−2, 2] such that u tends to remain constant along the curve while v varies, that is, the curve is orthogonal to the segment. For θ ≪ 1 the contact point is given by (2 − θ 2 , 0), that is, it is a point within the segment and very close to the point (2, 0). (d) As shown in the previous item, in the θ → 0 limit the coordinate u of the image in the (u, v) plane tends to (ρ + 1/ρ), which is in the interval

106

SOLUTIONS 7 [2, U1 ], while the corresponding coordinate v tends to zero. If we make ρ0 → 0 after that, we have that U1

=

ρ0 + 1/ρ0

→ ∞, so that in the θ → 0 limit the image is the real semi-axis [2, ∞). (e) We can simply calculate the derivative in the usual way, 1 dw(z) = 1 − 2, dz z so that the derivative is zero only for z 2 = 1, that is, for z = 1 and z = −1, which correspond to the points (1, 0) and (−1, 0) on the complex (u, v) plane. Problem 7. Starting from the known real derivatives, derive the formula for the complex derivative of the analytic function w(z) = ln(z). Show explicitly that this derivative gives the same result on all the infinitely many leaves of the Riemann surface of the logarithm function, thus mapping them all onto a function that has a Riemann surface with a single leaf, which is simply the complex plane except for one point. Complete Solution: The definition of the logarithmic function in terms of known real functions, including the discrimination of the leaves of the Riemann surface, by means of an integer n, in terms of the polar representation z = ρ exp(ıθ), is given by w(z) = ln(z) = ln(ρ) + ıθ + n(ı2π). We can choose as the direction for the variation of z either the radial direction, using dz = (dρ) exp(ıθ), or the angular direction, using dz = ıρ exp(ıθ)(dθ). In the first case we have for the derivative dw(z) d ln(ρ) = e−ıθ dz dρ −ıθ e = ρ 1 = . z

COMPLEX CALCULUS I: DIFFERENTIATION

107

Since this does not depend on n, it follows that the result is the same for all the leaves of the Riemann surface. Taking the derivative in the other way we have dw(z) dz

= = =

e−ıθ d(ıθ) ıρ dθ e−ıθ ρ 1 . z

As expected, the result is the same, and again the dependence on n vanishes.

108

SOLUTIONS 7

Solutions 8

Complex Calculus II: Integration We present here complete and commented solutions to all problems proposed in Chapter 8 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. For each function w(z) below, make an analysis of the possible singularities in order to determine whether or not one can use the CauchyGoursat theorem in order to determine that, on the unit circle C, I (a) w(z) =

w(z) dz = 0. C

z2 . z−3

(b) w(z) = z ez . (c) w(z) =

1 . z 2 + 2z + 2

(d) w(z) = sech(z). (e) w(z) = tan(z). (f) w(z) = ln(z + 2). 109

110

SOLUTIONS 8

Complete Solution: In each case, we list the singularities and determine the applicability or nor of the Cauchy-Goursat theorem for the integral over the unit circle. (a) Function: z2 . z−3 The function has only one singularity, a simple pole at z = 3. This point is outside the unit circle, therefore the theorem applies and the integral is zero. (b) Function: z ez . The function has no singularities, therefore the theorem applies and the integral is zero. (c) Function: 1 . z 2 + 2z + 2 We have for the roots of the polynomial in the denominator, by the Baskara formula, z±

√ 4−8 = 2 −2 ± 2ı = 2 = −1 ± ı. −2 ±

√ The numbers −1 ± ı have absolute value ρ = 2, and hence are outside the unit circle, so that the theorem applies and the integral is zero. (d) Function: sech(z).

COMPLEX CALCULUS II: INTEGRATION

111

In this case we need to recall that 1 , cosh(z) cosh(z) = cosh(x) cos(y) + ı sinh(x) sin(y). sech(z) =

Now we must determine where cosh(z) can be zero. For this, both the real part and the imaginary part should vanish. Since cosh(x) is never zero, from the real part we conclude that we must have cos(y) = 0. Due to this, we know that sin(y) 6= 0, so that, from the imaginary part, we should have sinh(x) = 0, which implies that x = 0. Thus, all zeros are on the imaginary axis. Their positions along the axis are given by cos(y) = 0, which implies that the zeros nearest to the origin are y = π/2 and y = −π/2. Since π/2 > 1, these zeros are outside the unit circle, so that the theorem applies and the integral is zero. (e) Function: tan(z). In this case we need to recall that sin(z) , cos(z) cos(z) = cos(x) cosh(y) − ı sin(x) sinh(y).

tan(z) =

By means of an analysis similar to that used in the previous item, we conclude that the zeros are only those of the function cos(x), with y = 0. The two zeros closest to the origin are therefore (π/2, 0) and (−π/2, 0). Again, since π/2 > 1, these zeros are outside the unit circle, so that the theorem applies and the integral is zero. (f) Function: ln(z + 2). The function has only one singularity, a branch point at z = −2, which is outside the unit circle. It follows that the theorem applies and the integral is zero. Note that the integral can be on any one of the infinite Riemann leaves of the logarithm, so long as it is entirely on a single leaf.

112

SOLUTIONS 8

Problem 2. Consider an arbitrary closed simple curve C on the (x, y) plane, with the condition that it do not pass through the point z = 0. Show that I dz = 0, 2 C z in the following cases.

(a) If the interior of the curve does not contain the origin z = 0. (b) If the interior of the curve contains the origin z = 0. Note: a closed simple curve is a closed curve that does not intersect itself. Complete Solution: We must calculate on simple closed curves the integral I dz , 2 C z

whose integrand has only one singularity at z = 0. Of course, the curve cannot pass exactly over the point z = 0, because in this case the integral would not be well defined. (a) If the curve does not contain the point z = 0, then the integrand is analytic on the curve and within it, so that the Cauchy-Goursat theorem applies and the integral is zero. (b) If the curve contains the point z = 0, then we can use the CauchyGoursat theorem in order to deform the curve until it coincides with the unit circle, without this changing the value of the integral. Then it suffices to calculate the value on the unit circle, using the polar representation z = ρ exp(ıθ), Z 2π I dz ρı eıθ dθ = 2 ρ2 e2ıθ 0 C z Z 2π ı = dθ e−ıθ ρ 0  ı −ıθ 2π ıe = ρ 0
−1 −ı2π = e − e0 ρ = 0,

113

COMPLEX CALCULUS II: INTEGRATION

due to the periodicity of the exponential with imaginary argument. It follows therefore that the integral is zero in this case as well. Problem 3. Consider an arbitrary closed simple curve C on the (x, y) plane, with the condition that it do not pass through the point z = 0. Show that I dz = 0, 3 C z in the following cases. (a) If the interior of the curve does not contain the origin z = 0. (b) If the interior of the curve contains the origin z = 0. Note: a closed simple curve is a closed curve that does not intersect itself. Complete Solution: We must calculate on simple closed curves the integral I dz , 3 C z whose integrand has only one singularity at z = 0. Of course, the curve cannot pass exactly over the point z = 0, because in this case the integral would not be well defined. (a) If the curve does not contain the point z = 0, then the integrand is analytic on the curve and within it, so that the Cauchy-Goursat theorem applies and the integral is zero. (b) If the curve contains the point z = 0, then we can use the CauchyGoursat theorem in order to deform the curve until it coincides with the unit circle, without this changing the value of the integral. Then it suffices to calculate the value on the unit circle, using the polar representation z = ρ exp(ıθ), I

C

dz z3

= =

Z

2π

dθ 0

ı ρ2

Z

0

2π

ρı eıθ ρ3 e3ıθ dθ e−2ıθ

114

SOLUTIONS 8  ı ı −2ıθ 2π e = ρ2 2 0
−1 −ı4π e − e0 = 2 2ρ = 0, due to the periodicity of the exponential with imaginary argument. It follows therefore that the integral is zero in this case as well.

Problem 4. Consider an arbitrary closed simple curve C on the (x, y) plane, with the condition that it do not pass through the point z = 1. Show that I dz = 0, 2 C (z − 1) in the following cases. (a) If the interior of the curve does not contain the point z = 1. (b) If the interior of the curve contains the point z = 1. Note: a closed simple curve is a closed curve that does not intersect itself. Complete Solution: We must calculate on simple closed curves the integral I dz . (z − 1)2 C If we make the transformation of variables z ′ = z − 1, then the condition on the curve, which is whether or not it contains the point z = 1, becomes the corresponding condition of whether or not it contains the point z ′ = 0, and the integral is simplified to I dz ′ , ′2 C z whose integrand has only one singularity at z ′ = 0. Of course, the curve cannot pass exactly over the point z ′ = 0, because in this case the integral would not be well defined.

115

COMPLEX CALCULUS II: INTEGRATION

(a) If the curve does not contain the point z ′ = 0, then the integrand is analytic on the curve and within it, so that the Cauchy-Goursat theorem applies and the integral is zero. (b) If the curve contains the point z ′ = 0, then we can use the CauchyGoursat theorem in order to deform the curve until it coincides with the unit circle, without this changing the value of the integral. Then it suffices to calculate the value on the unit circle, using the polar representation z ′ = ρ exp(ıθ), I

C

dz ′ z ′2

=

Z

=

ı ρ

2π

dθ

0

Z

0

2π

ρı eıθ ρ2 e2ıθ

dθ e−ıθ  2π

ı −ıθ ıe = ρ 0
−1 −ı2π e − e0 = ρ = 0, due to the periodicity of the exponential with imaginary argument. It follows therefore that the integral is zero in this case as well. Problem 5.

Consider the analytic function w(z) =

√

z.

(a) Calculate the integral I

w(z) dz, C

on the unit circle C. (b) Calculate the integral of this function over a curve that goes twice around the unit circle. (c) Examine the situation for the integrals over other circular contours centered at the origin, in order to determine whether it is possible to say something about the values of these integrals based on the results obtained here.

116

SOLUTIONS 8

Hint: use polar coordinates. Complete Solution: Using the polar representation z = ρ exp(ıθ), we have for the function √ z w(z) = √ ıθ/2 = ρe . (a) Let us calculate on the unit circle ρ = 1, the integral, I

w(z) dz = C

Z

2π

√ dθ ıρ eıθ ρ eıθ/2

0

Z

2π

dθ e3ıθ/2 0  −2ı 3ıθ/2 2π e = ı 3 0
2 3ıπ 0 e −e = 3 4 = − . 3 = ı

As one can see, the result is not zero. Note that when we go around the circle we change from one leaf of the Riemann surface of the function √ z to the other, so that this integral is not actually on a closed contour within the domain of the function. (b) Going around the unit circle twice, we have I

w(z) dz = C

Z

4π

√ dθ ıρ eıθ ρ eıθ/2

0

Z

4π

dθ e3ıθ/2 0  −2ı 3ıθ/2 4π e = ı 3 0
2 6ıπ 0 e −e = 3 = 0. = ı

We see that this time the result is zero. Note that when we go twice around the circle we change twice the leaf of the Riemann surface of the

COMPLEX CALCULUS II: INTEGRATION

117

√ function z, thus returning to the original leaf, so that this time the integral actually is over a closed contour, with regard to the domain of the function. √ (c) The change to circles with other radii simply includes a factor of ρ3 multiplying the integral, which does not introduce any divergence, and in fact makes the integral vanish for ρ → 0. On the other hand, if we make n turns on the unit circle, we obtain I

Z

w(z) dz =

n2π

√ dθ ıρ eıθ ρ eıθ/2

0

C

Z

n2π

dθ e3ıθ/2 0  −2ı 3ıθ/2 n2π e = ı 3 0
2 3nıπ 0 e −e = 3 2 = [(−1)n − 1] . 3 = ı

If n is even the result is zero, and if n is odd the result is −4/3. Therefore, the result is zero every time we return to the same point on the Riemann surface, exchanging leaves an even number of times, and thus closing the contour. Problem 6.

√ Consider the analytic function w(z) = 1/ z.

(a) Calculate the integral I

w(z) dz, C

on the unit circle C. (b) Calculate the integral of this function over a curve that goes twice around the unit circle. (c) Examine the situation for the integrals over other circular contours centered at the origin, in order to determine whether it is possible to say something about the values of these integrals based on the results obtained here.

118

SOLUTIONS 8

Hint: use polar coordinates. Complete Solution: Using the polar representation z = ρ exp(ıθ), we have for the function w(z) = =

1 √ z 1 −ıθ/2 . √ e ρ

(a) Let us calculate on the unit circle ρ = 1, the integral, I

w(z) dz =

Z

2π

0

C

Z

1 dθ ıρ eıθ √ e−ıθ/2 ρ

2π

dθ eıθ/2 0  2π ıθ/2 = ı(−2ı) e
0 = 2 eıπ − e0

= ı

= −4.

As one can see, the result is not zero. Note that when we go around the circle we change from one leaf of the Riemann surface of the function √ 1/ z to the other, so that this integral is not actually on a closed contour with regard to the domain of the function. (b) Going twice around the unit circle, we have I

C

w(z) dz =

Z

4π

0

Z

1 dθ ıρ eıθ √ e−ıθ/2 ρ

4π

dθ eıθ/2  4π ıθ/2 = ı(−2ı) e
0 = 2 e2ıπ − e0

= ı

0

= 0.

We see that this time the result is zero. Note that when we go twice around the circle we change twice the leaf of the Riemann surface of the

COMPLEX CALCULUS II: INTEGRATION

119

√ function 1/ z, thus returning to the original leaf, so that this time the integral actually is over a closed contour, with regard to the domain of the function. √ (c) The change to circles with other radii simply includes a factor of ρ multiplying the integral, which does not introduce any divergence, and in fact the integral vanishes for ρ → 0. On the other hand, if we make n turns on the unit circle, we obtain I

w(z) dz =

C

Z

n2π 0

Z

1 dθ ıρ eıθ √ e−ıθ/2 ρ

n2π

dθ eıθ/2 0  n2π = ı(−2ı) eıθ/2
0 nıπ 0 = 2 e −e = ı

= 2 [(−1)n − 1] .

If n is even the result is zero, and if n is odd the result is −4. So the result is zero every time we return to the same point on the Riemann surface, exchanging leaves an even number of times, and thus closing the contour. Problem 7.

Consider the analytic function w(z) = ln(z).

(a) Calculate the integrals I

w(z) dz, C

on the unit circle C. (b) Calculate the integral of this function over a curve that goes twice around the unit circle. (c) Examine the situation for the integrals over other circular contours centered at the origin, in order to determine whether it is possible to say something about the values of these integrals based on the results obtained here.

120

SOLUTIONS 8

Hint: use polar coordinates. Complete Solution: Using the polar representation z = ρ exp(ıθ), we have for the function w(z) = ln(z) = ln(ρ) + ıθ, where we indicate that we are initiating the integral on the central leaf of the Riemann surface, with n = 0. (a) Let us calculate on the unit circle ρ = 1, the integral, I

w(z) dz = C

Z

= ı

2π

dθ ıρ eıθ [ln(ρ) + ıθ] 0

2

2π

Z

dθ θ eıθ .

0

The integral can be made by parts, integrating the exponential and differentiating the factor θ. Doing this we obtain I

w(z) dz = −(−ı)θ e

ıθ

C

= ıθ eıθ

 2π 0

= ı2π e

ı2π

 2π

Z

+

0

−ı

Z

2π

dθ (−ı) eıθ 0

2π

dθ eıθ 0

− ı(−ı) e

= ı2π − eı2π + e0

ıθ

 2π 0

= ı2π.

As one can see, the result is not zero. Note that when we go around the circle we changed from a leaf of the Riemann surface of the function ln(z) to the next, so that this integral is not actually on a closed contour with regard to the domain of the function. (b) Going twice around the unit circle, we have

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COMPLEX CALCULUS II: INTEGRATION I

w(z) dz = C

Z

= ı

4π

dθ ıρ eıθ [ln(ρ) + ıθ] 0

2

4π

Z

dθ θ eıθ .

0

Integrating by parts we have I

w(z) dz = −(−ı)θ e

ıθ

C

= ıθ eıθ

 4π 0

= ı4π e

ı4π

 4π

Z

+

0

−ı

Z

4π

dθ (−ı) eıθ 0

4π

dθ eıθ 0

− ı(−ı) e

ıθ

 4π 0

= ı4π − eı4π + e0

= ı4π.

Again the result is not zero. Note that when we go twice around the circle we change twice from one leaf of the Riemann surface of the function ln(z) to the next, and therefore the contour again is not closed with regard to the domain of the function. (c) The modification to circles with other radii adds to the previous integral, for the case of the integration on the circle with a single turn, the integral of the real part of the function, Z

2π

dθ ıρ e ln(ρ) = ıρ ln(ρ) ıθ

0

Z

2π

dθ eıθ  2π ıθ = ıρ ln(ρ)(−ı) e 0
ı2π 0 = ρ ln(ρ) e − e 0

= 0.

Thus we see that the real part of the original integral never contributes to the final result. Clearly, this also applies to integrals that make any number of turns around the origin. Note that the factor in front of the integral is not divergent, since the limit of ρ ln(ρ) when ρ → 0 is in fact zero. On the other hand, if we make n turns on the unit circle, we obtain

122

SOLUTIONS 8 I

w(z) dz =

n2π

Z

dθ ıρ eıθ [ln(ρ) + ıθ]

0

C

= ı

2

Z

n2π

dθ θ eıθ . 0

Integrating by parts we have I

w(z) dz = −(−ı)θ e

ıθ

C

= ıθ eıθ

 n2π

+

0

 n2π 0

= ın2π e

ın2π

= ın2π − e

−ı

Z

Z

dθ (−ı) eıθ 0

n2π

dθ eıθ 0

− ı(−ı) e

ın2π

n2π

ıθ

0

+e

 n2π 0

= ın2π.

We see that the integral increases indefinitely to ı∞ if we make more and more turns around the origin in the positive direction. If we make an infinite number of turns in the negative direction, then the integral goes to −ı∞. This time there is no way to return to the same leaf of the Riemann surface, except if the total number of turns is zero, that is, if n = 0. Problem 8. Consider the proof of Green’s theorem in two dimensions that was presented in the text. Consider therefore an infinitesimal rectangle of dimensions dx and dy, a vector field (u, v) with Cartesian components u(x, y) and v(x, y), the line integral of (u dx + v dy) over the perimeter of the rectangle, and the surface integral 

∂u ∂v − ∂x ∂y



dx dy

over the area of the rectangle. In the case of the line integral, consider that each value of u and v is associated with a link, like those which form the perimeter of the rectangle, and represent these values as functions of the midpoint of each link. For example, in the first part of the integral we have u(x + dx/2, y), and so on.

123

COMPLEX CALCULUS II: INTEGRATION y (x, y + dy)

(x + dx, y + dy)

dy

(x, y)

(x + dx, y) dx

0

x

Figure 8.1: A rectangular plaquette with sites and links. (a) Show the central fact of the theorem, that is, that in this infinitesimal rectangle, with the perimeter positively oriented, (4) X

(u dx + v dy) =

links



∂v ∂u − ∂x ∂y



dx dy.

(b) Determine to which point one should associate each partial derivative in the most natural and symmetric way possible, and show that the expression 

∂u ∂v − ∂x ∂y



is naturally associated with the center of the rectangle, which is also called, in some circumstances, a plaquette. Complete Solution: We are to calculate the line and surface integration elements, involving the vector w ~ = (u, x) and the position ~z = (x, y), on the rectangle shown in Figure 8.1, made with four links of a Cartesian grid, or lattice. (a) In order to build the line integral element on the perimeter of the rectangle, we associate the vector components to the links, whose positions

124

SOLUTIONS 8 are represented by the midpoints of the links, as shown with crosses in Figure 8.1, so that we have, using the positive direction of the perimeter, (4) X

links

w ~ · d~z =

(4) X

(u dx + v dy)

links

= u (x + dx/2, y) dx + v (x + dx, y + dy/2) dy + −u (x + dx/2, y + dy) dx − v (x, y + dy/2) dy

= dy [v (x + dx, y + dy/2) − v (x, y + dy/2)] +

−dx [u (x + dx/2, y + dy) − u (x + dx/2, y)] v (x + dx, y + dy/2) − v (x, y + dy/2) + = dy dx dx u (x + dx/2, y + dy) − u (x + dx/2, y) −dx dy . dy

We now recognize in these two ratios, in the continuum limit, in which dx and dy both go to zero, the crossed partial derivatives of u and v, that we also associate with the midpoint of each link on which a derivative is taken, v (x + dx, y + dy/2) − v (x, y + dy/2) dx→0 dx ∂v = (x + dx/2, y + dy/2) , ∂x u (x + dx/2, y + dy) − u (x + dx/2, y) lim dy→0 dy ∂u (x + dx/2, y + dy/2) . = ∂y lim

As one can see, these partial derivatives end up associated to the center of the plaquette. We therefore have that our integration line element can be written in terms of a surface integration element, which establishes the central fact of the theorem, (4) X

(u dx + v dy)

links

= dy dx

v (x + dx, y + dy/2) − v (x, y + dy/2) + dx

COMPLEX CALCULUS II: INTEGRATION

125

u (x + dx/2, y + dy) − u (x + dx/2, y) −dx dy dy   ∂v ∂u = dx dy (x + dx/2, y + dy/2) − (x + dx/2, y + dy/2) . ∂y ∂y (b) As shown above, if we proceed as symmetrically as possible, the partial derivatives are both associated to the point (x + dx/2, y + dy/2), which is the center of the rectangle, or plaquette, as shown in the diagram of Figure 8.1. It follows that, if each vector component associated with a link is positioned at the midpoint of that link, and each derivative associated with a link is positioned at the midpoint of that link, then the surface integral element is associated to the central point of the plaquette, (4) X

links

Problem 9.



 ∂v ∂u (u dx + v dy) = dx dy − (x + dx/2, y + dy/2) . ∂y ∂y

Consider the integral of an analytic function given by Z B z n dz, I= A

for n a non-negative integer, between the points A = (0, 0) and B = (1, 1). (a) Calculate the integral along the path formed by a straight segment which links the point A directly to the point B. (b) Calculate the integral along the path formed by a straight segment from A to (1, 0) and another straight segment from (1, 0) to B. (c) Calculate the integral along the path formed by a straight segment from A to (0, 1) and another straight segment from (0, 1) to B. Complete Solution: In each case, the integrals in each segment can be parametrized in terms of either x or of y, where z = x + ıy. (a) On the path formed by a straight segment connecting the points A and B we have that x = y and therefore that z = x + ıx and dz = (1 + ı)dx, so that the integral I is given by

126

SOLUTIONS 8 Z

1

(1 + ı)dx [(i + ı)x]n 0 Z 1 n+1 = (1 + ı) dx xn 0 n+1  1 n+1 x = (1 + ı) n+1 0

I =

(1 + ı)n+1 . n+1

=

(b) On the path formed by a straight segment from A to (1, 0) and another straight segment from (1, 0) to B, we have that on the first segment z = x and dz = dx, and that on the second segment z = 1 + ıy and dz = ıdy, so that the integral I is given by

I = = = =

Z 1 ıdy (1 + ıy)n dx xn + 0 0   xn+1 1 (1 + ıy)n+1 1 + n+1 0 n+1 0

Z

1

(1 + ı)n+1 1 1 + − n+1 n+1 n+1 (1 + ı)n+1 . n+1

(c) On the path formed by a straight segment from A to (0, 1) and another straight segment from (0, 1) to B, we have that on the first segment z = ıy and dz = ıdy, and that on the second segment z = x + ı and dz = dx, so that the integral I is given by

I = = = =

Z

1

Z

1

dx (x + ı)n ıdy (ıy) + 0 0   (ıy)n+1 1 (x + ı)n+1 1 + n+1 0 n+1 0 n

(1 + ı)n+1 ın+1 ın+1 + − n+1 n+1 n+1 n+1 (1 + ı) . n+1

COMPLEX CALCULUS II: INTEGRATION

127

Problem 10. Consider the integral of a complex function, which is not analytic, given by Z B z ∗ dz, I= A

between the points A = (−1, 0) and B = (1, 0). (a) Calculate the integral along the path formed by a straight segment which links the point A directly to the point B. (b) Calculate the integral along the arc of the unit circle on the upper half-plane connecting the points A and B. (c) Calculate the integral along the arc of the unit circle on the lower halfplane connecting the points A and B. Complete Solution: In each case, the integrals in each segment or arc can be parametrized in terms of either x or of θ, where we have that either z = x + ıy or z = ρ exp(ıθ). (a) On the path formed by a straight segment connecting the points A and B we have that z = x and therefore that dz = dx and that z ∗ = x, so that the integral I is given by 1

Z

I =

dx x

−1  x2 1

=

2 = 0.

−1

(b) On the path formed by the arc of the unit circle that connects the points A and B on the upper half-plane we have that z = exp(ıθ) and therefore that dz = ı exp(ıθ)dθ and that z ∗ = exp(−ıθ), so that the integral I is given by I =

Z

= ı

0

ı eıθ dθ e−ıθ

π

Z

0

dθ π

128

SOLUTIONS 8 0

= ıθ

π

= −ıπ.

(c) On the path formed by the arc of the unit circle that connects the points A and B on the lower half-plane we have that z = exp(ıθ) and therefore that dz = ı exp(ıθ)dθ and that z ∗ = exp(−ıθ), so that the integral I is given by I =

Z

= ı

0

ı eıθ dθ e−ıθ

−π Z 0

= ıθ

dθ

−π 0

= ıπ.

−π

Problem 11. Consider the unit-side square and its diagonal going from the point (0, 0) to the point (1, 1) in a Cartesian coordinate system (x, y). Consider approximating this diagonal by a “staircase” that consists of horizontal and vertical segments, all the steps being equal, going from the point (0, 0) to the point (1, 1). Consider the limit in which the size of the steps goes to zero. (a) What is the limit of the distance between any point on the “staircase” and the diagonal of the square, when the size of the steps goes to zero? The “staircase” can be considered as a good representation of the diagonal in this limit, in terms of the points of the plane that make up each object? (b) What is the limit for the total length of the “staircase”, when the size of the steps goes to zero? The “staircase” can be considered as a good representation of the diagonal in this limit, if we now think in terms of the measure associated with each object? Complete Solution: If we have N steps, we will have 2N segments of the same length, half of them horizontal and the other half vertical. Each segment will have an end

129

COMPLEX CALCULUS II: INTEGRATION y 1

N

0 1/N

1 x

Figure 8.2: The “staircase” as an approximation of the diagonal of the square. located on the diagonal, and the other end away from it. One can draw the steps either above or beneath the diagonal, without this changing anything in the answers of the problem. The length of each segment is 1/N , as can be easily seen by drawing the elements involved, as shown in Figure 8.2. (a) For each segment, the point farthest from the diagonal is the end that is away from it. This distance is one half of the diagonal of the small √ square defined by the steps, and it is therefore 1/( 2N ). Since this distance goes to zero as N → ∞ and all other points are at distances shorter than this from the diagonal, it follows that the limit of the distance between an arbitrary point on the “staircase” and the diagonal of the square, when the size of the steps goes to zero, is zero. It follows therefore that all the points of the plane constituting the “staircase” converges to the set of points of the plane forming the diagonal of the square, and the two sets become identical in the limit. In this geometrical sense, in which we did not concern ourselves with measurements, we can therefore say that the “staircase” can be considered as a good representation of the diagonal in the N → ∞ limit. (b) Since each segment has length 1/N and there are 2N of them, the total length of all the chained segments is 2, which is equal to the sum of two sides of the square and that does not depend on N . It follows that the limit of this length when N → ∞ is equal to 2, and therefore it is not equal to the length of the diagonal of the square, which has the value √ 2.

130

SOLUTIONS 8 The value 2 is the linear measure associated with the √ set of points in the plane constituting the “staircase”, while the value 2 is the linear measure associated with the set of points of the plane that constitutes the diagonal of the square. Although the two sets of points become identical in the limit, the two measures do not coincide, so that in terms of the measure associated with each object the “staircase” cannot be considered a good representation of the diagonal in the N → ∞ limit. Extending the concepts that are at stake here to the case of curves in the plane, what we learn from this problem is that, in order for us to be able to properly represent the measure (that is, the length) of a simple closed curve in the plane, we must use a polygon whose vertices are all on the curve. Only in this way the total length of the polygon will have as its limit the length of the curve, in the limit where the size of the segments of the polygon goes to zero.

Problem 12. (Challenge Problem) Consider a closed simple curve C in the plane, the interior of which is the surface S. For simplicity, assume that the curve is differentiable. A certain maximum size is given, represented by the strictly positive real number ǫ, which can be chosen as small as needed. (a) Construct a set of rectangles and triangles, all with vertical and horizontal dimensions smaller than ǫ, which is almost entirely contained within the interior of the curve, in the sense that its boundary is a polygon with all its vertices located on the curve. (b) Given a continuous and differentiable vector field w ~ = (u, v), construct a discrete representation of the surface integral Z

S

~ ×w ∇ ~ da,

using this set of rectangles and triangles. Show that in the limit ǫ → 0 this discrete representation approaches the integral. (c) Given the same vector field w, ~ construct a discrete representation of the line integral I

C

~ w ~ · dℓ,

COMPLEX CALCULUS II: INTEGRATION

131

Figure 8.3: A regular discretization of the surface S. using the polygon which is the external boundary of the set of rectangles and triangles. Show that in the limit ǫ → 0 this discrete representation approaches the integral. (d) Use these constructions to elaborate a more complete proof of Green’s theorem, that is, in our vector language, of the result I

C

~ = w ~ · dℓ

Z

S

~ ×w ∇ ~ da.

Consider in detail the cancellations of integration elements, in order to transform the surface integral into a line integral. Complete Solution: (a) Cover the curve C and its interior S with a regular lattice formed of squares with sides smaller than ǫ, as illustrated in the diagram of Figure 8.3. We will now elaborate an algorithm for the construction of a lattice, that is, of a set of adjacent triangles and rectangles, which is contained within the surface S and whose boundary is a polygon with all of its vertices on the curve C. For each square, there are three disjoint possibilities: either it is completely outside the curve, or it is completely inside the curve, or it is cut by the curve. If the curve passes only

132

SOLUTIONS 8

111 000 000 111 000 111 000 111 000 111 000 111

111 000 0 0001 111 0 1 0001 111 0 0001 111 0 0001 111 0 000 111 0 0001 111 0 1

11 00 00 0011 11 00 11 0011 11 00 0011 11 00 0011 11 00 0011 11 00 0011 11 00

Figure 8.4: Characterization of the possibilities in the discretization of the curve C. through a single point of the square, we consider that it is completely inside or completely outside the curve, as appropriate. We do the same if the curve touches the square only at two vertices of one side of the square. In all other cases, since the curve is differentiable, we can choose ǫ sufficiently small, so that the curve always cuts the square exactly twice, which can occur only in one of the three distinct forms illustrated in the diagram of Figure 8.4. If a square is completely inside of the curve, we included it in our set of rectangles and triangles. If it is completely outside, we exclude it. If it is cut twice by the curve, then we interconnect the two points of intersection using a straight segment. This straight segment divides the square into two parts, one part being essentially all outside the curve and the other essentially all inside it, except for a small region contained between the segment and the curve. We can choose the region to keep in our set as follows: if one of the two regions is entirely outside of the curve, then we exclude it and keep the other region; if one of the two regions is entirely within the curve, then we keep it and exclude the other region. Our set now consists of squares and those regions that were retained, some of which already have a triangular shape. Those which are already triangles are included in our set without further modifications. The others can be divided by means of horizontal or vertical segments, in two or three disjoint parts, which consist of a triangle and one or two rectangles. Having made this division, we include all the other elements in our set, which now consists of squares (that is, a particular case of rectangle), rectangles and triangles. By construction, all these elements have dimensions smaller than ǫ, and its boundary is a polygon with all vertices on the curve, as requested.

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COMPLEX CALCULUS II: INTEGRATION

(b) Given a continuous and differentiable vector field w ~ = (u, v), we can build on each square or rectangle of our set the quantity 

 δv(x, y) δu(x, y) − δIS = δa , δx δy where δa = δxδy is the area element associated with the rectangle. On the triangles is not necessary to calculate in detail the contributions to the surface integral because, as noted in the text, the total area of these triangles tends to zero in the integration limit, because it fits within a strip with infinitesimal width along the perimeter of S, whose area goes therefore to zero. Since the surface integral is two-dimensional, the total contribution of this sub-domain of measure zero is zero. Adding the contributions of all rectangles of our set we therefore have a Riemann partition of the integral over the area S, IS =

X latt

 δv(x, y) δu(x, y) − . δa δx δy 

In the limit ǫ → 0 the dimensions δx and δy of all rectangles go to zero, and the union of all of them covers completely the domain of the integral, while the ratios of variations in the above formula tend to the corresponding partial derivatives, since u(x, y) and v(x, y) are differentiable, so that we have that lim IS

ǫ→0

= =

Z

da

ZS

S



∂v(x, y) ∂u(x, y) − ∂x ∂y



~ ×w ∇ ~ da.

(c) Given the same vector field w, ~ we can build on the boundary of each square or rectangle of our set the quantity δIC = u(x, y) δx + v(x + δx, y) δy − u(x, y + δy) δx − v(x, y) δy, which is a sum over all four sides of the rectangle where δx and δy are the length elements of the sides of the rectangle. As for the triangles, each one has a horizontal side, a vertical side and a sloping side. By construction, all the horizontal or vertical sides are strictly within the

134

SOLUTIONS 8

(x, y)

(x, y)

(x, y)

(x, y)

Figure 8.5: Classification of the triangles on the boundary of S. surface S, while all the sloping sides are part of the boundary of our lattice. The triangles may be oriented in one of four ways that are shown in the diagram of Figure 8.5. Using as an example a triangle with the first orientation, we can write on it the quantity ~ δIC = u(x, y) δx − v(x, y) δy + w ~ · δℓ, ~ is the displacement associated with the sloping side, with the where δℓ proper positive orientation for the triangle. A quantity of this type can be written for every triangle, varying only the application point of u(x, y) and v(x, y) and the orientation of each segment. For example, for the second type of orientation we have ~ δIC = u(x, y) δx + v(x + δx, y) δy + w ~ · δℓ, and so on. In all cases, the sloping sides, which are associated with the ~ are part of the outer boundary of the lattice, while the terms w ~ · δℓ, other two sides are internal. It turns out that, as noted in the text, all internal sides are traversed twice, once in each direction, so that the two identical contribution to the line integral have opposite signs, and therefore all cancel off. There remains therefore the outer boundary of the lattice, which is a polygon consisting of the sloping sides of the triangles and occasionally some side of a square, which happens to be located on a vertical or horizontal part of the curve. It follows that we have a discrete representation of the line integral, which is given by IC =

X

poly

~ w ~ · δℓ,

which is a sum over a polygon with all vertices on the curve. In the limit ǫ → 0 the dimensions of all the squares and triangles go to zero,

135

COMPLEX CALCULUS II: INTEGRATION

and therefore the sides of this polygon go to zero, so that the polygon approaches the curve with the correct measure, and hence we have that lim IC

ǫ→0

= lim

ǫ→0

=

I

C

X

poly

~ w ~ · δℓ

~ w ~ · dℓ.

(d) We started by showing that, in the limit ǫ → 0, the quantities δIC and δIS on one of the rectangles of our lattice are equal, just as was done in the text. We started by writing δIC in the following way, δIC = [v(x + δx, y) − v(x, y)] δy − [u(x, y + δy) − u(x, y)] δx. Since u(x, y) and v(x, y) are differentiable, these variations can be written in the limit ǫ → 0, in terms of the appropriate partial derivatives, δIC

δu(x, y) δv(x, y) δxδy − δyδx δx δy   δv(x, y) δu(x, y) = − δa, δx δy =

where δa = δxδy. This establishes that δIC = δIS in each of the rectangles of the lattice. If we add these quantities over all rectangles, we will have a corresponding equality for the two sums. X

δIC =

rect

X

δIS .

rect

It only remains for us to consider the triangles of our lattice. Since from the point of view of the surface integral they are a zero-measure subset, we can add the contributions of the triangles to the sum on the right-hand side, without changing its value in the limit. On the other hand, if these same contributions are added to the sum on the left-hand side, then it has as its limit the line integral. It follows that we can write that the equality holds in the limit for the sums over the entire lattice X latt

δIC =

X latt

δIS ,

136

SOLUTIONS 8 and therefore, taking the integration limit, we have Green’s theorem, that is, in our vector language, I

C

~ = w ~ · dℓ

Z

S

~ ×w ∇ ~ da.

Solutions 9

Complex Derivatives and Integrals We present here complete and commented solutions to all problems proposed in Chapter 9 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1.

Calculate the integral I C

zn dz, (z − z0 )2

for a non-negative integer n, on an arbitrary closed curve C that makes a single turn around the point z0 . Determine for which combinations of values of n and z0 the result is zero. Complete Solution: Let us recall that one of the Cauchy integral formulas, that for the first derivative, tells us that I f (z) 1 dz ⇒ f ′ (z0 ) = 2πı C (z − z0 )2 I f (z) dz = 2πıf ′ (z0 ), 2 C (z − z0 ) so long as the curve C contain z0 , so that we can use this relation in order to calculate the integral 137

138

SOLUTIONS 9

I

C

zn dz, (z − z0 )2

by simply making f (z) = z n . We have therefore I zn (n−1) dz = 2πınz0 . 2 C (z − z0 ) This vanishes if n = 0, for any value of z0 , or if z0 = 0 with n > 1. In the special case in which we have z0 = 0 with n = 1, the value is 2πı, because in this case we have the integral around a simple pole, I 1 dz = 2πı. C z Problem 2.

Calculate the integral √ I z dz, n C (z − z0 )

for a non-negative integer n, on an arbitrary closed curve C that makes a single turn around the point z0 and does not contain the origin. Calculate the z0 → 0 limit of the result obtained. How can we interpret the result in this limit? Hint: use the Cauchy integral formulas. Complete Solution: Let us recall the Cauchy integral formulas, which give the derivatives of an analytic function in terms of integrals, I f (z) n! n′ dz ⇒ f (z0 ) = 2πı C (z − z0 )n+1 I f (z) 2πı n′ f (z0 ), dz = n+1 n! C (z − z0 ) so long as the curve C contain z0 , where n ≥ 0. It follows that we can use this relation in order to calculate the integral √ I z dz, n (z − z 0) C

COMPLEX DERIVATIVES AND INTEGRALS

139

where n ≥ 0 and the contour does not contain the origin, which is the branch √ point of the square root, by simply adopting f (z) = z. Note that for n = 0 the integrand is analytic in C, so that in this case the integral is zero on the contour described, by the Cauchy-Goursat theorem. Hence, we can limit the values of n in the following discussion to n ≥ 1. We have therefore √ √ I z 2πı dn−1 z dz = (z0 ). n (n − 1)! dz C (z − z0 ) In the particular case n = 1 we have the function itself instead of a derivative, √ √ and hence the result is simply 2πı z0 . Of course, since the function z has a Riemann surface with two leaves, and since the contour is all on the same leaf due to the fact that the branch point is outside the contour, we have in fact two possible answers with opposite signs, one on each one of the two leaves. The same is true for the other values of n. What remains to be done here is to calculate all the derivatives and try to write a general form for them. Making a table of the first few cases we have √ √ 0′ z = z, √ 1′ 1 1 √ , z = 2 z √ 2′ 1 (−1) 1 z = √ 3, 2 2 z √ 3′ 1 (−1) (−3) 1 z = √ 5, 2 2 2 z √ 4′ 1 (−1) (−3) (−5) 1 z = √ 7, 2 2 2 2 z √ 5′ 1 (−1) (−3) (−5) (−7) 1 z = √ 9, 2 2 2 2 2 z √ 6′ 1 (−1) (−3) (−5) (−7) (−9) 1 z = √ 11 , 2 2 2 2 2 2 z ... = ... . Collecting the factors of (−1) in the numerator and of 2 in the denominator, separating a square root and a factor of (−1), and noting that there appear in the numerator the double factorials of odd numbers, we can write this in the following way, which displays clearly the regularity of almost all the factors in these results, √ (−1)0 (−1) 1 √ 0′ , z = − z 20 z0

140

SOLUTIONS 9 √ (−1)1 1 √ 1′ z = − z , 21 z 1 √ 2′ √ (−1)2 1!! 1 , z = − z 22 z2 √ 3′ √ (−1)3 3!! 1 z = − z , 23 z3 √ (−1)4 5!! 1 √ 4′ z = − z , 24 z4 √ 5′ √ (−1)5 7!! 1 , z = − z 25 z5 √ (−1)6 9!! 1 √ 6′ z = − z , 26 z6 ... = ... .

In order to facilitate the manipulation of the double factorials, we add additional factors so that they appear in all the cases, starting with 1!! = 1 in the first line, thus getting √ (−1)0 1!! 0 √ 0′ z = − z 0 z , 2 (1)(−1) √ (−1)1 3!! −1 √ 1′ z = − z 1 z , 2 (3)(1) √ (−1)2 5!! −2 √ 2′ z = − z 2 z , 2 (5)(3) √ (−1)3 7!! −3 √ 3′ z = − z 3 z , 2 (7)(5) √ (−1)4 9!! −4 √ 4′ z , z = − z 4 2 (9)(7) √ 5′ √ (−1)5 11!! −5 z , z = − z 5 2 (11)(9) √ 6′ √ (−1)6 13!! −6 z , z = − z 6 2 (13)(11) ... = ... . We now see that, in the case of the derivative of order k, the double factorial which appears is (2k + 1)!!, and that the two additional factors in the denominator are (2k + 1) and (2k − 1), which are never zero and whose product is  2 (2k) − 1 . Therefore, we can already write the general case, for k ≥ 0, √

z

k′

√ (−1)k (2k + 1)!! −k = − z k z 2 [(2k)2 − 1]

141

COMPLEX DERIVATIVES AND INTEGRALS =

√ (−1)(k+1) (2k + 1)!! 1 . z 2k [(2k)2 − 1] zk

The square root that appears multiplying these results clearly shows that they all have a Riemann surface with two leaves. We can write the result solely in terms of factorials instead of double factorials, multiplying and dividing by (2k)!!, in order to obtain √

z

k′

√ (−1)(k+1) (2k + 1)!!(2k)!! 1 z 2k (2k)!! [(2k)2 − 1] zk √ (−1)(k+1) (2k + 1)! 1 . = z (2k) 2 k! [(2k)2 − 1] z k =

Applying this at z0 , for k = n − 1, with n ≥ 1, we have √ I (−1)n (2n − 1)! 1 2πı √ z z dz = 0 (2n−2) n 2 (n−1) (n − 1)! 2 (n − 1)! [(2n − 2) − 1] z C (z − z0 ) 0 √ πı(−1)n n2 (2n − 1)! 1 z0 (2n−3) = 2 (n!)2 (2n − 1)(2n − 3) z (n−1) 0

=

√

πı(−1)n n(2n)! 1 z0 (2n−2) . 2 (n−1) 2 (n!) (2n − 1)(2n − 3) z 0

Note that, since the factors in denominator never vanish, this result is certainly well defined for n ≥ 1. If we make z0 → 0, the contour C, which must contain z0 but cannot contain the origin, must pass through an increasingly narrow region between 0 and z0 . This means that in this limit the contour is forced to approach indefinitely the two singularities, the pole at z0 and the branch point at 0. As a consequence, it is reasonable to expect divergences in these results, in this kind of limit. However, the case n = 1 is different from the others, because in this case we have that √ I √ z dz = lim 2πı z0 lim z0 →0 z0 →0 C z − z0 = 0, which happens due to the fact that, although the square root has a point of singularity at the origin, it is not a pole but a branch point at which the function itself exists and is zero, although it is not analytic. For all other values of n the derivatives of the square root are involved, which are in fact divergent at the origin, so that in this case we have

142

SOLUTIONS 9 √

z dz z0 →0 C (z − z0 )n √ πı(−1)n n(2n)! 1 = lim z0 (2n−2) 2 (n−1) z0 →0 2 (n!) (2n − 1)(2n − 3) z 0 1 πn(2n)! lim (n−1/2) , = ı(−1)n (2n−2) 2 2 (n!) (2n − 1)(2n − 3) z0 →0 z lim

I

0

where for n > 1 the numerical fraction displayed is a positive real number. Writing z0 in polar form, in terms of ρ0 and θ0 , we have that √ I z lim dz z0 →0 C (z − z0 )n πn(2n)! e−ı(n−1/2)θ lim 2(2n−2) (n!)2 (2n − 1)(2n − 3) z0 →0 ρ(n−1/2) 0 1 πn(2n)! n −ı(n−1/2)θ lim (n−1/2) , = ı(−1) e (2n−2) 2 ρ →0 2 (n!) (2n − 1)(2n − 3) 0 ρ

= ı(−1)n

0

where the remaining limit goes to real positive infinity, and we are assuming, for simplicity, that the limit is taken in the complex plane with θ constant, that is, along straight lines through the origin. The result is therefore infinite in magnitude, with a certain phase, which depends on the angle θ, √ I z dz = ı(−1)n eı(−n+1/2)θ ∞ lim z0 →0 C (z − z0 )n =

=

eıπ/2 e−ıπn eı(−n+1/2)θ ∞ eı(−n+1/2)(π+θ) ∞,

where we represented by ∞ the product of the divergent limit and the positive real quantity multiplying it. Note that the limit depends on θ, and if it changes during the limit the limit can, for example, spiral out to infinity, instead of going along straight lines. Problem 3.

Calculate the integral I sin(z) dz, 2 C (z − z0 )

on an arbitrary closed curve C that makes a single turn around the point z0 . Determine for which values of z0 the result is zero.

COMPLEX DERIVATIVES AND INTEGRALS

143

Hint: use the Cauchy integral formulas. Complete Solution: Let us recall that one of the Cauchy integral formulas, that for the first derivative, tells us that I f (z) 1 ′ dz ⇒ f (z0 ) = 2πı C (z − z0 )2 I f (z) dz = 2πıf ′ (z0 ), 2 (z − z ) 0 C so long as the curve C contain z0 , so that we can use this relation in order to calculate the integral I sin(z) dz, (z − z0 )2 C by simply making f (z) = sin(z). We have therefore I sin(z) dz = 2πı cos(z0 ). (z − z0 )2 C This vanishes where cos(z0 ) vanishes, and as we have seen before, the only zeros of cos(z0 ) are the real zeros that we already know quite well. Thus, the result of the integral is zero for z0 =

π + nπ, 2

where n is an arbitrary integer. Problem 4.

Calculate the integral I sin(z) dz, 3 C (z − z0 )

on an arbitrary closed curve C that makes a single turn around the point z0 . Determine for which values for z0 the result is zero. Answer: −ıπ sin(z0 ). Hint: use the Cauchy integral formulas. Complete Solution:

144

SOLUTIONS 9

Let us recall that one of the Cauchy integral formulas, that for the second derivative, tells us that I f (z) 2 ′′ dz ⇒ f (z0 ) = 2πı C (z − z0 )3 I f (z) dz = πıf ′′ (z0 ), 3 (z − z ) 0 C so long as the curve C contain z0 , so that we can use this relation in order to calculate the integral I sin(z) dz, (z − z0 )3 C by simply making f (z) = sin(z). We have therefore I sin(z) dz = −πı sin(z0 ). 3 C (z − z0 ) This vanishes where sin(z0 ) vanishes, and as we have seen before, the only zeros of sin(z0 ) are the real zeros that we already know quite well. Thus, the result of the integral is zero for z0 = nπ, where n is an arbitrary integer. Problem 5.

Prove by finite induction the Newton binomial formula, n

(a + b) =

n X k=0

n! ak bn−k , k!(n − k)!

where a and b are any complex numbers. Make sure that the proof holds for complex numbers. Complete Solution: We start by showing that the Newton binomial formula is valid for the initial case n = 1, that is trivial. Putting n = 1 in (a + b)n =

n X k=0

n! ak bn−k k!(n − k)!

145

COMPLEX DERIVATIVES AND INTEGRALS we obtain 1

(a + b)

=

1 X k=0

1! ak b1−k k!(1 − k)!

⇒

1 1! a0 b1−0 + a1 b1−1 0!(1 − 0)! 1!(1 − 1)! = a0 b1 + a1 b0

a+b =

= b + a, which is nothing more than a simple identity that is clearly true for both real numbers and complex numbers. Let us now assume the result for the case n − 1, n−1

(a + b)

=

n−1 X k=0

(n − 1)! ak bn−1−k , k!(n − 1 − k)!

and show that this implies the case n. In order to do this, we multiply both sides of this formula by (a + b), thus obtaining (a + b)n = (a + b)

n−1 X k=0

n−1 X

=

k=0

(n − 1)! ak bn−1−k k!(n − 1 − k)! n−1

X (n − 1)! (n − 1)! ak+1 bn−1−k + ak bn−k . k!(n − 1 − k)! k!(n − 1 − k)! k=0

In order to show that the result holds for complex numbers, it suffices to note that everything that is being done here consists of arithmetic operations of the field. In the second sum the factor ak bn−k is already in the form we want. In the first sum we make the transformation of variables k′ = k + 1, k = k′ − 1, so that this factor takes the correct form, thus obtaining n

(a + b) =

n X

k ′ =1

n−1

X (n − 1)! (n − 1)! k ′ n−k ′ + b a ak bn−k , ′ ′ (k − 1)! (n − k )! k!(n − 1 − k)! k=0

where we can now change the name k′ back to k. The last term of the first sum and the first term of the second sum must be considered separately, but for all the other terms the two sums can now be united in a single sum, thus resulting in

146

SOLUTIONS 9 (a + b)n (n − 1)! 0 n (n − 1)! an bn−n + a b + (n − 1)!(n − n)! 0!(n − 1)!   n−1 X 1 1 + ak bn−k + (n − 1)! (k − 1)!(n − k)! k!(n − 1 − k)! k=1   n−1 X n−k k n 0 0 n + ak bn−k = a b +a b + (n − 1)! k!(n − k)! k!(n − k)!

=

k=1

n−1 X (n − 1)! (k + n − k) ak bn−k + an b0 . = a0 bn + k!(n − k)! k=1

As one can see, we get a factor of n that completes the factorial of n in the numerator. It is easy to verify that the first and last terms can be included in the sum as the terms k = 0 and k = n respectively, so that we in fact have the result n X n! ak bn−k . (a + b)n = k!(n − k)! k=0

Since all this derivation did not use anything other than the operations and properties of the fields, and since they all hold for complex numbers as well as for real numbers, it follows that this result holds for all complex numbers a and b, and for any natural number n.

Problem 6. Prove by finite induction the Cauchy integral formula for the n-th derivative of an analytic function, I f (z) n! n′ f (z0 ) = dz, 2πı C (z − z0 )n+1

that is, assume the formula for the case n − 1 and show that this implies that it holds for the case n.

Hints: do not separate the expression (z − z0 ) during the calculations; write δz = z1 − z0 , recalling that δz will be eventually taken to zero; and write the difference z − z1 as z − z1 = (z − z0 ) − δz. Complete Solution: We will prove the general Cauchy formula for the n-th derivative of an analytic function f (z), by means of finite induction. This means that we have the result for the case n = 0, which is the Cauchy formula

COMPLEX DERIVATIVES AND INTEGRALS

1 f (z0 ) = 2πı

I

C

147

f (z) dz, z − z0

and that we will assume the validity of the case n − 1, I (n − 1)! f (z) (n−1)′ f (z0 ) = dz, n 2πı C (z − z0 ) from which we will prove the case n. In order to do this, we use this formula to calculate f (n−1)′ (z) at two points, z0 and z1 , and set up the ratio which, in the limit (z1 − z0 ) → 0, is the definition of n-th derivative f n′ (z), δf (n−1)′ (z) δz

= = =

f (n−1)′ (z1 ) − f (n−1)′ (z0 ) z1 − z0  I  (n − 1)! 1 1 − f (z) dz 2πı(z1 − z0 ) C (z − z1 )n (z − z0 )n I (z − z0 )n − (z − z1 )n (n − 1)! f (z) dz. 2πı(z1 − z0 ) C (z − z1 )n (z − z0 )n

We will now write everything in terms of δz = (z1 − z0 ), using the fact that we can write that (z − z1 ) = (z − z0 ) − δz, δf (n−1)′ (z) δz I (n − 1)! (z − z0 )n − [(z − z0 ) − δz]n = f (z) dz 2πıδz C (z − z0 )n [(z − z0 ) − δz]n (n − 1)! = × 2πıδz I (z − z )n − Pn n! k k n−k 0 k=0 k!(n−k)! (−1) (δz) (z − z0 ) f (z) dz, × (z − z0 )n [(z − z0 ) − δz]n C where we used in the numerator the Newton binomial formula, applied to [(z − z0 ) − δz]n , n

[(z − z0 ) − δz] =

n X k=0

n! (−1)k (δz)k (z − z0 )n−k . k!(n − k)!

We now observe that, in the numerator of the resulting formula, the first term cancels with the term k = 0 of the sum, so that only the terms k ≥ 1 of this sum remain, and we therefore have

148

SOLUTIONS 9 δf (n−1)′ (z) δz =

(n − 1)! 2πıδz

=

(n − 1)! 2πı

I − Pn k=1

n! k k n−k k!(n−k)! (−1) (δz) (z − z0 ) f (z) dz (z − z0 )n [(z − z0 ) − δz]n C I Pn n! k+1 (δz)k−1 (z − z )n−k 0 k=1 k!(n−k)! (−1) f (z) dz, n n (z − z0 ) [(z − z0 ) − δz] C

where we canceled a factor of δz. Let us now observe that, in the denominator, the δz → 0 limit produces a finite and non-zero result, whereas in the numerator the same limit produces zero for all terms of the sum except for the term k = 1, so that we can write that f n′ (z0 ) = = = = = =

δf (n−1)′ (z) δz→0 δz I Pn n! k+1 (δz)k−1 (z − z )n−k 0 (n − 1)! k=1 k!(n−k)! (−1) lim f (z) dz n n 2πı δz→0 C (z − z0 ) [(z − z0 ) − δz] Pn I n! k+1 (δz)k−1 (z − z )n−k 0 (n − 1)! k=1 k!(n−k)! (−1) lim f (z) dz n n 2πı (z − z0 ) [(z − z0 ) − δz] C δz→0 I n! 2 n−1 (n − 1)! 1!(n−1)! (−1) (z − z0 ) lim f (z) dz n n 2πı C δz→0 (z − z0 ) [(z − z0 ) − δz] I 1 n(n − 1)! lim f (z) dz n 2πı C δz→0 (z − z0 )[(z − z0 ) − δz] I n! f (z) dz, 2πı C (z − z0 )n+1 lim

where we can pass the limit into the integral because the integration contour C is fixed, it does not change at all when we take the δz → 0 limit, and thus we prove that I n! f (z) f n′ (z0 ) = dz, 2πı C (z − z0 )n+1 as was our intention. Since we already know that the case n = 0 holds, and since we have just shown that the case n − 1 implies the case n, it follows by finite induction that the result holds for all values of n. Problem 7. Consider an analytic function f (z). Suppose that we differentiate it a certain number n of times, and then integrate it the same number

COMPLEX DERIVATIVES AND INTEGRALS

149

of times in the sense of indefinite integration or primitivization, thus obtaining an analytic function g(z). Write the most general possible form of the difference g(z) − f (z). Complete Solution: Let us start by examining the problem in the case n = 1. The analytic function g(z) which is the primitive of the derivative of the analytic function f (z) may differ from f (z) by a complex constant C0 , which is the integration constant, so that in this case we can have, in the most general case possible, g(z) − f (z) = C0 . If we have n = 2 and we take two derivatives followed by two integrations, the two first derivatives, g ′ (z) and f ′ (z) also differ in this way, say by a constant C1 , g ′ (z) − f ′ (z) = C1 , but when we make the second integration on g′ (z), the term C1 is integrated to C1 z, and a new integration constant appears, which we may call C0 . Thus, in this case the most general possible difference is a first degree polynomial, that is, g(z) − f (z) = C0 + C1 z. Further, examining the case n = 3, in this case the three integrations produce, on the way back, in the first step a constant C2 , in the second step the combination C1 + C2 z, and in the third step the combination C0 + C1 z + C2 z 2 /2, where we now have three arbitrary integration constants. It is clear now that the generalization to the case n produces a polynomial of order n − 1, that is, g(z) − f (z) = C0 + C1 z +

Cn−1 n−1 C2 2 C3 3 C4 3 z + z + z + ... + z . 2 3! 4! (n − 1)!

It is interesting to note that the set of functions which appear naturally in this process is the set of function z n /n! that we got here. However, since all integration constants are arbitrary, we can write, in a completely equivalent way, that the most general possible difference is given by g(z) − f (z) =

n−1 X

Ak z k ,

k=0

for n arbitrary complex constants Ak , k ∈ {0, . . . , n − 1}.

150

SOLUTIONS 9

Solutions 10

Complex Inequalities and Series We present here complete and commented solutions to all problems proposed in Chapter 10 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. integer N ,

Starting from the discrete triangle inequality for an arbitrary

N N X X |zn |, zn ≤ n=1

n=1

show that the corresponding inequality for complex integrals over finite integration contours C holds, Z Z f (z) dz ≤ |f (z)| |dz| , C

C

so long as the two integrals involved exist.

Hint: use the Riemann definition of the integrals. Complete Solution: According to the definition of a complex Riemann integral, we divide the curve C in N segments of equal length |δz|, and then we write that 151

152

SOLUTIONS 10

Z

f (z) dz = C

N X

lim

δz → 0 n=1 N →∞

f (zn ) δzn ,

where each zn is within the corresponding segment described by δzn . Using the triangle inequalities for these finite sums we have now N N X X |f (zn ) δzn | f (zn ) δzn ≤ n=1

=

n=1 N X

n=1

|f (zn )| |δzn |.

Since this holds for any value of N , it also holds in the limit in which N goes to infinity, so long as the sums converge, so that we can write that Z N X f (z) dz = lim f (z ) δz n n δz → 0 C N → ∞ n=1 N X f (zn ) δzn = lim δz → 0 N →∞

≤

=

lim

n=1

N X

δz → 0 n=1 N →∞

Z

C

|f (zn )| |δzn |

|f (z)| |dz| ,

where we use once again, in this last passage, the definition of the Riemann integral. So long as the integrals exist, that is, so long as the indicated limits exist, we therefore obtain the result Z Z f (z) dz ≤ |f (z)| |dz| . C

C

Problem 2. Starting from the Cauchy integral formulas that give the derivatives of an analytic function f (z) in terms of contour integrals, I n! f (z) n′ f (z0 ) = dz, 2πı C (z − z0 )n+1

COMPLEX INEQUALITIES AND SERIES

153

where the integration contour C contains z0 , show that the inequalities n′ f (z0 ) ≤ n! |f | r0n C

hold, where the indicated average is taken on a circular contour C of radius r0 centered at z0 . Complete Solution: Starting from the Cauchy integral formulas that give the derivatives of an analytic function f (z) in terms of contour integrals, n! f (z0 ) = 2πı n′

I

C

f (z) dz, (z − z0 )n+1

and taking C as a circle of radius r0 centered at z0 , on which we have that |z − z0 | = r0 and dz = r0 ı exp(ıθ) dθ, we can write that I n′ f (z) f (z0 ) = n! dz 2πı n+1 C (z − z0 ) I n! f (z) ≤ dz n+1 2π C (z − z0 ) I |f (z)| n! |dz| = 2π C |z − z0 |n+1 Z n! 2π |f (z)| r0 dθ = 2π 0 r0n+1   Z 2π n! 1 = |f (z)| dθ r0n 2π 0 n! |f | , = r0n C that is, we conclude that in fact the inequalities n′ f (z0 ) ≤ n! |f | , r0n C

hold, where the indicated average is taken on the circular contour C of radius r0 centered at z0 .

154

SOLUTIONS 10

Problem 3. Consider the Taylor series of the functions exp(x), sin(x) and cos(x), for real x, around x = 0. (a) Write each one of the series as an infinite sum of a general term. (b) Write the series of exp(ıx), with x real, separating its real and imaginary parts. (c) Identify the real and imaginary parts of the series, and write the resulting relation between the functions involved. Answer: exp(ıx) = cos(x) + ı sin(x).

Complete Solution: Taking the appropriate derivatives and applying at the point x = 0, it is not difficult to verify that the series are as follows, 1 1 1 1 2 x + x3 + x4 + x5 + . . . , 2 3! 4! 5! 1 3 1 5 1 7 1 9 sin(x) = x − x + x − x + x + ... , 3! 5! 7! 9! 1 2 1 4 1 6 1 8 cos(x) = 1 − x + x − x + x + ... . 2 4! 6! 8!

exp(x) = 1 + x +

(a) Systematizing the above sequences of coefficient, we can write the series in explicit form,

exp(x) = sin(x) =

∞ X 1 n x , n!

n=0 ∞ X k=0

cos(x) =

(−1)k x2k+1 , (2k + 1)!

∞ X (−1)k k=0

(2k)!

x2k .

(b) We have for the series of exp(ıx), separating terms with even n from those with odd n,

155

COMPLEX INEQUALITIES AND SERIES ∞ X 1 n n ı x exp(ıx) = n! n=0

=

∞ X k=0

=

∞ X k=0

=

∞

1 2k 2k X 1 ı x + ı2k+1 x2k+1 (2k)! (2k + 1)! k=0

∞

X
k
k 1 1 ı2 x2k + ı ı2 x2k+1 (2k)! (2k + 1)! k=0

∞ X (−1)k x2k + ı x2k+1 , (2k)! (2k + 1)!

∞ X (−1)k k=0

k=0

where we used the fact that ı2 = −1, and the sums are now explicitly real. (c) As discussed above, we have that

exp(ıx) =

∞ X (−1)k k=0

(2k)!

x

2k

∞ X (−1)k +ı x2k+1 , (2k + 1)! k=0

where the real and imaginary parts are written explicitly. We now identify the first series as the Taylor series of cos(x), and the second as the Taylor series of sin(x). Since the series are faithful representations of the respective functions, we conclude that holds between them the identity exp(ıx) = cos(x) + ı sin(x), which is the Euler formula. Problem 4.

Consider the function f (z) = cos(z).

(a) Expand the function around the point z = π/2. (b) Identify the coefficients of the terms of the resulting series and write the trigonometric identity that follows from them. Answer: cos(z) = sin(π/2 − z).

156

SOLUTIONS 10

Complete Solution: We will expand in a power series around z0 = π/2 the function w(z) = cos(z). (a) Calculating the first few derivatives we have w0′ (z) = cos(z), w1′ (z) = − sin(z),

w2′ (z) = − cos(z),

w3′ (z) = sin(z),

w4′ (z) = cos(z), w5′ (z) = − sin(z), ...

... .

Systematizing the answers separately for the cases of even n and odd n, we can write that w(2k)′ (z) = (−1)k cos(z), w(2k+1)′ (z) = (−1)k+1 sin(z), for k ∈ {0, 1, 2, 3, . . . , ∞}. Applying this at z0 = π/2 we have w(2k)′ (π/2) = (−1)k cos(π/2) = 0, w

(2k+1)′

(π/2) = (−1)k+1 sin(π/2) = (−1)k+1 .

We have therefore for the Taylor series of the function around π/2,

w(z) =

∞ X (−1)k+1 (z − π/2)2k+1 (2k + 1)! k=0 ∞ X

cos(z) = −

k=0

⇒

(−1)k (z − π/2)2k+1 . (2k + 1)!

COMPLEX INEQUALITIES AND SERIES

157

(b) If we compare the result above with the series of sin(z) around z = 0, sin(z) =

∞ X (−1)k z 2k+1 , (2k + 1)! k=0

we verify that the coefficients and the powers involved are the same, so that it follows that cos(z) = − sin(z − π/2) or, to put it in a more familiar form for the case in which we have a real z, cos(z) = sin(π/2 − z). Problem 5.

Consider the function f (z) = sinh(z).

(a) Expand the function around the point z = ıπ. (b) Identify the coefficients of the terms of the resulting series and write the identity between functions that follows from them. Answer: sinh(z) = sinh(ıπ − z). Complete Solution: We will expand in a power series around z0 = ıπ the function w(z) = sinh(z). First of all, let us recall that the hyperbolic functions are given by the series cosh(z) =

∞ X k=0

sinh(z) =

∞ X k=0

1 z 2k , (2k)! 1 z 2k+1 , (2k + 1)!

and that for purely imaginary arguments it follows therefore that cosh(ıy) = =

∞ X

k=0 ∞ X k=0

1 2k 2k ı y (2k)! (−1)k 2k y (2k)!

= cos(y),

158

SOLUTIONS 10

sinh(ıy) =

∞ X

1 ı2k+1 y 2k+1 (2k + 1)!

k=0 ∞ X

= ı

k=0

(−1)k y 2k+1 (2k + 1)!

= ı sin(y). (a) Calculating the first few derivatives we have w0′ (z) = sinh(z), w1′ (z) = cosh(z), w2′ (z) = sinh(z), w3′ (z) = cosh(z), w4′ (z) = sinh(z), w5′ (z) = cosh(z), ...

... .

Systematizing the answers separately for the cases of even n and odd n, we can write that w(2k)′ (z) = sinh(z), w(2k+1)′ (z) = cosh(z), for k ∈ {0, 1, 2, 3, . . . , ∞}. Applying this at z0 = ıπ we have w(2k)′ (π/2) = sinh(ıπ) = ı sin(π) = 0, w

(2k+1)′

(π/2) = cosh(ıπ) = cos(π) = −1.

We have therefore for the Taylor series of the function around ıπ, w(z) =

∞ X

(−1) (z − ıπ)2k+1 (2k + 1)!

k=0 ∞ X

sinh(z) = −

k=0

⇒

1 (z − ıπ)2k+1 . (2k + 1)!

COMPLEX INEQUALITIES AND SERIES

159

(b) If we compare the result above with the series of sinh(z) around z = 0, which was given before, we verify that the coefficients and the powers involved are the same, so that it follows that sinh(z) = − sinh(z − ıπ), that is, we have that sinh(z) = sinh(ıπ − z). Problem 6. (Challenge Problem) For each of the following functions, use its Taylor series around z = 0 in order to explicitly write the real and imaginary parts of the function in explicit form, in terms of other known functions. (a) sin(z). Answer: sin(x) cosh(y) + ı cos(x) sinh(y). (b) cos(z). Answer: cos(x) cosh(y) − ı sin(x) sinh(y). (c) sinh(z). Answer: sinh(x) cos(y) + ı cosh(x) sin(y). (d) cosh(z). Answer: cosh(x) cos(y) + ı sinh(x) sin(y). Hints: it will be necessary to use the Newton binomial formula, and to figure out how to make changes of summation variables in infinite double sums with two indices. Complete Solution: Let us begin by recalling that the functions listed are given by the series sin(z) = cos(z) =

∞ X (−1)k z 2k+1 , (2k + 1)!

k=0 ∞ X k=0

sinh(z) =

∞ X k=0

cosh(z) =

∞ X k=0

(−1)k 2k z , (2k)!

1 z 2k+1 , (2k + 1)! 1 z 2k . (2k)!

160

SOLUTIONS 10

Moreover, in all cases it will be necessary to use the Newton binomial formula for (x + ıy)n , separately for the cases of even n and odd n, (x + ıy)n =

n X l=0

= (x + ıy)2k = (x + ıy)2k+1 =

n X

n! xn−l (ıy)l l!(n − l)!

l=0

ıl n! xn−l y l l!(n − l)!

2k X

ıl (2k)! x2k−l y l , l!(2k − l)!

l=0 2k+1 X l=0

⇒

ıl (2k + 1)! x2k+1−l y l , l!(2k + 1 − l)!

where we wrote separately the cases n = 2k and n = 2k + 1. In each one of these cases, it is necessary to separate the sum over l in two, one for even l and another for odd l, in order to allow us to separate the real and imaginary parts. Let us do first l = 2j and then l = 2j + 1. Since l ranges from 0 to n, that is, from 0 to 2k or from 0 to 2k + 1, the terms with even l range from 0 to 2k, regardless of what the parity of n may be. Thus we see that in this case we have l = 2j with j going from 0 to k. However, the terms with odd l go from 1 to 2k + 1 if n is odd, but only from 1 to 2k − 1 if n its even. Thus, if either l or n are odd, we have l = 2j + 1 with j going from 0 to k, but if l is odd and n even, we have l = 2j + 1 with j going from 0 to k − 1. This means, in particular, that in the latter case the value k = 0 is not included. We therefore have, for even n,

2k

(x + ıy)

2k X ıl (2k)! x2k−l y l = l!(2k − l)!

=

l=0 k X j=0

+

ı2j (2k)! x2k−2j y 2j + (2j)!(2k − 2j)!

k−1 X j=0

=

k X j=0

ı2j+1 (2k)! x2k−2j−1 y 2j+1 (2j + 1)!(2k − 2j − 1)!

(−1)j (2k)! x2k−2j y 2j + (2j)!(2k − 2j)!

COMPLEX INEQUALITIES AND SERIES k−1 X

+ı

j=0

161

(−1)j (2k)! x2k−2j−1 y 2j+1 , (2j + 1)!(2k − 2j − 1)!

with real and imaginary parts properly separated, and in which the second term exists only for k ≥ 1. For odd n we have 2k+1 X

(x + ıy)2k+1 =

l=0 k X

=

j=0

+

ıl (2k + 1)! x2k+1−l y l l!(2k + 1 − l)!

ı2j (2k + 1)! x2k+1−2j y 2j + (2j)!(2k + 1 − 2j)!

k X j=0

k X

=

j=0

+ı

ı2j+1 (2k + 1)! x2k+1−2j−1 y 2j+1 (2j + 1)!(2k + 1 − 2j − 1)!

(−1)j (2k + 1)! x2k+1−2j y 2j + (2j)!(2k + 1 − 2j)!

k X j=0

(−1)j (2k + 1)! x2k−2j y 2j+1 , (2j + 1)!(2k − 2j)!

where the real and imaginary parts are properly separated. We can write the final form of these two expressions in terms of the difference (k − j), 2k

(x + ıy)

=

k X j=0

+ı

(−1)j (2k)! x2(k−j) y 2j + (2j)![2(k − j)]!

k−1 X j=0

(x + ıy)2k+1 =

k X j=0

+ı

(−1)j (2k)! x2(k−j)−1 y 2j+1 , (2j + 1)![2(k − j) − 1]!

(−1)j (2k + 1)! x2(k−j)+1 y 2j + (2j)![2(k − j) + 1]!

k X j=0

(−1)j (2k + 1)! x2(k−j) y 2j+1 . (2j + 1)![2(k − j)]!

(a) sin(z): in this case we have that sin(z) =

∞ X (−1)k z 2k+1 , (2k + 1)! k=0

162

SOLUTIONS 10 and hence we have only odd powers of z, and therefore we use the version of the binomial formula odd n, thus obtaining ∞ k X (−1)k X (−1)j (2k + 1)! x2(k−j)+1 y 2j + sin(z) = (2k + 1)! (2j)![2(k − j) + 1]! j=0

k=0

+ı

∞ X k=0

=

k (−1)k X (−1)j (2k + 1)! x2(k−j) y 2j+1 (2k + 1)! (2j + 1)![2(k − j)]!

∞ X k X k=0 j=0

+ı

j=0

(−1)k+j x2(k−j)+1 y 2j + (2j)![2(k − j) + 1]!

∞ X k X k=0 j=0

(−1)k+j x2(k−j) y 2j+1 . (2j + 1)![2(k − j)]!

Let us now consider how to make a change of variables in these sums, in order to switch to using the variable l = k − j, that is, to eliminate k by means of k = l + j, since now k appears only in this particular combination. The problem is to determine the set of the pairs (j, l) which runs through exactly the same set of possibilities as (j, k), with the limits specified for j and k. In order to do this, let us observe the diagram in Figure 10.1, where axes are placed with the values of k, j and l, and where the points that correspond to each term of the double series are marked. The description of this set of points in the (j, k) coordinate system corresponds to going through all the points along vertical segments, ranging from j = 0 to the diagonal line which represents the equation j = k. In this case we can have any value of k ∈ [0, ∞), and values of j ∈ [0, k]. Moreover, the same set of points can be traversed along the horizontal lines, each interpreted as a copy of the l axis, as shown in the figure, where we see that l = k − j. In this case, j can take any value in [0, ∞), and for each value of j we have that l ∈ [0, ∞), so that in the (j, l) coordinate system neither coordinate is limited. It follows that we can write sin(z) =

∞ X ∞ X

(−1)l+2j x2l+1 y 2j + (2j)!(2l + 1)!

l=0 j=0 ∞ X ∞ X

+ı

l=0 j=0

(−1)l+2j x2l y 2j+1 (2j + 1)!(2l)!

163

COMPLEX INEQUALITIES AND SERIES j 7 6 5 4 0

1

2

3

4

5

3

4

5

6

7

k

3

l

2 1 0

1

2

Figure 10.1: Diagram depicting the change of summation variables in the double series, in the case of the sine.  # ∞ ∞ l X X 1 (−1) x2l+1  y 2j  + = (2l + 1)! (2j)! j=0 l=0  # ∞ "∞ X X (−1)l 1 x2l  y 2j+1  , +ı (2l)! (2j + 1)! "

l=0

j=0

where we see that, because of the change of variables, it was possible to separate each double series into the product of two simple series. We can now identify each of the four series as the Taylor series of known functions. Those in which the terms exchange sign represent trigonometric functions, and the others represent hyperbolic functions, so that we have sin(z) = sin(x) cosh(y) + ı cos(x) sinh(y), just as was previously derived by other means, using the Euler formula

164

SOLUTIONS 10 instead of the series representation.

(b) cos(z): in this case we have that

cos(z) =

∞ X (−1)k k=0

(2k)!

z 2k ,

and hence we have only even powers of z, and therefore we use the version of the binomial formula for even n, noting that the second term of this formula exists only for k ≥ 1, and thus obtaining cos(z) =

∞ k X (−1)k X k=0

+ı

(2k)!

k−1 ∞ X (−1)k X k=1

=

∞ X k X k=0 j=0

+ı

j=0

(−1)j (2k)! x2(k−j) y 2j + (2j)![2(k − j)]!

(2k)!

j=0

(−1)j (2k)! x2(k−j)−1 y 2j+1 (2j + 1)![2(k − j) − 1]!

(−1)k+j x2(k−j) y 2j + (2j)![2(k − j)]!

∞ X k−1 X k=1 j=0

(−1)k+j x2(k−j)−1 y 2j+1 . (2j + 1)![2(k − j) − 1]!

In the first term the same transformation of variables that we used before can be used. However, in the second term we have that j ranges from 0 to k − 1, instead of k, and that k ≥ 1. Thus, in this term we make the change of variables l = k − j − 1, that is, k = l + j + 1, which is illustrated by the diagram of Figure 10.2. Since this diagram is similar to the previous one, and considering the new variation intervals which are shown, it is not difficult to see that in this case once more the (j, l) coordinate system variables start at 0 and are not limited. It follows that we can write ∞ X ∞ X (−1)l+2j 2l 2j x y + cos(z) = (2j)!(2l)! l=0 j=0 ∞ X ∞ X

+ı

l=0 j=0

(−1)l+2j+1 x2l+1 y 2j+1 (2j + 1)!(2l + 1)!

165

COMPLEX INEQUALITIES AND SERIES j 7 6 5 4 0

1

2

3

4

5

4

5

6

7

8

k

3

l

2 1 0

1

2

3

Figure 10.2: Diagram depicting the change of summation variables in the double series, in the case of the cosine.

=

"

∞ X (−1)l

# ∞ X x2l 



1 y 2j  + (2l)! (2j)! j=0 l=0  # ∞ "∞ X X (−1)l 1 x2l+1  y 2j+1  , −ı (2l + 1)! (2j + 1)! l=0

j=0

where we see that, because of the change of variables, it was possible to separate each double series into the product of two simple series. We can now identify each of the four series as the Taylor series of known functions. Those in which the terms exchange sign represent trigonometric functions, and the others represent hyperbolic functions, so that we have cos(z) = cos(x) cosh(y) − ı sin(x) sinh(y), just as was previously derived by other means, using the Euler formula

166

SOLUTIONS 10 instead of the series representation.

(c) sinh(z): in this case we have that sinh(z) =

∞ X k=0

1 z 2k+1 , (2k + 1)!

and hence we have only odd powers of z, and therefore we use the version of the binomial formula for odd n, thus obtaining sinh(z) =

∞ X k=0

+ı

k X (−1)j (2k + 1)! 1 x2(k−j)+1 y 2j + (2k + 1)! (2j)![2(k − j) + 1]! j=0

∞ X k=0

=

∞ X k X k=0 j=0

+ı

k X (−1)j (2k + 1)! 1 x2(k−j) y 2j+1 (2k + 1)! (2j + 1)![2(k − j)]! j=0

(−1)j x2(k−j)+1 y 2j + (2j)![2(k − j) + 1]!

∞ X k X k=0 j=0

(−1)j x2(k−j) y 2j+1 . (2j + 1)![2(k − j)]!

In this case the change of variables in each sum is exactly the same as that used in the case of the function sin(z), and therefore it follows that we can write sinh(z) =

∞ X ∞ X

(−1)j x2l+1 y 2j + (2j)!(2l + 1)!

l=0 j=0 ∞ ∞ X X

(−1)j x2l y 2j+1 (2j + 1)!(2l)! l=0 j=0  # ∞ "∞ X X (−1)j 1 = x2l+1  y 2j  + (2l + 1)! (2j)! j=0 l=0  # ∞ "∞ X (−1)j X 1 x2l  y 2j+1  , +ı (2l)! (2j + 1)! +ı

l=0

j=0

where we see that, because of the change of variables, it was possible to separate each double series into the product of two simple series.

COMPLEX INEQUALITIES AND SERIES

167

We can now identify each of the four series as the Taylor series of known functions. Those in which the terms exchange sign represent trigonometric functions, and the others represent hyperbolic functions, so that we have sinh(z) = sinh(x) cos(y) + ı cosh(x) sin(y), just as was previously derived by other means, using the Euler formula instead of the series representation. (d) cosh(z): in this case we have that cosh(z) =

∞ X k=0

1 z 2k , (2k)!

and hence we have only even powers of z, and therefore we use the version of the binomial formula for even n, noting that the second term of this formula exists only for k ≥ 1, and thus getting cosh(z) =

∞ X k=0

+ı

k 1 X (−1)j (2k)! x2(k−j) y 2j + (2k)! (2j)![2(k − j)]! j=0

∞ X k=1

=

k ∞ X X k=0 j=0

+ı

k−1 1 X (−1)j (2k)! x2(k−j)−1 y 2j+1 (2k)! (2j + 1)![2(k − j) − 1]! j=0

(−1)j x2(k−j) y 2j + (2j)![2(k − j)]!

∞ X k−1 X k=1 j=0

(−1)j x2(k−j)−1 y 2j+1 . (2j + 1)![2(k − j) − 1]!

In this case the change of variables in each sum is exactly the same as that used in the case of the function cos(z), and therefore it follows that we can write ∞ X ∞ X (−1)j cosh(z) = x2l y 2j + (2j)!(2l)! l=0 j=0 ∞ X ∞ X

+ı

l=0 j=0

(−1)j x2l+1 y 2j+1 (2j + 1)!(2l + 1)!

168

SOLUTIONS 10  # ∞ ∞ j X X (−1) 2j  1 x2l  y + = (2l)! (2j)! j=0 l=0  "∞ # ∞ j X X 1 (−1) +ı x2l+1  y 2j+1  , (2l + 1)! (2j + 1)! "

l=0

j=0

where we see that, because of the change of variables, it was possible to separate each double series into the product of two simple series. We can now identify each of the four series as the Taylor series of known functions. Those in which the terms exchange sign represent trigonometric functions, and the others represent hyperbolic functions, so that we have cosh(z) = cosh(x) cos(y) + ı sinh(x) sin(y), just as was previously derived by other means, using the Euler formula instead of the series representation.

Solutions 11

Series, Limits and Convergence We present here complete and commented solutions to all problems proposed in Chapter 11 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. Consider an arbitrary series whose terms are all positive real numbers, such as S∞ =

∞ X k=0

|ak |.

Show that, if this series is limited from above, that is, if there is a positive real number SM such that S∞ ≤ SM , then the series necessarily converges to some positive real number less than or equal to SM . Hint: the statement S∞ ≤ SM actually means that all the partial sums SN of this series are limited in this way, for all N . Complete Solution: If we consider the partial sums of the series S∞ , SN =

N X k=0

|ak |,

then the condition of the existence of an upper limit SM is that SN ≤ SM for all N . Thus, if the upper limit exists, then the entire sequence of partial 169

170

SOLUTIONS 11

sums is contained within the closed interval [0, SM ], that is, 0 ≤ SN ≤ SM for all N . We should now consider the special cases in which the limit of the sequence is 0 (which only happens if all terms are zero) or SM . In any of these two cases, the limit exists and hence the convergence of the sequence is established. In any other case, both the complete sequence and its limit, if it exists, must be contained in the open interval (0, SM ). Now, since the series is a sum of positive numbers, when we add more numbers to the sum it can never decrease, so that the sequence of real numbers SN is an increasing sequence, and therefore ordered. Thus, if this sequence goes above the point SM /2 at some stage, then it can no longer be below this point later on. If the limit of the sequence is the point SM /2 then the convergence of the sequence is established. In any other case, if it goes above the point SM /2 then there is a certain value N1 of N such that for N above this value all the rest of the sequence is contained in the interval (SM /2, SM ). We therefore have two remaining possibilities, one and only one of which must be true: either the sequence never goes above the point SM /2, in which case it is all contained in the interval (0, SM /2), or it goes above the point SM /2, in which case the part of the sequence above N1 is all contained in the interval (SM /2, SM ). In any of these two cases, the part of the sequence for N > N1 is always contained in an open interval of length SM /2. This implies, of course, that if there is a limit S∞ then it is also contained in this interval, and that the distance from each point of the sequence to this limit is smaller than SM /2, that is, we have the condition that SM . 2 If we repeat this whole argument, starting from this point, we conclude that either the sequence converges to one end of the intervals involved, or there is a value N2 of N such that N > N1 =⇒ |S∞ − SN |
Nk =⇒ |S∞ − SN | < k . 2 N > N2 =⇒ |S∞ − SN |
Nk(ǫ) =⇒ |S∞ − SN |
0, gn′ (0) = (−1)n−1 (n − 1)!, so that the Maclaurin series of the function is g(z) =

∞ X (−1)n−1

n=1

n

zn,

which is, in fact, exactly the same series that we obtained before. Problem 5. Consider the following differential equation for a real function f (x), with k 6= 0, ∂2 f (x) − k2 f (x) = 0. ∂x2 Assume that f (x) can be faithfully represented by a power series,

197

REPRESENTATION OF FUNCTIONS BY SERIES

f (x) =

∞ X

an xn ,

n=0

substitute this power series in the equation, differentiating it term by term, and manipulate the summation indices in order to write the equation as a certain power series equaled to zero. Determine which should be the relations between the coefficients an of the series above so that it is indeed a solution of the equation. Pay particular attention to the first two coefficients. Determine in this way two different functions that are solutions of the equation and identify these functions. Complete Solution: We will solve, using the representation of functions by power series, the equation ∂2 f (x) − k2 f (x) = 0, ∂x2 where k 6= 0. This technique is known as the Frobenius method, and it aims at finding one or more solutions of the equation that have the property of being regular at the origin, that is, they do not have singularities at this point. Let us assume for f (x) the form of a power series around x = 0, f (x) =

∞ X

an xn ,

n=0

let us assume that this series is convergent in an open neighborhood around x = 0, and therefore uniformly convergent, so that it can be differentiated term-by-term. Substituting the series in the equation we have ∂2 f (x) − k2 f (x) = ∂x2 =

∞ ∞ X ∂2 X n 2 an x n an x − k ∂x2 n=0 n=0

∞ X

n=2

= 0.

n(n − 1)an x

n−2

−k

2

∞ X

an x n

n=0

In order to be able to compare the powers in the two terms, and write the resulting equation as a single power series, let us make a transformation in

198

SOLUTIONS 12

the summation variable in the first term, using m = n − 2, that is, n = m + 2, in order to obtain ∂2 f (x) − k2 f (x) = ∂x2 =

∞ X

(m + 2)(m + 1)am+2 xm − k2

n=0



m=0 ∞ X

= 0.

∞ X

an x n

n=0

 (n + 2)(n + 1)an+2 − k2 an xn

In the resulting equation we have a powers series equaled to zero. Since the series of the identically zero function is the only power series in which all the coefficients are zero, and since there is a bijection between the functions and the power series, the series above must be the one associated with the identically zero function, that is, all its coefficients must be zero. Another way of saying the same thing is to note that the set of powers is a complete basis for the space of all functions f (x) which are regular at the origin, so that the only way in which a linear combination of elements of this basis can be zero is that all its coefficients be zero. One way or another, it follows that (n + 2)(n + 1)an+2 − k2 an = 0

2

⇒

(n + 2)(n + 1)an+2 = k an an+2 =

⇒ k2 an , (n + 2)(n + 1)

for all values of n. What this establishes is what is called a recurrence relation for the coefficients an of the series. Note, however, that the first two coefficients, a0 and a1 , never appear on the left-hand side of this equation, and therefore they are not determined by it. It follows therefore that we have two constants that remain indeterminate, that is, they are arbitrary, as is to be expected for the general solution of an ordinary differential equation of the second order. Writing explicitly the first few equations in this recurrence relation we have a2 = a3 = a4 =

k2 a0 , (2)(1) k2 a1 , (3)(2) k2 a2 , (4)(3)

REPRESENTATION OF FUNCTIONS BY SERIES

199

k2 a3 , (5)(4) k2 a6 = a4 , (6)(5) k2 a7 = a5 , (7)(6) ... = ... . a5 =

Note that each coefficients is related with the one two steps behind, such that a0 determines all the coefficients with even n, and a1 all the coefficients with odd n. We therefore have two separate sequences of equations, each associated with one of the two arbitrary constants. We can solve these steptwo recursion relations, thus finding all coefficients in terms of a0 and a1 , by iterating the relations, that is, by substituting each one in the later ones until all coefficients have been written in terms of the first. Doing this for the even case as an example, we have a2 = a4 = a6 = a8 = ... = a2 = a4 = a6 = a8 = ... = a2 = a4 =

k2 a0 , (2)(1) k2 a2 , (4)(3) k2 a4 , (6)(5) k2 a6 , (8)(7) ... ⇒ k2 a0 , (2)(1) k4 (4)(3)(2)(1) k4 (6)(5)(4)(3) k4 (8)(7)(6)(5) ... ⇒ k2 a0 , (2)(1) k4 (4)(3)(2)(1)

a0 , a2 , a4 ,

a0 ,

200

SOLUTIONS 12 k6 a0 , (6)(5)(4)(3)(2)(1) k6 a2 , (8)(7)(6)(5)(4)(3) ... ⇒ k2 a0 , (2)(1) k4 a0 , (4)(3)(2)(1) k6 a0 , (6)(5)(4)(3)(2)(1) k8 a0 , (8)(7)(6)(5)(4)(3)(2)(1) ... .

a6 = a8 = ... = a2 = a4 = a6 = a8 = ... =

Therefore, it is easy to induce the general solution, which is, for n = 2j, k2j a0 , (2j)!

a2j =

thereby determining all coefficients with even n in terms of a0 . In an entirely analogous way, we have for the odd case a2j+1 =

k2j a1 . (2j + 1)!

It follows that we obtained in this way two independent solutions which can now be written explicitly in terms of power series. In order to generate one of the two solutions, we choose a1 = 0 and keep a0 arbitrary, which makes all the coefficients with odd n equal to zero, thereby generating a series with only even powers, and thus an even solution. In order to generate another solution we choose a0 = 0 and keep a1 arbitrarily, thereby generating both an odd series and an odd solution, f1 (x) =

∞ X k2j a0 x2j , (2j)! j=0

f2 (x) =

∞ X j=0

f1 (x) = a0

k2j a1 x2j+1 (2j + 1)!

∞ X j=0

1 (kx)2j , (2j)!

⇒

REPRESENTATION OF FUNCTIONS BY SERIES

f2 (x) =

201

∞ 1 a1 X (kx)2j+1 , k (2j + 1)! j=0

where we can divide by k so long as this constant is not zero, as is explicited in the proposition of the problem. We now recognize the two resulting power series as the Taylor series of the functions cosh(kx) and sinh(kx), respectively. It follows that the general solution of the equation is given by f (x) = α cosh(kx) + β sinh(kx), where α and β are two arbitrary real constants. Note that, although this problem was formulated in the real context, since we are dealing with power series, all that we did here holds in exactly the same way in the complex context. Problem 6. (Challenge Problem) Consider the question of the calculation of the sucessive derivatives of the product of two analytic functions, f1 (z) and f2 (z), that is, derivatives of f (z) = f1 (z)f2 (z), where f1 (z) and f2 (z) are known only in terms of their expressions in series around a reference point z0 . (a) Using the Leibniz rule, write the first, second and third derivatives of f (z) in terms of the derivatives of f1 (z) and f2 (z). (b) From this kind of experimentation, induce a general formula for the n-th derivative of f (z), in terms of a sum involving the appropriate combinatorial factors. (c) Prove by finite induction the formula that was induced, that is, assume that the formula is true for the case n − 1 and show that, as a consequence of this, the case n also holds. (d) Consider now the product f (z) of the two functions, represented by the product of the Taylor series of f1 (z) and of f2 (z). Collect the terms with a definite power n of z − z0 and write a general formula for the coefficient of the power n. (e) Use the results to prove that the Taylor series of the function f (z) that is the product of f1 (z) and f2 (z) is obtained as the product of the two series.

202

SOLUTIONS 12

Complete Solution: Let us consider that f (z) = f1 (z)f2 (z), where each of these functions may be represented by a Taylor series around a certain point z0 , f (z) = f1 (z) = f2 (z) =

∞ X f n′ (z0 )

n=0 ∞ X

n=0 ∞ X

n!

(z − z0 )n ,

f1n′ (z0 ) (z − z0 )n , n!

f2n′ (z0 ) (z − z0 )n . n! n=0

(a) Let us calculate the first few derivatives of the product f (z), using the Leibniz rule repeatedly. By doing this and using primes to indicate the derivatives with respect to z, we obtain f 0′ (z) = f10′ (z)f20′ (z), f 1′ (z) = f11′ (z)f20′ (z) + f10′ (z)f21′ (z), f 2′ (z) = f12′ (z)f20′ (z) + 2f11′ (z)f21′ (z) + f10′ (z)f22′ (z), f 3′ (z) = f13′ (z)f20′ (z) + 2f12′ (z)f21′ (z) + 2f11′ (z)f22′ (z) + f10′ (z)f23′ (z), f 4′ (z) = f14′ (z)f20′ (z) + 3f13′ (z)f21′ (z) + 6f12′ (z)f22′ (z) + +3f11′ (z)f23′ (z) + f10′ (z)f24′ (z), ...

... .

As one can see, in general the result is a sum of bilinear terms in derivatives of f1 (z) and f2 (z), and the orders of the derivatives have a structure similar to that of the powers in the Newton binomial formula. The coefficients are also those that appear in the binomial expansion. (b) Assuming that the structure is in fact the same as that of the formula of the binomial expansion, we can induce that the general formula is n′

f (z) =

n X i=0

n! (n−i)′ f (z)f2i′ (z). i!(n − i)! 1

(c) The formula proposed above reduces, in the case n = 0, simply to f (z) = f1 (z)f2 (z), and in the case n = 1 to

REPRESENTATION OF FUNCTIONS BY SERIES

203

f ′ (z) = f1′ (z)f2 (z) + f1 (z)f2′ (z), which we already know to be true. Let us assume now that this formula applies to the case n − 1, f (n−1)′ (z) =

n−1 X i=0

(n − 1)! (n−1−i)′ f (z)f2i′ (z), i!(n − 1 − i)! 1

and simply take an additional derivative on each side, using for this the Leibniz rule once again. Doing this we have f n′ (z) = =

n−1 X

i=0 n−1 X

i (n − 1)! h (n−i)′ (n−1−i)′ (i+1)′ f1 (z)f2i′ (z) + f1 (z)f2 (z) i!(n − 1 − i)! (n − 1)! (n−i)′ f1 (z)f2i′ (z) + i!(n − 1 − i)!

i=0 n−1 X

+ =

i=0 n−1 X i=0

+

(n − 1)! (n−i)′ f1 (z)f2i′ (z) + i!(n − 1 − i)!

n X j=1

=

(n − 1)! (n−1−i)′ (i+1)′ f (z)f2 (z) i!(n − 1 − i)! 1

(n − 1)! (n−j)′ f (z)f2j′ (z) (j − 1)!(n − j)! 1

f1n′ (z)f20′ (z)

+

n−1 X i=1

+

n−1 X i=1

(n − 1)! (n−i)′ f1 (z)f2i′ (z) + i!(n − 1 − i)!

(n − 1)! (n−i)′ f (z)f2i′ (z) + f10′ (z)f2n′ (z) (i − 1)!(n − i)! 1

= f1n′ (z)f20′ (z) + f10′ (z)f2n′ (z) +  n−1 X  (n − 1)! (n − 1)! (n−i)′ + + f (z)f2i′ (z) i!(n − i − 1)! (i − 1)!(n − i)! 1 =

i=1 n′ f1 (z)f20′ (z) n−1 X

+

i=1

+ f10′ (z)f2n′ (z) +

 (n − 1)!i (n−i)′ (n − 1)!(n − i) + f (z)f2i′ (z) i!(n − i)! i!(n − i)! 1

204

SOLUTIONS 12 = f1n′ (z)f20′ (z) + f10′ (z)f2n′ (z) + n−1 X (n − 1)!(n − i + i) (n−i)′ + f1 (z)f2i′ (z) i!(n − i)! i=1

= f1n′ (z)f20′ (z) +

n−1 X i=1

=

n X i=0

n! (n−i)′ f (z)f2i′ (z) + f10′ (z)f2n′ (z) i!(n − i)! 1

n! (n−i)′ f (z)f2i′ (z), i!(n − i)! 1

where we made the transformation of variables j = i + 1, i = j − 1 in one of the sums, subsequently returning the variable name to i. Thus we see that the case n − 1 does in fact imply the case n, and therefore that it is true in general, by finite induction, that f n′ (z) =

n X i=0

n! (n−i)′ f (z)f2i′ (z). i!(n − i)! 1

(d) Representing each of the two factor-functions of the product by its series, we have for the product-function f (z) = f1 (z)f2 (z) ∞ ∞ X X f2n2 ′ (z0 ) f1n1 ′ (z0 ) (z − z0 )n1 (z − z0 )n2 = n ! n ! 1 2 n =0 n =0 1

=

2

∞ X ∞ X f1n1 ′ (z0 )f2n2 ′ (z0 ) (z − z0 )n1 +n2 . n1 !n2 !

n1 =0 n2 =0

It is now necessary to collect all the terms in which the total power n1 + n2 is a constant, n1 + n2 = n, where n is a new variable, which clearly varies from 0 to ∞, being equal to 0 only if n1 = 0 = n2 . Given a certain constant value for n, the smallest possible value for n1 is 0, in which case n2 = n, and the highest possible value for n1 is n, in which case n2 = 0. Thus, given a constant value for n, there are n + 1 possible pairs of values for n1 and n2 . We can index these values by the variable n2 , which we now call i, and which varies from 0 to n. In this case, we have for n1 the corresponding values n − i. Thus, the sums can now be described by the pair of variables i and n, instead of n1 and n2 . We therefore have, changing n2 to i, n1 to n − i and n1 + n2 to n,

REPRESENTATION OF FUNCTIONS BY SERIES

f (z) =

∞ X n (n−i)′ X f (z0 )f i′ (z0 ) 2

1

(n − i)!i!

n=0 i=0

205

(z − z0 )n .

(e) We can write the result of the previous item as follows, " n # ∞ X X n! 1 (n−i)′ n i′ (z − z0 ) f (z0 )f2 (z0 ) , f (z) = n! i!(n − i)! 1 n=0

i=0

where we multiplied and divided by n! and moved out of the sum on i some quantities that do not depend on i. We recognize now in the second sum the n-th derivative of the product-function f (z), as it was previously derived, applied at the point z0 . We therefore have the Taylor series of f (z), f (z) =

∞ X f (n)′ (z0 ) n=0

n!

(z − z0 )n ,

which proves that the product of the two Taylor series, that of f1 (z) and that of f2 (z), in fact produces the Taylor series of f (z), which has its coefficients related to those of the two factor-series by the Leibniz rule. Problem 7. (Challenge Problem) Assuming that f2 (z) does not have any zeros in the common region of analyticity of f1 (z) and f2 (z), initially show that if f (z) is the ratio of the two analytic functions f1 (z) and f2 (z), f (z) =

f1 (z) , f2 (z)

then the Taylor series S(z) of f (z) is such that S2 (z)S(z) = S1 (z), where S1 (z) is the Taylor series of f1 (z) and S2 (z) is the Taylor series of f2 (z), all with respect to the same reference point z0 . Prove this fact without using the explicit expression of the coefficients of S(z) in terms of the coefficients S1 (z) and S2 (z). Hint: use the result for the product of two analytic functions, which was proved before.

206

SOLUTIONS 12

Then consider the question of the calculation of the derivatives of f (z) where f1 (z) and f2 (z) are known only in terms of their expressions in series. The idea is to try to repeat the inductive-deductive scheme that was used in the corresponding deduction for the product of two analytic functions, and to use the result to explicitly show how the Taylor series of f (z) is in fact obtained from the Taylor series of f1 (z) and f2 (z). Hint: consider simplifying the problem by first trying to solve the case where f1 (z) = 1, that is, first try to solve the problem of finding the Taylor series of the multiplicative inverse of a function, f (z) =

1 , f2 (z)

which naturally means that f (z)f2 (z) = 1. After that one may consider combining this solution with that of the problem of the product, which was solved before. Warning: this problem has a very difficult combinatorial part and, for the time being, a truly complete solution is not available. Complete Solution: (A bit incomplete) Let us now consider the problem of the series in the case of the division, where we have the series of f1 (z) and f2 (z), and we want to determine the ratio of the two series, assuming that f2 (z) has no zeros within the region of interest, f (z) =

f1 (z) . f2 (z)

One objective of this problem is to show that the series which results from the division of the series of f1 (z) and f2 (z) is in fact the Taylor series of the ratio-function f (z). This is a fact that we can prove based on what we already know about the case of the multiplication. Let S(z) be the series of f (z), S1 (z) the series of f1 (z) and S2 (z) the series of f2 (z). Note that, by the definition of the division, determining the ratio-series S(z) =

S1 (z) S2 (z)

REPRESENTATION OF FUNCTIONS BY SERIES

207

means determining, given the series S1 (z) and S2 (z), a series S(z) such that S(z)S2 (z) = S1 (z), where the multiplication of the series should be done by collecting powers, as we did in the case of the multiplication. From the case of the multiplication we already know that the fact that f (z)f2 (z) = f1 (z) is equivalent to the fact that S(z)S2 (z) = S1 (z). Hence, we can elaborate a proof by reductio ad absurdum. If there are functions f (z), f1 (z) and f2 (z), each with its Taylor series, S(z), S1 (z) and S2 (z), such that S(z) 6=

S1 (z) , S2 (z)

it follows, from the definition of the division, that S(z) does not satisfy the corresponding product, namely that S(z)S2 (z) 6= S1 (z). However, for the corresponding functions it is certainly true that f1 (z) f2 (z) f (z)f2 (z) = f1 (z), f (z) =

⇒

which, as we know from the case of the multiplication, implies that S(z)S2 (z) = S1 (z), contradicting the previous conclusion. It follows that there can be no functions f (z), f1 (z) and f2 (z), each with its Taylor series, S(z), S1 (z) and S2 (z), such that f (z) =

f1 (z) , f2 (z)

S(z) 6=

S1 (z) . S2 (z)

while

It follows therefore, by reductio ad absurdum, that it is always true that

208

SOLUTIONS 12 f (z) = S(z) =

f1 (z) ⇒ f2 (z) S1 (z) , S2 (z)

that is, that the power series obtained by the ratio of two Taylor series is always the Taylor series of the ratio-function. The other objective of the problem is the determination of the coefficients of the ratio-series S(z) in terms of the coefficients of the series S1 (z) and S2 (z). This is a considerably more complex combinatorial problem than that of the case of the multiplication. Given the above result, we can determine these coefficients in two ways: on the one hand, we can simply take directly multiple derivatives of the function f (z) =

f1 (z) , f2 (z)

subsequently applying the results at z0 in order to obtain the coefficients of S(z); on the other hand, we can start from S1 (z) and S2 (z), and solve the problem of dividing one series by the other, thereby determining the coefficients of S(z). In what follows we will explore this second possibility, for two reasons: first, because it is the less immediate and familiar alternative; second, because there is the possibility that it is in fact the simplest one. If we consider the function h(z) which is the multiplicative inverse of f2 (z), we can reduce this problem to an earlier problem, that of the multiplication of two series, since 1 ⇒ f2 (z) f (z) = f1 (z)h(z). h(z) =

Since this inversion is a particular case of the division, we already know that the series obtained by the inversion of the series of f2 (z) is in fact the Taylor series of h(z). In addition, we already know the formula that gives us the coefficients of the product-series. Let us therefore focus our efforts on explicitly getting the coefficients of the series of h(z), that is, on the inversion problem. If we have the series h(z) = f2 (z) =

∞ X hn′ (z0 )

n=0 ∞ X

n!

(z − z0 )n ,

f2n′ (z0 ) (z − z0 )n , n! n=0

REPRESENTATION OF FUNCTIONS BY SERIES

209

the first of which is our unknown, while the second is known, we can write the expression h(z)f2 (z) = 1 as follows, using the series to represent the functions, 1 = h(z)f2 (z) ∞ ∞ X X f2n2′ (z0 ) hn0 ′ (z0 ) n0 (z − z0 ) (z − z0 )n2 = n0 ! n2 ! =

n0 =0 ∞ X ∞ X

n0 =0 n2 =0

n2 =0

hn0 ′ (z0 )f2n2 ′ (z0 ) n0 !n2 !

(z − z0 )n0 +n2 .

Just as we did before in the case of the product of two series, we can now collect terms with constant n0 + n2 . Thus, the sums can now be described by the pair of variables i = n2 and n = n0 + n2 , instead of n0 and n2 . We therefore have, changing n2 to i, n0 to n − i and n0 + n2 to n, ∞ X n X h(n−i)′ (z0 )f i′ (z0 )

2 (z − z0 )n (n − i)!i! n=0 i=0 " n # ∞ X X n! 1 (z − z0 )n h(n−i)′ (z0 )f2i′ (z0 ) . = n! i!(n − i)! n=0

1 =

i=0

Comparing powers of (z −z0 ) on both sides of this equation, we conclude that for n = 0 this implies simply that 1 = h(z0 )f2 (z0 ), that is, h(z0 ) = 1/f2 (z0 ). For all other cases we have that the coefficients of the series on the right-hand side of the expression above must vanish, that is, we have that n X i=0

n! h(n−i)′ (z0 )f2i′ (z0 ) = 0, i!(n − i)!

for n ≥ 1. This is an infinite set of equations, determining the derivatives of h(z) at the point z0 in terms of the derivatives of f2 (z) at the same point. However, note that the unknowns, that is, the various derivatives of h(z), of orders 0 to n, applied at the point z0 , are mixed in these equations. In order to solve these equations for h(n−i)′ (z0 ), it is necessary to proceed iteratively using, to start the process, the already known case n = 0. It is because of this fact that this problem is considerably more complex than the corresponding problem in the case of the multiplication. Writing the first few cases we have h1′ (z0 )f20′ (z0 ) + h0′ (z0 )f21′ (z0 ) = 0, h2′ (z0 )f20′ (z0 ) + 2h1′ (z0 )f21′ (z0 ) + h0′ (z0 )f22′ (z0 ) = 0,

210

SOLUTIONS 12 h3′ (z0 )f20′ (z0 ) + 3h2′ (z0 )f21′ (z0 ) + 3h1′ (z0 )f22′ (z0 )+ +h0′ (z0 )f23′ (z0 ) = 0, h4′ (z0 )f20′ (z0 ) + 4h3′ (z0 )f21′ (z0 ) + 6h2′ (z0 )f22′ (z0 )+ +4h1′ (z0 )f23′ (z0 ) + h0′ (z0 )f24′ (z0 ) = 0, ...

... ,

where from now on, in order to simplify the expressions, we will cease to indicate explicitly that all functions and derivatives are applied at z0 , making this fact implicit. Isolating the derivative of h(z) of highest order in each case, and including again the case n = 0 we have h0′ = h1′ = h2′ = h3′ = h4′ = ...

1 , f2 1 − f2 1 − f2 1 − f2 1 − f2 ... .



 h0′ f21′ ,



 3h2′ f21′ + 3h1′ f22′ + h0′ f23′ ,

 

 2h1′ f21′ + h0′ f22′ ,

 4h3′ f21′ + 6h2′ f22′ + 4h1′ f23′ + h0′ f24′ ,

Substituting now the known value h0′ everywhere, we have h0′ = h1′ = h2′ = h3′ = h4′ = ...

1 , f2 1 − 2 f21′ , f2   1 1 2′ − f2 , 2h1′ f21′ + f2 f2   1 1 3′ 2′ 1′ 1′ 2′ − f , 3h f2 + 3h f2 + f2 f2 2   1 1 4′ − f2 , 4h3′ f21′ + 6h2′ f22′ + 4h1′ f23′ + f2 f2 ... ,

where we now have the value of h1′ determined. Substituting this value everywhere we have

REPRESENTATION OF FUNCTIONS BY SERIES h0′ = h1′ = h2′ = h3′ = h4′ = ...

211

1 , f2 1 − 2 f21′ , f2 1 1  1′ 2 2 3 f2 − 2 f22′ , f2 f2   1 3′ 1 1 1′ 2′ 2′ 1′ − f , 3h f2 − 3 2 f2 f2 + f2 f2 2 f2   1 4′ 1 1′ 3′ 1 3′ 1′ 2′ 2′ f , 4h f2 + 6h f2 − 4 2 f2 f2 + − f2 f2 2 f2 ... ,

where we now have the value of h2′ determined. Substituting this value everywhere we have h0′ = h1′ = h2′ = h3′ = h4′ =

...

1 , f2 1 − 2 f21′ , f2 1 1  2 2 3 f21′ − 2 f22′ , f2 f2 1  3 1 1 −6 4 f21′ + 6 3 f22′ f21′ − 2 f23′ , f2 f2 f2    1 1 1  2 2 4h3′ f21′ + 12 3 f21′ f22′ − 6 2 f22′ + − f2 f2 f2  1 4′ 1 1′ 3′ f , − 4 2 f2 f2 + f2 2 f2 ... ,

where we now have the value of h3′ determined. Substituting this value everywhere we have 1 , f2 1 = − 2 f21′ , f2 1 1  1′ 2 = 2 3 f2 − 2 f22′ , f2 f2   1 1 1 3 = −6 4 f21′ + 6 3 f22′ f21′ − 2 f23′ , f2 f2 f2

h0′ = h1′ h2′ h3′

212

SOLUTIONS 12 1  1′ 2 2′ 1 1  1′ 4 f f2 f2 + 8 3 f21′ f23′ + − 36 2 5 4 f2 f2 f2 1 1  2′ 2 +6 3 f2 − 2 f24′ , f2 f2 ... ,

h4′ = 24

...

where we now have the value of h4′ determined. The problem that remains open is to systematize and generalize this type of procedure to the general case.

Solutions 13

Convergence Criteria and Proofs We present here complete and commented solutions to all problems proposed in Chapter 13 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. Use the ratio-limit test in order to show that the Maclaurin series of each of the following functions is convergent, and determine in each case the radius of the convergence disk. (a) w(z) = cos(z). (b) w(z) = sin(z). (c) w(z) = cosh(z). (d) w(z) = sinh(z). (e) w(z) = ln(1 + z). Complete Solution: (a) The Maclaurin series of w(z) = cos(z) is given by cos(z) =

∞ X (−1)k k=0

213

(2k)!

z 2k .

214

SOLUTIONS 13 Taking the ratio of the absolute values of two successive terms we have |tk+1 | |tk |

= =

(2k)!|z|2k+2 (2k + 2)!|z|2k |z|2 . (2k + 2)(2k + 1)

Given an arbitrary finite z, this ratio goes to zero in the k → ∞ limit, and is therefore smaller than 1. It follows that the series converges on the whole complex plane, and that the radius of the convergence disk centered at z0 = 0 is infinite. (b) The Maclaurin series of w(z) = sin(z) is given by sin(z) =

∞ X (−1)k z 2k+1 . (2k + 1)! k=0

Taking the ratio of the absolute values of two successive terms we have |tk+1 | |tk |

= =

(2k + 1)!|z|2k+3 (2k + 3)!|z|2k+1 |z|2 . (2k + 3)(2k + 2)

Given an arbitrary finite z, this ratio goes to zero in the k → ∞ limit, and is therefore smaller than 1. It follows that the series converges on the whole complex plane, and that the radius of the convergence disk centered at z0 = 0 is infinite. (c) The Maclaurin series of w(z) = cosh(z) is given by cosh(z) =

∞ X k=0

1 z 2k . (2k)!

Taking the ratio of the absolute values of two successive terms we have |tk+1 | |tk |

= =

(2k)!|z|2k+2 (2k + 2)!|z|2k |z|2 . (2k + 2)(2k + 1)

215

CONVERGENCE CRITERIA AND PROOFS

Given an arbitrary finite z, this ratio goes to zero in the k → ∞ limit, and is therefore smaller than 1. It follows that the series converges on the whole complex plane, and that the radius of the convergence disk centered at z0 = 0 is infinite. (d) The Maclaurin series of w(z) = sinh(z) is given by sin(z) =

∞ X

1 z 2k+1 . (2k + 1)!

k=0

Taking the ratio of the absolute values of two successive terms we have |tk+1 | |tk |

(2k + 1)!|z|2k+3 (2k + 3)!|z|2k+1 |z|2 . (2k + 3)(2k + 2)

= =

Given an arbitrary finite z, this ratio goes to zero in the k → ∞ limit, and is therefore smaller than 1. It follows that the series converges on the whole complex plane, and that the radius of the convergence disk centered at z0 = 0 is infinite. (e) The Maclaurin series of w(z) = ln(1 + z) is given by ln(1 + z) =

∞ X (−1)n+1

n=1

n

zn .

Taking the ratio of the absolute values of two successive terms we have |tk+1 | |tk |

= =

n|z|n+1 (n + 1)|z|n n |z|. n+1

The ratio involving n goes to 1 from below in the n → ∞ limit. It follows that the series converges provided that |z| < 1. We have therefore that the radius of the convergence disk centered at z0 = 0 is equal to 1, which is consistent with the fact that the function has a singularity at z = −1.

216

SOLUTIONS 13

Problem 2.

Consider the series of positive real numbers S∞ =

∞ X

n=0

|an |,

where an are complex numbers. (a) Write in symbolic mathematical language, involving a positive real number ǫ, the convergence condition of the series. (b) Write in symbolic mathematical language the condition that the limit of |an | be zero when n goes to infinity. (c) Show that if the series converges, then we do have this value for the limit, that is, lim |an | = 0.

n→∞

Hint: one can try using the method of reductio ad absurdum, but this time a constructive proof is simpler. Complete Solution: (a) If we consider the partial sums of the series,

SN =

N X

n=0

|an |,

then we have that the series is given by the limit S∞ = lim SN . N →∞

The convergence condition is therefore the expression of the existence of this limit, that is, it is the condition that, given any real number ǫ1 > 0, there is a value N (ǫ1 ) of N such that N > N (ǫ1 ) =⇒ |S∞ − SN | < ǫ1 .

CONVERGENCE CRITERIA AND PROOFS

217

(b) The condition that |an | should go to zero for large values of n can be expressed as the limit lim |an | = 0.

n→∞

The expression of the existence of this limit is the condition that, given any real number ǫ2 > 0, there is a value n(ǫ2 ) of n such that n > n(ǫ2 ) =⇒ |an | < ǫ2 . (c) If we consider that the difference of two successive partial sums of the series, SN +1 −SN , is exactly the term |aN +1 | of the series, and therefore is positive, we can write that |aN +1 | = |SN +1 − SN |

= |(SN +1 − S∞ ) − (SN − S∞ )|

≤ |SN +1 − S∞ | + |SN − S∞ |,

where we used the triangle inequality. Let us now assume that the series is convergent and show that this implies that the N → ∞ limit of |aN | is zero. Thus, we are given a strictly positive real number ǫ2 , and we need to show that there is an N (ǫ2 ) that satisfies the condition related to this limit. Let us start by choosing for ǫ1 the value ǫ2 /2. Since the series converges, for any value of ǫ1 , including this one, there is an N (ǫ1 ) such that N > N (ǫ1 ) implies that |SN − S∞ | < ǫ1 . Since this holds for any N > N (ǫ1 ), it also applies to the value N + 1, so that we have |SN − S∞ | < ǫ1 ,

|SN +1 − S∞ | < ǫ1 ,

and adding these two inequalities we therefore obtain the inequality |SN +1 − S∞ | + |SN − S∞ | < 2ǫ1 .

218

SOLUTIONS 13 As we saw earlier, the left-hand side of this inequality is greater than or equal to |aN +1 |, while 2ǫ1 = ǫ2 , so that we can write |aN +1 | ≤ |SN +1 − S∞ | + |SN − S∞ | < ǫ2 . This is true for any value of N larger than the value N (ǫ1 ) which we know to exist due to the convergence of the series. Since N > N (ǫ1 ) implies that N + 1 > N (ǫ1 ) + 1, we have that the above inequality is valid for any value of N that is greater than N (ǫ2 ) = N (ǫ1 ) + 1. In summary, we have shown that, given a strictly positive real number ǫ2 , there is a value of N (ǫ2 ), the value N (ǫ1 ) + 1, such that N > N (ǫ2 ) =⇒ |aN | < ǫ2 , as we wanted to prove. Note that this proof works equally well for the series without the absolute values, that is, for the series given by the partial sums ′ SN =

N X

an .

n=0 ′ ′ This is true because in this case we have that SN +1 − SN = aN +1 , and if we take absolute values on both sides we therefore have that ′ ′ |SN +1 − SN | = |aN +1 |, which is exactly the relation we need to use for the proof. The rest of the proof proceeds without any modification.

Problem 3.

Consider the series of positive real numbers S∞ =

∞ X

n=0

|an |,

where an are complex numbers. (a) Write in symbolic mathematical language, involving a positive real number ǫ, the convergence condition of the series. (b) Show that if the series converges, then there is a real number A such that |an | < A, for all n. Hint: use the method of reductio ad absurdum.

CONVERGENCE CRITERIA AND PROOFS

219

Complete Solution: (a) If we consider the partial sums of the series,

SN =

N X

n=0

|an |,

then we have that the series is given by the limit S∞ = lim SN . N →∞

The convergence condition is therefore the expression of the existence of this limit, that is, it is the condition that, given any real number ǫ > 0, there is a value N (ǫ) of N such that N > N (ǫ) =⇒ |S∞ − SN | < ǫ. (b) Assuming that the series converges, let us consider the negation of the thesis, that is, let us assume that there is no real number A such that |an | < A, for all n. It follows that, given an arbitrary strictly positive real number ǫ, there is at least one term N (ǫ) of the series such that aN (ǫ) ≥ ǫ,

because otherwise there would be a real number A, the number A = ǫ, such that |an | < A, for all n. Thus, given the number 1, there is a term N (1) such that aN (1) ≥ 1,

given the number 2, there is a term N (2) such that aN (2) ≥ 2,

and so on for all strictly positive integers n. Thus, there is an infinite subset of terms of the series that has the property that

220

SOLUTIONS 13

aN (n) ≥ n,

for n ∈ {1, 2, 3, . . .}, and adding the two sides of this inequality for n going from 1 to N , we have that N N X X aN (n) ≥ n.

n=1

n=1

The sum on the right-hand side of this inequality is the sum of an arithmetic progression which has a well-known value, N X aN (n) ≥ N (N + 1) . 2 n=1

Since the limit of this expression for N → ∞ is infinite, we see that there is a subset of terms of the series whose sum diverges to infinity. Since all terms of the series are positive, this sum of part of the terms is certainly less than or equal to the sum of the whole series, and it follows therefore that the complete series diverges to infinity, which contradicts the hypothesis. Thus, it cannot be true that there is no real number A such that |an | < A, for all n, and it thus follows that there is such a number A. To make this argument a little more precise, given the partial sum N X aN (n) ,

n=1

which has a finite number of terms, we can consider the maximum among the integers N (n), for n = 1, . . . , N , which we will denote as NM . Consider the partial sum SNM of the series. Since it is a sum of terms in ascending order, it follows that it contains all the terms of the sum above. Since all terms are positive, it is therefore greater than or equal to the sum above, that is, we can write that S NM

N X aN (n) ≥ N (N + 1) . ≥ 2 n=1

221

CONVERGENCE CRITERIA AND PROOFS

Thus, given an arbitrary integer N , one can always find a partial sum SNM of the series, where NM is some function of N , such that S NM ≥

N (N + 1) . 2

That is, the partial sums of the series grow without limit and therefore the series does not converge. Note that because all terms of the series are positive, it follows that NM increases with N . That is, as we increase N , the corresponding partial sums SNM are progressively further along into the sequence of partial sums. Problem 4.

Consider the Maclaurin series of the complex function p w(z) = z + ǫ2 ,

where ǫ is a strictly positive real number.

(a) Write the series as an infinite sum of a general term, involving factorials, double factorials and powers. (b) Use the ratio-limit test in order to establish the convergence of the series, and calculate the radius of the convergence disk. (c) Consider the limit in which ǫ → 0. What is the convergence domain of the series in this case? Complete Solution: (a) Taking the first few successive derivatives of w(z) = plying at z0 = 0, we obtain w0′ (0) = +ǫ 11 w1′ (0) = + 2 ǫ 11 1 2′ w (0) = − 2 2 ǫ3 113 1 w3′ (0) = + 2 2 2 ǫ5 1135 1 w4′ (0) = − 2 2 2 2 ǫ7

√ z + ǫ2 and ap-

222

SOLUTIONS 13 11357 1 2 2 2 2 2 ǫ9 113579 1 w6′ (0) = − 2 2 2 2 2 2 ǫ11 ... ... .

w5′ (0) = +

Systematizing these results we can induce the general term of the series, and write therefore that ∞ X zk (−1)k+1 (2k + 1)!! . w(z) = 2k (2k + 1)(2k − 1)k! ǫ2k−1 k=0

It is not difficult to verify that this general expression reproduces all the particular cases listed above. (b) Calculating the ratio of the absolute values of two successive terms we have |tk+1 | |tk |

= = = =

|z|k+1 ǫ2k−1 2k (2k + 3)!!(2k + 1)(2k − 1)k! 2k+1 (2k + 1)!!(2k + 3)(2k + 1)(k + 1)! |z|k ǫ2k+1 (2k + 3)!!(2k − 1)k! |z| 2(2k + 3)!!(k + 1)! ǫ2 (2k − 1) |z| 2(k + 1) ǫ2 k − 1/2 |z| . k + 1 ǫ2

The k → ∞ limit of the ratio involving k is 1, this limit being reached from below. It follows that the k → ∞ limit of the ratio of successive terms is less than 1, so that the series is convergent, provided that |z| < ǫ2 . Thus, the convergence disk centered at z0 = 0 has radius equal to ǫ2 . (c) In the limit ǫ → 0 the convergence disk tends to have zero radius and hence reduces to a single point, the point z = 0. In this case the series converges at a single point, its point of reference z0 = 0, despite the fact that the function has a singularity at that point.

Solutions 14

Laurent Series and Residues We present here complete and commented solutions to all problems proposed in Chapter 14 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. value of z0 .

Consider the function f (z) = cos(z)/(z − z0 ), for an arbitrary

(a) Write the Taylor series of cos(z) around z0 . (b) Write the Laurent series of f (z) around z0 . (c) Determine the radii of the convergence ring of this series. (d) Determine the value of the residue of f (z) at the point z0 . Answer: cos(z0 ). Complete Solution: (a) The derivatives of even order of the function cos(z) are the cos(z) itself with alternating signs, and those of odd order are sin(z), also with alternating signs. Therefore, we have for the Taylor series of cos(z) around z0 cos(z) =

∞ X (−1)k cos(z0 ) k=0

(2k)!

223

(z − z0 )2k +

224

SOLUTIONS 14

−

∞ X (−1)k sin(z0 ) k=0

= cos(z0 )

(2k + 1)!

∞ X (−1)k

(2k)!

k=0 ∞ X

− sin(z0 )

k=0

(z − z0 )2k+1

(z − z0 )2k +

(−1)k (z − z0 )2k+1 . (2k + 1)!

Note that another way to obtain this result is to use the trigonometric identity relative to the cosine of the sum of two angles, cos(z) = cos(z0 + z − z0 )

= cos(z0 ) cos(z − z0 ) − sin(z0 ) sin(z − z0 ),

using then the usual Maclaurin series of cos(z) and sin(z), applied at the point z − z0 . (b) Using this series expansion in the formula of f (z), and separating the first two terms of the sums we have

f (z) =

∞

X (−1)k cos(z0 ) − sin(z0 ) + cos(z0 ) (z − z0 )2k−1 + z − z0 (2k)! k=1 ∞ X

− sin(z0 )

k=1

(−1)k (z − z0 )2k . (2k + 1)!

This is the Laurent series of the function f (z) around z0 . (c) Since cos(z) is analytic throughout the complex plane, the only singularity of f (z) is the simple pole at the point z0 . It follows that the convergence ring of the Laurent series is the whole complex plane except for the point z0 . We have therefore that the inner radius of the convergence ring is zero and that the outer radius is infinite. (d) The residue of f (z) at the point z0 is the coefficient of the term of the Laurent series that contains the power (z − z0 )−1 , and hence the value of the residue is cos(z0 ).

225

LAURENT SERIES AND RESIDUES

Problem 2. Consider the function f (z) = ln(z)/(z − z0 ), for a z0 whose real part is strictly positive. Consider just the n = 0 leaf of the Riemann surface of the logarithm. (a) Write the Taylor series of ln(z) around z0 . (b) Write the Laurent series of f (z) around z0 . (c) Determine the radii of the convergence ring of this series. (d) Determine the value of the residue of f (z) at the point z0 . Answer: ln(z0 ). Complete Solution: (a) Since the derivatives of the logarithm function always have the same value on all the leaves of the Riemann surface, the leaf only needs to be chosen in fact for the first term of the series. Since only the n = 0 leaf should be considered, the first term is simply ln(z0 ). The derivatives of the function ln(z) are negative powers with alternating signs and factorials, so that we have for the Taylor series of ln(z) around z0 ln(z) = ln(z0 ) +

∞ X (−1)k+1 k=1

kz0k

(z − z0 )k .

Observe how for z0 = 1 this reduces to the usual and more familiar series of the logarithm. (b) Using this series expansion in the formula of f (z), and separating the first two terms of the sum, we have ∞

X (−1)k+1 ln(z0 ) 1 (z − z0 )k−1 . f (z) = + + z − z0 z0 kz0k k=2

This is the Laurent series of the function f (z) around z0 . (c) Since the logarithm function has a single point of singularity at z = 0, the outer radius of the convergence ring can only be extended to the value |z0 |. On the other hand, the value of the inner radius is zero.

226

SOLUTIONS 14 Thus, the convergence ring is the disk of radius |z0 | centered at z0 , with the exclusion of the center, that is, of the point z0 itself. Note that the limitation that the real part of z0 be strictly positive is not really necessary, it just avoids the singular point z = 0 and simplifies the analysis because it leaves this convergence ring entirely contained in the n = 0 leaf of the Riemann surface.

(d) The residue of f (z) at the point z0 is the coefficient of the term of the Laurent series that contains the power (z − z0 )−1 , and hence the value of the residue is ln(z0 ). Problem 3. Consider the function f (z) = 1/Pn (z), where Pn (z) is a polynomial of degree n, Pn (z) = a0 + a1 z + a2 z 2 + . . . + an−1 z n−1 + an z n , with complex coefficients a0 , . . . , an , where an 6= 0 and n > 0, and where the coefficients are such that the polynomial has n distinct complex roots z1 , . . . , zn . (a) Write the first three terms of the Laurent series of f (z) around zn . (b) Determine the radii of the convergence ring of this series. (c) Determine the value of the residue of f (z) at the point zn . Answer: n−1 1 1 Y . an zn − zk k=1

Complete Solution: We are considering the function f (z) = 1/Pn (z), where Pn (z) is a polynomial of degree n, Pn (z) = a0 + a1 z + a2 z 2 + . . . + an−1 z n−1 + an z n , with complex coefficients a0 , . . . , an , where an 6= 0 and n > 0, these coefficients being such that the polynomial has n distinct complex roots z1 , . . . , zn , so that it can be factored as

227

LAURENT SERIES AND RESIDUES

Pn (z) = an

n Y

(z − zk ).

k=1

It follows that f (z) has exactly n singularities, which are simple poles at the positions z1 , . . . , zn . (a) In order to develop a representation in series of f (z) around zn , we begin by noting that, since all the roots of Pn (z) are distinct and therefore f (z) has n distinct simple poles, the function w(z) = (z − zn )f (z) z − zn = Pn (z) 1 = Pn−1 (z) is analytic at zn , and therefore has a Taylor series around that point. We will use the fact that the polynomial Pn−1 (z) can be written as Pn−1 (z) = an

n−1 Y k=1

(z − zk ).

Calculating the coefficients of the first three terms of the Taylor series of w(z) around zn we have w0′ (zn ) = w1′ (zn ) = w2′ (zn ) =

1 , Pn−1 (zn ) ′ −Pn−1 (zn ) , 2 Pn−1 (zn ) ′ ′′ (z ) 2[Pn−1 (zn )]2 − Pn−1 (zn )Pn−1 n , 3 (z ) Pn−1 n

so that we have for the three initial terms of the Taylor series of w(z) w(z) =

P ′ (zn ) 1 − n−1 2 (z ) (z − zn ) + Pn−1 (zn ) Pn−1 n ′′ (z ) ′ (zn )]2 − Pn−1 (zn )Pn−1 2[Pn−1 n (z − zn )2 + 3 (z ) 2Pn−1 n +... .

+

228

SOLUTIONS 14 It follows that we have for the first three terms of the Laurent series of f (z) f (z) =

P ′ (zn ) 1 1 − n−1 2 (z ) + Pn−1 (zn ) z − zn Pn−1 n

′ ′′ (z ) 2[Pn−1 (zn )]2 − Pn−1 (zn )Pn−1 n (z − zn ) + 3 2Pn−1 (zn ) +... .

+

(b) Since all the roots of the polynomial Pn (z) are distinct, and therefore all the poles of f (z) are simple, there is a non-zero distance |zk − zn | from the pole zn to each of the poles zk , with k ∈ {1, . . . , n − 1}. We denote the least of these n − 1 distances by |zm − zn |, where zm is the position of the corresponding pole. We can extend the convergence disk of the Taylor series of w(z) around zn only until we reach this singularity, which is the one nearest to zn . Thus, the convergence disk of this series has radius |zm − zn |. It follows that the convergence ring of the Laurent series of f (z) has outer radius given by |zm − zn | and zero inner radius, consisting of the disk of radius |zm − zn | centered at zn , with the exclusion of the center, that is, of the point zn itself. (c) The residue of f (z) at the point zn is the coefficient of the term of the Laurent series that contains the power (z − zn )−1 , and hence the value of the residue is given by n−1 1 1 Y 1 = . Pn−1 (zn ) an zn − zk k=1

Problem 4.


Consider the function f (z) = sin z 2 /z 2 , for z 6= 0.


(a) Write the Taylor series of sin z 2 around z = 0.

(b) Write a series based on the series obtained above, which represents the function f (z). (c) Determine the convergence ring or disk of this series, noting that the series of sin(z) converges on the entire complex plane. (d) Calculate the limit of f (z) when z → 0.

229

LAURENT SERIES AND RESIDUES

(e) How should we define the value of f (z) at z = 0, in such a way that it becomes analytic at that point? Complete Solution: Let us recall that the Taylor series of sin(z) around z = 0 is ∞ X (−1)k z 2k+1 . sin(z) = (2k + 1)! k=0

(a) Simply composing the series of sin(z) with the function z 2 we have the Taylor series ∞
X (−1)k z 4k+2 . sin z 2 = (2k + 1)! k=0


(b) Considering now the function f (z) = sin z 2 /z 2 , for z 6= 0, we can write for it the series ∞ X (−1)k z 4k . f (z) = (2k + 1)! k=0

(c) Note that this series has no negative powers of z, and therefore is a Taylor series and not a Laurent series. Thus, the series automatically extends the domain of analyticity of the function, relative to the initial definition, which cannot be used at z = 0. Since the series of sin(z) is convergent on the whole complex plane, it follows that this series is also convergent on the entire complex plane. (d) If we calculate the limit z → 0 using the series to represent the function, it immediately follows that only the first term is non-zero, and therefore that we have lim f (z) =

z→0

∞ X (−1)k z 4k z→0 (2k + 1)!

lim

k=0

∞ X (−1)k lim z 4k = (2k + 1)! z→0 k=0

= 1.

230

SOLUTIONS 14

(e) If we define the function at z = 0 by the continuity criterion, that is, f (0) = 1, which is the value returned by the Taylor series at that point, it follows that the function is represented by the series throughout the complex plane, and since a convergent power series converges to an analytic function, it follows that the resulting function is analytic on the whole complex plane.

Solutions 15

Calculation of Integrals by Residues We present here complete and commented solutions to all problems proposed in Chapter 15 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1.

Calculate by residues the following integral, Z ∞ 1 dx. 2 + 2x + 2 x −∞

Consider the following steps. (a) Factor completely the polynomial in the denominator. (b) Determine how to close the contour and what are the relevant singularities. (c) Show that the integral over the additional part of the contour, used to close it, vanishes. (d) Calculate the relevant residues and use the residue theorem in order to find the value of the integral. Answer: π. Complete Solution: 231

232

SOLUTIONS 15

We must calculate the asymptotic integral Z ∞ 1 dx. I= 2 −∞ x + 2x + 2 (a) We use the Baskara formula in order to find the roots of the polynomial, √ −2 ± 4 − 8 x± = 2√ −2 ± ı 4 = 2 = −1 ± ı, so that using a complex variable z in place of x, the polynomial can be factored as z 2 + 2z + 2 = [z − (−1 + ı)][z − (−1 − ı)], and therefore the integral can be written as Z ∞ 1 dz, I= −∞ [z − (−1 + ı)][z − (−1 − ı)] which still extends over the real axis in the complex z plane. (b) We can close the contour using an arc that goes to infinity, both in the upper half-plane and in the lower half-plane. If we use the arc in the upper half-plane, the relevant singularity is z = −1 + ı. If we use the arc in the lower half-plane, the relevant singularity is z = −1 − ı. In what follows we will use the first alternative. (c) The additional integral IR is defined on a semi-circle C of radius R, with θ going from 0 to π, in the limit in which we make R → ∞. Using z = R exp(ıθ) and thus dz = ız dθ on the arc, we have for this integral Z 1 dz 2 IR = z + 2z + 2 ZCπ 1 = dθ ız 2 z + 2z + 2 Z0 π 1 dθ ıR eıθ 2 ı2θ = R e + 2R eıθ + 2 0 Z π ı 1 = . dθ eıθ ı2θ R 0 e + 2 eıθ /R + 2/R2

CALCULATION OF INTEGRALS BY RESIDUES

233

We see that in the R → ∞ limit the integrand is simplified, so that we have Z 1 ı π dθ eıθ ı2θ lim IR = lim R→∞ R→∞ R 0 e + 2 eıθ /R + 2/R2 Z ı π 1 = lim dθ eıθ ı2θ R→∞ R 0 e Z 1 ı π dθ ıθ . = lim R→∞ R 0 e Since the exponential with imaginary argument has unit absolute value, the remaining angular integral is the integral of a limited function over a finite domain, and therefore is finite and independent of R. It follows that the limit vanishes due to the factor of R in the denominator, and therefore we have lim IR = 0.

R→∞

Another way to see this, which is often useful, is to calculate the absolute value of the integral in this limit, Z 1 1 π ıθ dθ e lim |IR | = lim ı2θ ıθ 2 R→∞ R→∞ R e + 2 e /R + 2/R 0 Z 1 π ıθ 1 ≤ lim dθ e ı2θ R→∞ R 0 | e + 2 eıθ /R + 2/R2 | Z 1 1 π dθ ı2θ = lim R→∞ R 0 |e | Z 1 π = lim dθ R→∞ R 0 π = lim R→∞ R = 0. Since the absolute value of IR is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that once again we end up with lim IR = 0.

R→∞

234

SOLUTIONS 15

(d) The contour is being traversed in the positive direction, and the residue ξ⊕ of the relevant pole can be calculated by the limit z − (−1 + ı) [z − (−1 + ı)][z − (−1 − ı)] 1 = lim z→(−1+ı) z − (−1 − ı) 1 = −1 + ı + 1 + ı 1 . = 2ı

ξ⊕ =

lim

z→(−1+ı)

It follows that we have for the integral I = 2πıξ⊕ = π.

Problem 2.

Consider the following integral, to be calculated by residues, Z ∞ x2 dx. x4 + 5x2 + 4 0

Consider the following steps.

(a) Discover how to extend the integral to the interval (−∞, ∞). (b) Factor completely the polynomial in the denominator. (c) Determine how to close the contour in the complex plane and what are the relevant singularities. (d) Show that the integral over the additional part of the contour, used to close it, vanishes. (e) Calculate the relevant residues and use the residue theorem in order to find the value of the integral. Answer: π/6. Complete Solution: We must calculate the asymptotic integral Z ∞ x2 I= dx. x4 + 5x2 + 4 0

CALCULATION OF INTEGRALS BY RESIDUES

235

(a) Since both the numerator and the denominator are even functions of x, the integrand is even, and therefore if we extend the integral to the whole real axis, then we will have exactly doubled the result. Therefore, we can write for the integral 1 I= 2

Z

∞

−∞

x2 dx. x4 + 5x2 + 4

(b) We use the Baskara formula in order to find the roots of the quadratic polynomial in x2 , x2±

= = =

−5 ±

√

25 − 16 √2 −5 ± 9 2 −5 ± 3 . 2

As one can see, the two results are negative, because we have the possibilities −8/2 = −4 and −2/2 = −1. It follows that the roots for x are all imaginary, x = ±2ı and x = ±ı, so that using a complex variable z in place of x, the polynomial can be factored as z 4 + 5z 2 + 4 = (z − 2ı)(z + 2ı)(z − ı)(z + ı), so that the integral can be written as 1 I= 2

Z

∞ −∞

z2 dz, (z − 2ı)(z + 2ı)(z − ı)(z + ı)

which still extends over the real axis in the complex z plane. (c) We can close the contour using an arc that goes to infinity, both in the upper half-plane and in the lower half-plane. If we use the arc in the upper half-plane, the relevant singularities are z = ı and z = 2ı. If we use the arc in the lower half-plane, the relevant singularities are z = −ı and z = −2ı. In what follows we will use the first alternative.

236

SOLUTIONS 15

(d) The additional integral IR is defined on a semi-circle of radius R, with θ going from 0 to π, in the limit in which we make R → ∞. Using z = R exp(ıθ) and thus dz = ız dθ on the arc, we have for this integral π

z2 z 4 + 5z + 4 0 Z π z2 dθ ız 4 = z + 5z + 4 0 Z π R2 eı2θ dθ ıR eıθ 4 ı4θ = R e + 5R2 eı2θ + 4 0 Z π ı eı3θ = . dθ ı4θ R 0 e + 5 eı2θ /R2 + 4/R4

IR =

Z

dz

We see that in the limit R → ∞ the integrand is simplified, so that calculating the absolute value of the integral in this limit, we have lim |IR | =

R→∞

≤ = = = =

Z eı3θ 1 π . dθ ı4θ lim R→∞ R e + 5 eı2θ /R2 + 4/R4 0 ı3θ Z e 1 π lim dθ ı4θ R→∞ R 0 | e + 5 eıθ /R2 + 4/R4 | Z π 1 1 lim dθ ı4θ R→∞ R 0 |e | Z 1 π dθ lim R→∞ R 0 π lim R→∞ R 0.

Since the absolute value of IR is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim IR = 0.

R→∞

(e) The contour is being traversed in the positive direction, and the residues ξ1 and ξ2 of the relevant poles of the integrand can be calculated by means of the limits

CALCULATION OF INTEGRALS BY RESIDUES

237

z 2 (z − ı) ξ1 = lim z→ı (z − 2ı)(z + 2ı)(z − ı)(z + ı)

ξ2

z2 = lim z→ı (z − 2ı)(z + 2ı)(z + ı) −1 = (−ı)(3ı)(2ı) ı , = 6 z 2 (z − 2ı) = lim z→2ı (z − 2ı)(z + 2ı)(z − ı)(z + ı) z2 = lim z→2ı (z + 2ı)(z − ı)(z + ı) −4 = (4ı)(ı)(3ı) −ı . = 3

It follows that we have for the integral 1 (2πı)(ξ1 + ξ2 ) 2  ı ı = πı − 6 3  1 2 = −π − 6 6 π = . 6

I =

Problem 3.

Consider the following integral, to be calculated by residues, Z ∞ x2 dx. 4 2 −∞ x + 3x + 2

Consider the following steps. (a) Factor completely the polynomial in the denominator. (b) Determine how to close the contour in the complex plane and what are the relevant singularities. (c) Show that the integral over the additional part of the contour, used to close it, vanishes.

238

SOLUTIONS 15

(d) Calculate the relevant residues and use the residue theorem in order to find the value of the integral. √
Answer: π 2 − 1 . Complete Solution: We must calculate the asymptotic integral Z ∞ x2 dx. I= 4 2 −∞ x + 3x + 2 (a) We use the Baskara formula in order to find the roots of the quadratic polynomial in x2 , x2±

= = =

−3 ±

√

9−8

2 √ −3 ± 1 2 −3 ± 1 . 2

As one can see, the two results are negative, because we have the possibilities −4/2 = −2 √ and −2/2 = −1. It follows that the roots for x are all imaginary, x = ± 2ı and x = ±ı, so that using a complex variable z in place of x, the polynomial can be factored as  √  √  z 4 + 3z 2 + 2 = z − 2ı z + 2ı (z − ı)(z + ı), so that the integral can be written as I=

Z

∞

−∞

z2 √
√
dz, z − 2ı z + 2ı (z − ı)(z + ı)

which still extends over the real axis in the complex z plane. (b) We can close the contour using an arc that goes to infinity, both in the upper half-plane and in the lower half-plane. If we use the arc √ in the upper half-plane, the relevant singularities are z = ı and z = 2ı. If we use the arc in √ the lower half-plane, the relevant singularities are z = −ı and z = − 2ı. In what follows we will use the first alternative.

CALCULATION OF INTEGRALS BY RESIDUES

239

(c) The additional integral IR is defined on a semicircle of radius R, with θ going from 0 to π, in the limit in which we make R → ∞. Using z = R exp(ıθ) and thus dz = ız dθ on the arc, we have for this integral π

z2 z 4 + 3z + 2 0 Z π z2 dθ ız 4 = z + 3z + 2 0 Z π R2 eı2θ dθ ıR eıθ 4 ı4θ = R e + 3R2 eı2θ + 2 0 Z π ı eı3θ = . dθ ı4θ R 0 e + 3 eı2θ /R2 + 2/R4

IR =

Z

dz

We see that in the limit R → ∞ the integrand is simplified, so that calculating the absolute value of the integral in this limit, we have lim |IR | =

R→∞

≤ = = = =

Z eı3θ 1 π . dθ ı4θ lim R→∞ R e + 3 eı2θ /R2 + 2/R4 0 ı3θ Z e 1 π lim dθ ı4θ R→∞ R 0 | e + 3 eıθ /R2 + 2/R4 | Z π 1 1 lim dθ ı4θ R→∞ R 0 |e | Z 1 π dθ lim R→∞ R 0 π lim R→∞ R 0.

Since the absolute value of IR is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim IR = 0.

R→∞

(d) The contour is being traversed in the positive direction, and the residues ξ1 and ξ2 of the relevant poles of the integrand can be calculated by means of the limits

240

SOLUTIONS 15 z 2 (z − ı) √
√
ξ1 = lim z→ı z − 2ı z + 2ı (z − ı)(z + ı)

ξ2

z2 √
√
= lim z→ı z − 2ı z + 2ı (z + ı) −1 √
√
= 1 − 2 ı 1 + 2 ı(2ı) −ı =  √ 2 12 − 2 2 ı = (2 − 1)2 ı = , 2 √
z 2 z − 2ı √
√
= lim √ z→ 2ı z − 2ı z + 2ı (z − ı)(z + ı) =

= =

= =

z2 √
z→ 2ı z + 2ı (z − ı)(z + ı) −2 √
√
√
2 2 ı 2−1 ı 2+1 ı −ı  √ √ 2 2 2 − 12 √ −ı 2 2(2 − 1) √ −ı 2 . 2 lim √

It follows that we have for the integral I = (2πı)(ξ1 + ξ2 ) √ ! ı 2ı = 2πı − 2 2  √  = −π 1 − 2  √ 2−1 . = π Problem 4.

Calculate by residues the following real asymptotic integral,

CALCULATION OF INTEGRALS BY RESIDUES

Z

∞

−∞

x2

241

sin(x) dx. + 4x + 5

Consider the following steps. (a) Write sin(x) in terms of the complex exponentials exp(±ıx). Note that we will have two integrals to calculate. (b) Factor completely the polynomial in the denominator. (c) Determine how to close the contour in each case, and what are the relevant singularities. (d) Show that the integral over the additional parts of the contours, used to close them, vanish. (e) Calculate the relevant residues and use the residue theorem in order to find the value of the original integral. Answer: −π sin(2)/e. Complete Solution: We must calculate the asymptotic integral Z ∞ sin(x) dx. I= 2 −∞ x + 4x + 5 (a) Writing the integral in terms of a complex variable z, and decomposing the sine into complex exponentials, we have I

∞

sin(z) dz + 4z + 5 −∞ Z ∞ 1 eız − e−ız dz = 2 −∞ 2ı z + 4z + 5 Z Z eız e−ız 1 ∞ 1 ∞ dz − dz, = 2ı −∞ z 2 + 4z + 5 2ı −∞ z 2 + 4z + 5

=

Z

z2

where the integrals still extend over the real axis in the complex z plane.

242

SOLUTIONS 15

(b) We use the Baskara formula in order to find the roots of the polynomial, −4 ±

√

16 − 20 2 √ −4 ± ı 4 = 2 = −2 ± ı,

x± =

so that, using the complex variable z in place of x, the polynomial can be factored as z 2 + 4z + 5 = [z − (−2 + ı)][z − (−2 − ı)], and therefore the integral can be written as 1 1 I⊕ − I⊖ 2ı Z 2ı eız 1 ∞ dz + = 2ı −∞ [z − (−2 + ı)][z − (−2 − ı)] Z 1 ∞ e−ız − dz, 2ı −∞ [z − (−2 + ı)][z − (−2 − ı)]

I =

where the two integrals still run over the real axis in the complex z plane. (c) Each of the two integrals has to be closed in a different way. Since we have that z = x+ıy, it follows that for the first integral the exponential appearing in the numerator is eız =

eı(x+ıy)

=

eıx e−y ,

wherein the first exponential, with an imaginary argument, is a limited function, and the second goes to zero if y → ∞, but diverges if y → −∞. Thus we see that this first integral must be closed by an arc in the upper half-plane, where y > 0, and that in this case the relevant singularity is at z = −2 + ı. For the other integral the situation is reversed due to the inversion of the sign in the argument of the exponential, and therefore it will have to be closed by the lower half-plane, and the relevant singularity is at z = −2 − ı.

CALCULATION OF INTEGRALS BY RESIDUES

243

(d) Let us start with the first of the two integrals. The additional integral I⊕R is defined on a semicircle of radius R, with θ going from 0 to π, in the limit in which we make R → ∞. Using z = R exp(ıθ) and thus dz = ız dθ on the arc, we have for this integral π

eız z 2 + 4z + 5 Z0 π eız dθ ız 2 = z + 4z + 5 0 Z π eıR[cos(θ)+ı sin(θ)] dθ ıR eıθ 2 ı2θ = R e + 4R eıθ + 5 0 Z π ı eıR cos(θ) e−R sin(θ) = . dθ eıθ ı2θ R 0 e + 4 eıθ /R + 5/R2

I⊕R =

Z

dz

We see that in the limit R → ∞ the integrand is simplified, so that we have, calculating the absolute value of the integral in this limit, lim |I⊕R | =

R→∞

≤ = = ≤ = =

Z ıR cos(θ) e−R sin(θ) e 1 π dθ eıθ ı2θ lim ıθ 2 R→∞ R 0 e + 4 e /R + 5/R Z eıR cos(θ) e−R sin(θ) 1 π lim dθ eıθ ı2θ R→∞ R 0 | e + 4 eıθ /R + 5/R2 | Z π e−R sin(θ) 1 dθ lim R→∞ R 0 | eı2θ | Z π 1 lim dθ e−R sin(θ) R→∞ R 0 Z 1 π lim dθ R→∞ R 0 π lim R→∞ R 0.

In these passages we replaced sin(θ) by its minimum value within the integration interval, [0, π], in order to maximize the real exponential. Note that in this interval the sine is always positive, and that its minimum value is zero. Since the absolute value of I⊕R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have

244

SOLUTIONS 15

lim I⊕R = 0.

R→∞

For the second integral everything takes place analogously but with θ ranging from π to 2π, so that we have 2π

e−ız z 2 + 4z + 5 π Z 2π e−ız dθ ız 2 = z + 4z + 5 π Z 2π e−ıR[cos(θ)+ı sin(θ)] dθ ıR eıθ 2 ı2θ = R e + 4R eıθ + 5 π Z 2π ı e−ıR cos(θ) eR sin(θ) = . dθ eıθ ı2θ R π e + 4 eıθ /R + 5/R2

I⊖R =

Z

dz

Considering now the R → ∞ limit of the absolute value of the integral, lim |I⊖R | =

R→∞

≤ = = ≤ = =

Z e−ıR cos(θ) eR sin(θ) 1 2π ıθ dθ e lim R→∞ R π eı2θ + 4 eıθ /R + 5/R2 Z e−ıR cos(θ) eR sin(θ) 1 2π lim dθ eıθ ı2θ R→∞ R π | e + 4 eıθ /R + 5/R2 | Z eR sin(θ) 1 2π dθ lim R→∞ R π | eı2θ | Z 2π 1 lim dθ eR sin(θ) R→∞ R π Z 1 2π lim dθ R→∞ R π π lim R→∞ R 0.

In these passages we replaced sin(θ) by its maximum value within the integration interval, [π, 2π], in order to maximize the real exponential. Note that in this interval the sine is always negative, and its maximum value is zero. Since the absolute value of I⊖R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute

CALCULATION OF INTEGRALS BY RESIDUES

245

value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim I⊖R = 0.

R→∞

(e) The contour of the integral I⊕ is being traversed in the positive direction, and the residue ξ⊕ of the relevant pole in this case can be calculated by the limit [z − (−2 + ı)] eız z→(−2+ı) [z − (−2 + ı)][z − (−2 − ı)] eız = lim z→(−2+ı) z − (−2 − ı) eı(−2+ı) = −2 + ı + 2 + ı e−2ı−1 = . 2ı

ξ⊕ =

lim

But the contour of the integral I⊖ is being traversed in the negative direction, and the residue ξ⊖ of the relevant pole in this case can be calculated by the limit [z − (−2 − ı)] e−ız z→(−2−ı) [z − (−2 + ı)][z − (−2 − ı)] e−ız = lim z→(−2−ı) z − (−2 + ı) e−ı(−2−ı) = −2 − ı + 2 − ı − e2ı−1 = . 2ı

ξ⊖ =

lim

It follows that we have for the original integral 1 1 I⊕ − I⊖ 2ı 2ı 1 1 2πıξ⊕ − (−2)πıξ⊖ = 2ı 2ı − e2ı−1 e−2ı−1 +π = π 2ı 2ı

I =

246

SOLUTIONS 15 π e2ı − e−2ı e 2ı π = − sin(2). e

= −

Problem 5.

Calculate by residues the following integral, where a > 0, Z ∞ cos(ax) dx. x2 + 1 0

Consider the following steps. (a) Discover how to extend the integral to the interval (−∞, ∞). (b) Consider another integral similar to this extended integral, but with cos(ax) replaced by sin(ax). What is its value? (c) Combine the two integrals in order to write a third one, whose calculation is simpler. (d) Determine how to close the contour and what are the relevant singularities. Consider the relevance of the sign of a. (e) Show that the integral over the additional part of the contour, used to close it, vanishes. (f) Calculate the relevant residues and use the residue theorem in order to find the value of the integral. Answer: π exp(−|a|)/2. Complete Solution: We must calculate the asymptotic integral Z ∞ cos(ax) I= dx. x2 + 1 0 (a) Since both the numerator and the denominator are even functions of x, the integrand is even, and therefore if we extend the integral to the whole real axis, then we will have exactly twice the values as the result. Thus we can write for the integral 1 I= 2

Z

∞

−∞

cos(ax) dx. x2 + 1

CALCULATION OF INTEGRALS BY RESIDUES

247

(b) Replacing the cosine by a sine in the integral above, I′ =

1 2

Z

∞ −∞

sin(ax) dx, x2 + 1

we have the integral of an odd function over a symmetric domain, and the result is therefore zero, that is, I ′ = 0. (c) Since we have I ′ = 0, we can write that I = I +ıI ′ , and we can therefore write for I Z Z 1 ∞ ı sin(ax) 1 ∞ cos(ax) dx + dx I = 2 −∞ x2 + 1 2 −∞ x2 + 1 Z 1 ∞ cos(ax) + ı sin(ax) dx = 2 −∞ x2 + 1 Z 1 ∞ eıax dx. = 2 −∞ x2 + 1 Writing the integral in terms of a complex variable z we have I = =

Z 1 ∞ eıaz dz 2 −∞ z 2 + 1 Z eıaz 1 ∞ dz, 2 −∞ (z − ı)(z + ı)

where the integral still extends over the real axis in the complex z plane, and where we factored the polynomial in the denominator, using its roots z = ±ı. (d) The integral has to be closed from above or from below, depending on the sign of a. Since we have z = x + ıy, it follows that the exponential appearing in the numerator is eıaz =

eı(ax+ıay)

=

eıax e−ay ,

where the first exponential, with imaginary argument, is a limited function when y → ±∞. However, the behavior of the second exponential depends on the sign of a. On the one hand, when a > 0 it goes to zero if y → ∞, but diverges if y → −∞. On the other hand, if a < 0 the

248

SOLUTIONS 15 situation is reversed. Thus, we see that for a > 0 the integral must be closed by an arc in the upper half-plane, where y > 0, and in this case the relevant singularity is at z = ı. As for the case a < 0, it will have to be closed by an arc in the lower half-plane and the relevant singularity is at z = −ı. Note that for a = 0 one can close the contour in any of these two ways, and therefore it is not necessary to consider this case separately. It can be included in either of the two other cases, so that we can consider as our two alternatives a ≥ 0 and a ≤ 0.

(e) Let us start with the first of the two integrals, that is, the case a ≥ 0. The additional integral I⊕R is defined on a semicircle of radius R, with θ going from 0 to π, in the limit in which we make R → ∞. Using z = R exp(ıθ) and thus dz = ız dθ on the arc, we have for this integral Z π eıaz dz 2 I⊕R = z +1 Z0 π eıaz dθ ız 2 = z +1 0 Z π eıaR[cos(θ)+ı sin(θ)] = dθ ıR eıθ R2 eı2θ + 1 0 Z π ı eıaR cos(θ) e−aR sin(θ) . = dθ eıθ R 0 eı2θ + 1/R2 We see that in the R → ∞ limit the integrand is simplified, so that we have, calculating the absolute value of the integral in this limit, Z ıaR cos(θ) e−aR sin(θ) e 1 π dθ eıθ lim |I⊕R | = lim R→∞ R→∞ R 0 eı2θ + 1/R2 Z π eıaR cos(θ) e−aR sin(θ) 1 ≤ lim dθ eıθ R→∞ R 0 | eı2θ + 1/R2 | Z π 1 e−aR sin(θ) = lim dθ R→∞ R 0 | eı2θ | Z π 1 = lim dθ e−aR sin(θ) R→∞ R 0 Z 1 π dθ ≤ lim R→∞ R 0 π = lim R→∞ R = 0.

CALCULATION OF INTEGRALS BY RESIDUES

249

In these passages we replaced sin(θ) by its minimum value within the integration interval, [0, π], in order to maximize the real exponential. Note that in this interval the sine is always positive, and that its minimum value is zero. Since the absolute value of I⊕R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim I⊕R = 0.

R→∞

For the other case, that is, for a ≤ 0, everything takes place similarly, but with θ ranging from π to 2π, so that we have 2π

eıaz z2 + 1 π Z 2π eıaz dθ ız 2 = z +1 π Z 2π eıaR[cos(θ)+ı sin(θ)] dθ ıR eıθ = R2 eı2θ + 1 π Z 2π eıaR cos(θ) e−aR sin(θ) ı . dθ eıθ = R π eı2θ + 1/R2

I⊖R =

Z

dz

Considering now the R → ∞ limit of the absolute value of the integral, lim |I⊖R | =

R→∞

≤ = = ≤ = =

Z ıaR cos(θ) e−aR sin(θ) 1 2π e lim dθ eıθ R→∞ R π eı2θ + 1/R2 Z 2π eıaR cos(θ) e−aR sin(θ) 1 lim dθ eıθ R→∞ R π | eı2θ + 1/R2 | Z 2π e−aR sin(θ) 1 dθ lim R→∞ R π | eı2θ | Z 2π 1 dθ e−aR sin(θ) lim R→∞ R π Z 1 2π lim dθ R→∞ R π π lim R→∞ R 0.

250

SOLUTIONS 15 In these passages we replace sin(θ) by its maximum value within the integration interval, [π, 2π], in order to maximize the real exponential. Note that in this interval the sine is always negative, and that its maximum value is zero. Since the absolute value of I⊖R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim I⊖R = 0.

R→∞

(f) When a ≥ 0 the integration contour is being traversed in the positive direction, and the residue ξ⊕ of the relevant pole in this case can be calculated by the limit (z − ı) eıaz ξ⊕ = lim z→ı (z − ı)(z + ı) eıaz = lim z→ı z + ı eıaı = ı+ı e−a = . 2ı However, when a ≤ 0 the integration contour is being traversed in the negative direction, and the residue ξ⊖ of the relevant pole in this case can be calculated by the limit (z + ı) eıaz z→−ı (z − ı)(z + ı) eıaz = lim z→−ı z − ı e−ıaı = −ı − ı ea = − . 2ı

ξ⊖ =

lim

It follows that we have for the original integral in the case a ≥ 0,

CALCULATION OF INTEGRALS BY RESIDUES

251

1 (2πı)ξ⊕ 2   e−a = (πı) 2ı −a e , = π 2

I =

while in the case a ≤ 0 we have 1 (−2πı)ξ⊖ 2  a e = (−πı) − 2ı ea = π . 2

I =

Note that the two results in fact coincide in the case a = 0. We can write the result for any value of a, in terms of |a|, thus obtaining I= Problem 6.

π e−|a| . 2

Calculate by residues the following real asymptotic integral, Z ∞ cos(x) dx. 2 −∞ x + 4x + 5

Consider the following steps. (a) Write cos(x) in terms of the complex exponentials exp(±ıx). Note that we will have two integrals to calculate. (b) Factor completely the polynomial in the denominator. (c) Determine how to close the contour in each case, and what are the relevant singularities. (d) Show that the integrals over the additional parts of the contours, used to close them, vanish. (e) Calculate the relevant residues and use the residue theorem in order to find the value of the original integral. Answer: π cos(2)/e.

252

SOLUTIONS 15

Complete Solution: We must calculate the asymptotic integral Z ∞ cos(x) I= dx. 2 −∞ x + 4x + 5 (a) Writing the integral in terms of a complex variable z, and decomposing the cosine into complex exponentials, we have ∞

cos(z) dz + 4z + 5 −∞ Z ∞ 1 eız + e−ız = dz 2 −∞ 2 z + 4z + 5 Z Z eız 1 ∞ e−ız 1 ∞ dz + dz, = 2 −∞ z 2 + 4z + 5 2 −∞ z 2 + 4z + 5

I =

Z

z2

where the integral still extends over the real axis in the complex z plane. (b) We use the Baskara formula in order to find the roots of the polynomial, √

16 − 20 2 √ −4 ± ı 4 = 2 = −2 ± ı,

x± =

−4 ±

so that, using the complex variable z in place of x, the polynomial can be factored as z 2 + 4z + 5 = [z − (−2 + ı)][z − (−2 − ı)], and therefore the integral can be written as I

1 1 I⊕ + I⊖ 2Z 2 1 ∞ eız = dz + 2 −∞ [z − (−2 + ı)][z − (−2 − ı)] Z 1 ∞ e−ız + dz. 2 −∞ [z − (−2 + ı)][z − (−2 − ı)]

=

CALCULATION OF INTEGRALS BY RESIDUES

253

(c) Each of the two integrals has to be closed in a different way. Since we have z = x + ıy, it follows that for the first integral the exponential appearing in the numerator is eız =

eı(x+ıy)

=

eıx e−y ,

wherein the first exponential, with an imaginary argument, is a limited function, and the second goes to zero if y → ∞, but diverges if y → −∞. Thus, we see that this first integral must be closed by an arc in the upper half-plane, where y > 0, and that in this case the relevant singularity is at z = −2 + ı. For the other integral the situation is reversed due to the inversion of the sign in the argument of the exponential, and therefore it will have to be closed by the lower half-plane, and the relevant singularity is at z = −2 − ı. (d) Let us start with the first of the two integrals. The additional integral I⊕R is defined on a semicircle of radius R, with θ going from 0 to π, in the limit in which we make R → ∞. Using z = R exp(ıθ) and thus dz = ız dθ on the arc, we have for this integral Z π eız dz 2 I⊕R = z + 4z + 5 Z0 π eız dθ ız 2 = z + 4z + 5 0 Z π eıR[cos(θ)+ı sin(θ)] dθ ıR eıθ 2 ı2θ = R e + 4R eıθ + 5 0 Z π ı eıR cos(θ) e−R sin(θ) = . dθ eıθ ı2θ R 0 e + 4 eıθ /R + 5/R2 We see that in the R → ∞ limit the integrand is simplified, so that we have, calculating the absolute value of the integral in this limit, Z ıR cos(θ) e−R sin(θ) e 1 π lim |I⊕R | = lim dθ eıθ ı2θ ıθ 2 R→∞ R→∞ R 0 e + 4 e /R + 5/R Z eıR cos(θ) e−R sin(θ) 1 π dθ eıθ ı2θ ≤ lim R→∞ R 0 | e + 4 eıθ /R + 5/R2 | Z π 1 e−R sin(θ) = lim dθ R→∞ R 0 | eı2θ |

254

SOLUTIONS 15 = ≤ = =

Z 1 π dθ e−R sin(θ) lim R→∞ R 0 Z 1 π lim dθ R→∞ R 0 π lim R→∞ R 0.

In these passages we replaced sin(θ) by its minimum value within the integration interval, [0, π], in order to maximize the real exponential. Note that in this interval the sine is always positive, and that its minimum value is zero. Since the absolute value of I⊕R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim I⊕R = 0.

R→∞

For the second integral everything takes place analogously but with θ ranging from π to 2π, so that we have 2π

e−ız + 4z + 5 π Z 2π e−ız = dθ ız 2 z + 4z + 5 π Z 2π e−ıR[cos(θ)+ı sin(θ)] dθ ıR eıθ 2 ı2θ = R e + 4R eıθ + 5 π Z 2π e−ıR cos(θ) eR sin(θ) ı . dθ eıθ ı2θ = R π e + 4 eıθ /R + 5/R2

I⊖R =

Z

dz

z2

Considering now the R → ∞ limit of the absolute value of the integral, Z −ıR cos(θ) eR sin(θ) 1 2π e lim |I⊖R | = lim dθ eıθ ı2θ ıθ 2 R→∞ R→∞ R π e + 4 e /R + 5/R Z e−ıR cos(θ) eR sin(θ) 1 2π dθ eıθ ı2θ ≤ lim R→∞ R π | e + 4 eıθ /R + 5/R2 | Z 2π 1 eR sin(θ) = lim dθ R→∞ R π | eı2θ |

CALCULATION OF INTEGRALS BY RESIDUES

255

Z 1 2π dθ eR sin(θ) R→∞ R π Z 1 2π ≤ lim dθ R→∞ R π π = lim R→∞ R = 0. =

lim

In these passages we replaced sin(θ) by its maximum value within the integration interval, [π, 2π], in order to maximize the real exponential. Note that in this interval the sine is always negative, and that its maximum value is zero. Since the absolute value of I⊖R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim I⊖R = 0.

R→∞

(e) The contour of the integral I⊕ is being traversed in the positive direction, and the residue ξ⊕ of the relevant pole in this case can be calculated by the limit [z − (−2 + ı)] eız z→(−2+ı) [z − (−2 + ı)][z − (−2 − ı)] eız = lim z→(−2+ı) z − (−2 − ı) eı(−2+ı) = −2 + ı + 2 + ı e−2ı−1 . = 2ı

ξ⊕ =

lim

However, the contour of the integral I⊖ is being traversed in the negative direction, and the residue ξ⊖ of the relevant pole in this case can be calculated by the limit [z − (−2 − ı)] e−ız z→(−2−ı) [z − (−2 + ı)][z − (−2 − ı)] e−ız = lim z→(−2−ı) z − (−2 + ı)

ξ⊖ =

lim

256

SOLUTIONS 15 e−ı(−2−ı) −2 − ı + 2 − ı − e2ı−1 . 2ı

= =

It follows that we have for the original integral I = = = = =

1 1 I⊕ + I⊖ 2 2 1 1 2πıξ⊕ + (−2)πıξ⊖ 2 2 e−2ı−1 − e2ı−1 πı − πı 2ı 2ı π e2ı + e−2ı e 2 π cos(2). e

Problem 7. Consider the following real asymptotic integral, to be calculated by residues, Z ∞ cos(x + 1) dx. 2 −∞ x + 4x + 5 Consider the following steps. (a) Determine and describe one or more closed contours in the complex plane that can be used to compute this integral. (b) Show that the integrals over the additional parts of the contours, used to close the original contour over the real line, vanish. (c) Calculate the residues of the relevant singularities and use the residue theorem in order to find the value of the integral. Answer: π cos(1)/e. Complete Solution: We must calculate the asymptotic integral Z ∞ cos(x + 1) dx. I= 2 + 4x + 5 x −∞

CALCULATION OF INTEGRALS BY RESIDUES

257

Writing the integral in terms of a complex variable z, and decomposing the cosine into complex exponentials, we have Z ∞ cos(z + 1) dz I = 2 + 4z + 5 z −∞ Z ∞ 1 eı(z+1) + e−ı(z+1) = dz z 2 + 4z + 5 −∞ 2 Z Z eı(z+1) 1 ∞ e−ı(z+1) 1 ∞ dz + dz, = 2 −∞ z 2 + 4z + 5 2 −∞ z 2 + 4z + 5 where the integral still extends over the real axis in the complex z plane. We now use the Baskara formula in order to find the roots of the polynomial, √ −4 ± 16 − 20 x± = 2 √ −4 ± ı 4 = 2 = −2 ± ı, so that, using the complex variable z in place of x, the polynomial can be factored as z 2 + 4z + 5 = [z − (−2 + ı)][z − (−2 − ı)], and therefore the integral can be written as 1 1 I⊕ + I⊖ 2 2 Z 1 ∞ eı(z+1) dz + = 2 −∞ [z − (−2 + ı)][z − (−2 − ı)] Z e−ı(z+1) 1 ∞ dz, + 2 −∞ [z − (−2 + ı)][z − (−2 − ı)]

I =

where the two integrals still extend over the real axis in the complex z plane. (a) Each of the two integrals has to be closed in a different way. Since we have z = x + ıy, it follows that for the first integral, the exponential appearing in the numerator is eı(z+1) = =

eı(x+1+ıy) eı(x+1) e−y ,

258

SOLUTIONS 15 wherein the first exponential, with an imaginary argument, is a limited function, and the second goes to zero if y → ∞, but diverges if y → −∞. Thus, we see that this first integral must be closed by an arc in the upper half-plane, where y > 0, and that in this case the relevant singularity is at z = −2 + ı. For the other integral the situation is reversed due to the inversion of the sign in the argument of the exponential, and therefore it will have to be closed by the lower half-plane, and the relevant singularity is at z = −2 − ı.

(b) Let us start with the first of the two integrals. The additional integral I⊕R is defined on a semicircle of radius R, with θ going from 0 to π, in the limit in which we make R → ∞. Using z = R exp(ıθ) and thus dz = ız dθ on the arc, we have for this integral Z π eı(z+1) dz 2 I⊕R = z + 4z + 5 0 Z π eı(z+1) dθ ız 2 = z + 4z + 5 0 Z π eı{R[cos(θ)+ı sin(θ)]+1} dθ ıR eıθ 2 ı2θ = R e + 4R eıθ + 5 0 Z π ı eı[R cos(θ)+1] e−R sin(θ) = . dθ eıθ ı2θ R 0 e + 4 eıθ /R + 5/R2 We see that in the R → ∞ limit the integrand is simplified, so that we have, calculating the absolute value of the integral in this limit, Z ı[R cos(θ)+1] e−R sin(θ) e 1 π dθ eıθ ı2θ lim |I⊕R | = lim R→∞ R→∞ R 0 e + 4 eıθ /R + 5/R2 Z eı[R cos(θ)+1] e−R sin(θ) 1 π ≤ lim dθ eıθ ı2θ R→∞ R 0 | e + 4 eıθ /R + 5/R2 | Z π 1 e−R sin(θ) = lim dθ R→∞ R 0 | eı2θ | Z π 1 = lim dθ e−R sin(θ) R→∞ R 0 Z 1 π dθ ≤ lim R→∞ R 0 π = lim R→∞ R = 0.

CALCULATION OF INTEGRALS BY RESIDUES

259

In these passages we replace sin(θ) by its minimum value within the integration interval, [0, π], in order to maximize the real exponential. Note that in this interval the sine is always positive, and that its minimum value is zero. Since the absolute value of I⊕R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim I⊕R = 0.

R→∞

For the second integral everything takes place analogously but with θ ranging from π to 2π, so that we have I⊖R = =

Z

2π

dz

π

Z

2π

dθ ız

π

=

Z

=

ı R

e−ı(z+1) z 2 + 4z + 5 e−ı(z+1) z 2 + 4z + 5

2π

dθ ıR eıθ

π

Z

π

2π

dθ eıθ

e−ı{R[cos(θ)+ı sin(θ)]+1} R2 eı2θ + 4R eıθ + 5 e−ı[R cos(θ)+1] eR sin(θ) . eı2θ + 4 eıθ /R + 5/R2

Considering now the R → ∞ limit of the absolute value of the integral, lim |I⊖R | =

R→∞

≤ = = ≤ = =

Z −ı[R cos(θ)+1] eR sin(θ) e 1 2π dθ eıθ ı2θ lim R→∞ R π e + 4 eıθ /R + 5/R2 Z e−ı[R cos(θ)+1] eR sin(θ) 1 2π lim dθ eıθ ı2θ R→∞ R π | e + 4 eıθ /R + 5/R2 | Z 2π eR sin(θ) 1 dθ lim R→∞ R π | eı2θ | Z 2π 1 dθ eR sin(θ) lim R→∞ R π Z 1 2π lim dθ R→∞ R π π lim R→∞ R 0.

260

SOLUTIONS 15 In these passages we replace sin(θ) by its maximum value within the integration interval, [π, 2π], in order to maximize the real exponential. Note that in this interval the sine is always negative, and its maximum value is zero. Since the absolute value of I⊖R is positive and is bounded from above by zero in the limit, it follows that the limit of the absolute value is zero. If the absolute value tends to zero, it is necessary that the number also tend to zero, so that we have lim I⊖R = 0.

R→∞

(c) The contour of the integral I⊕ is being traversed in the positive direction, and the residue ξ⊕ of the relevant pole in this case can be calculated by the limit ξ⊕ = = = =

[z − (−2 + ı)] eı(z+1) z→(−2+ı) [z − (−2 + ı)][z − (−2 − ı)] lim

eı(z+1) z→(−2+ı) z − (−2 − ı) eı(−1+ı) lim

−2 + ı + 2 + ı e−ı−1 . 2ı

However, the contour of the integral I⊖ is being traversed in the negative direction, and the residue ξ⊖ of the relevant pole in this case can be calculated by the limit ξ⊖ = = = =

[z − (−2 − ı)] e−ı(z+1) z→(−2−ı) [z − (−2 + ı)][z − (−2 − ı)] lim

e−ı(z+1) z→(−2−ı) z − (−2 + ı) lim

e−ı(−1−ı) −2 − ı + 2 − ı − eı−1 . 2ı

It follows that we have for the original integral

CALCULATION OF INTEGRALS BY RESIDUES 1 1 I⊕ + I⊖ 2 2 1 1 2πıξ⊕ + (−2)πıξ⊖ 2 2 − eı−1 e−ı−1 − πı πı 2ı 2ı π eı + e−ı e 2 π cos(1). e

I = = = = = Problem 8.

261

Calculate by residues the following integral, Z

2π 0

4 dθ. 5 + 4 sin(θ)

Consider the following steps. (a) Write sin(θ) in terms of the complex exponentials exp(±ıθ). (b) Change variables in the integral, from θ to z = exp(ıθ). For this purpose write dθ in terms of dz. Also determine what is the integration contour in the complex z plane. (c) Write the transformed integral as the integral of a rational function, that is, the ratio of two polynomials on z. (d) Factor completely the polynomial in the denominator. (e) Determine which are the relevant singularities. (f) Calculate the relevant residues and use the residue theorem in order to find the value of the integral. Answer: 8π/3. Complete Solution: We will calculate by residues the real integral I=

Z

2π

dθ 0

4 . 5 + 4 sin(θ)

262

SOLUTIONS 15

(a) Writing the sine in terms of complex exponentials we have sin(θ) = =

 1  ıθ e − e−ıθ 2ı   1 1 z− , 2ı z

where the last form is valid provided that it be restricted to the unit circle in complex z plane, with z = ρ exp(ıθ) and ρ = 1. (b) Writing z in the polar representation, we have z = ρ eıθ dz = ız dθ,

⇒

so that we have dθ =

1 dz. ız

This holds independently of the value of ρ but, as we saw above, only for ρ = 1 the sine can be written simply in terms of z, so that the integration contour in the complex z plane is the unit circle. (c) Making in the integral the transformations indicated above, we obtain 2π

1 1 ız 5 − 2ı(z − 1/z) 0 Z 2π 1 1 dz = 4 ı 5z − 2ı(z 2 − 1) I0 1 dz = 4 , 5ız + 2z 2 − 2 C

I = 4

Z

ız dθ

where the closed contour C is the unit circle. (d) It is now necessary to factor the polynomial in the denominator. Using the Baskara formula it is not difficult to see that the two roots are −2ı and −ı/2, so that we have for the integral I 1 dz 2 I = 4 2z + 5ız − 2 IC 1 dz . = 4 2(z + 2ı)(z + ı/2) C

CALCULATION OF INTEGRALS BY RESIDUES

263

(e) Only the pole given by z = −ı/2 lies within the integration contour, so that only its residue ξ will contribute to the integral. (f) The relevant residue of the integrand can be calculated very simply through the limit (z + ı/2) 2(z + 2ı)(z + ı/2) 1 = lim z→−ı/2 2(z + 2ı) 1 = −ı + 4ı 1 = . 3ı

ξ =

lim

z→−ı/2

It follows that the value of the integral is given, without any difficulty, by I = 4(2πı)ξ 8π . = 3 Problem 9.

Calculate by residues the following integral, Z π 1 dθ. 2 −π 1 + sin (θ)

Consider the following steps. (a) Write sin2 (θ) in terms of the complex exponentials exp(±ıθ). (b) Change variables in the integral, from θ to z = exp(ıθ). For this purpose write dθ in terms of dz. Also determine what is the integration contour in the complex z plane. (c) Write the transformed integral as the integral of a rational function, that is, the ratio of two polynomials on z. (d) Factor completely the polynomial in the denominator. (e) Determine which are the relevant singularities.

264

SOLUTIONS 15

(f) Calculate the relevant residues and use the residue theorem in order to find the value of the integral. √ Answer: π 2. Complete Solution: We will calculate by residues the real integral Z π 1 dθ I= . 1 + sin2 (θ) −π (a) Writing the sine in terms of complex exponentials we have sin(θ) = =

 1  ıθ e − e−ıθ 2ı   1 1 z− , 2ı z

where the last form is valid provided that it be restricted to the unit circle in complex z plane, with z = ρ exp(ıθ) and ρ = 1. We have therefore for the square of the sine   1 2 −1 z− sin (θ) = 4 z   1 1 2 = − z −2+ 2 . 4 z 2

(b) Writing z in the polar representation, we have z = ρ eıθ dz = ız dθ,

⇒

so that we have dθ =

1 dz. ız

This holds independently of the value of ρ but, as we saw above, only for ρ = 1 the sine can be written simply in terms of z, so that the integration contour in the complex z plane is the unit circle.

CALCULATION OF INTEGRALS BY RESIDUES

265

(c) Making in the integral the transformations indicated above, we obtain π

4 1 2 ız 4 − (z − 2 + 1/z 2 ) −π Z π 1 z ız dθ 2 2 = 4ı z z − 6 + 1/z 2 I−π z dz 4 = 4ı , z − 6z 2 + 1 C

I =

Z

ız dθ

where the closed contour C is the unit circle. (d) It now necessary to factor the polynomial in the denominator. Using the Baskara formula for the quadratic polynomial on z 2 it √ is not difficult to see that the two roots are given in terms of z 2 by 3 ± 2p2. It follows √ that the four roots of the polynomial are given by z = ± 3 ± 2 2, so that we have for the integral I = 4ı = 4ı

I

dz

IC C

×

z4

dz ×

z − 6z 2 + 1 z

 p p p p √  √  √  √  z− 3+2 2 z + 3+2 2 z− 3−2 2 z + 3−2 2

.

(e) Aspone can see, all four poles are on the real axis. If we take the square of √ √ ± 3 ±√2 2, we obtain 3 ± 2 2, and taking the square again we obtain 17 ± 12 2. It is thus clear that the cases in which the remaining sign is positive fall outside the unit circle. Calculating the√numbers involved in the other case in an approximate way, we have 12 2 ≈ 16.97, which makes it clear that in this case the two roots are inside the unit circle. We therefore conclude that only the four poles contribute to the p two of √ integral, those given by z = ± 3 − 2 2. (f) The relevant residues of the integrand can be calculated very simply through the limit

ξ± =

z→±

lim √

 p √  z z∓ 3−2 2

p p √
  √  √  √  2 z− 3−2 2 z+ 3−2 2 3−2 2 z − 3 + 2 2

266

SOLUTIONS 15 z p √
  √  z→± z2 − 3 + 2 2 z± 3−2 2 p √ ± 3−2 2 =  √
  p √
√  3−2 2 − 3+2 2 ±2 3 − 2 2

=

= =

lim √

√ 3−2 2



1 √ √
2 3−2 2−3−2 2 −1 √ , 8 2

which ends up being, as we see, the same result for the two poles, ξ⊕ = ξ⊖

1 = − √ . 8 2

It follows that the value of the integral is given, without greater difficulty, by I = 4ı(2πı)(ξ⊕ + ξ⊖ )   (−1) = (−8π) 2 √ 8 2 √ = π 2. Problem 10. Consider the complex function w(z) = 1/z, and the unit circle in the complex z plane. (a) Calculate the closed-contour integral of w(z) on the complete unit circle. (b) Calculate the open integral of w(z) on the upper semicircle of the unit circle, that is, the integral over the circle between the points (−1, 0) and (1, 0). (c) Calculate the Cauchy principal value of the closed-contour integral of w(z) on the upper semicircle of the unit circle, closed by the real line segment between the points (1, 0) and (−1, 0). (d) Show that the complex version of the criterion for the Cauchy principal value is equivalent to the real version of the criterion, that is, to the use of the symmetric limit defined in the text for the calculation of the real integral between −1 and 1.

CALCULATION OF INTEGRALS BY RESIDUES

267

Complete Solution: We must calculate several definite integrals of the complex function w(z) = 1/z, on curves in the complex plane. (a) Let us calculate the integral of the function on the unit circle of the complex z plane, I

dz

C

1 = 2πı, z

which can be obtained immediately in several different ways, for example using the Cauchy integral formula, 1 f (z0 ) = 2πı

I

dz

f (z) , z − z0

with f (z) = 1 for all z, and z0 = 0. (b) Let us calculate the open integral of w(z) on the upper semicircle C⊕ (1) of the unit circle, that is, the integral over the unit circle between the points (1, 0) and (−1, 0). Using z = ρ exp(ıθ), with ρ = 1 and θ going from 0 to π, we have dz = ız dθ, and therefore IC⊕ (1) =

Z

= ı

π

dθ ız 0

Z

1 z

π

dθ 0

= ıπ.

(c) Let us calculate the Cauchy principal value of the closed-contour integral of w(z) on the contour C formed by upper semicircle of the unit circle, closed by the real line segment between the points (−1, 0) and (1, 0). The residue at the pole is 1, and since we are calculating the principal value, we take half of its value, I =

I

C

= πı.

dz

1 z

268

SOLUTIONS 15

(d) Let us consider the upper unit semicircle C⊕ (1) and a small circle of radius ρ centered at the origin, formed by the upper semicircle C⊕ (ρ) and by the lower semicircle C⊖ (ρ). By its definition in the complex plane, the Cauchy principal value is the average of the following two integrals, I⊕ = I⊖ = I = =

=

Z −ρ Z Z 1 1 1 1 1 dz + dx − dx , dz + z z z z C⊕ (ρ) −1 ρ C⊕ (1) Z Z −ρ Z Z 1 1 1 1 1 dz + dz + dx + dx ⇒ z z z z C⊕ (1) C⊖ (ρ) −1 ρ I⊕ + I⊖ 2 Z Z −ρ Z 1 1 1 1 1 dz + 2 2 dx + 2 dx + 2 z z z C⊕ (1) −1 ρ ! Z Z 1 1 dz − + dz z z C⊖ (ρ) C⊕ (ρ) Z −ρ Z 1 Z 1 1 1 dx + dx + dz + z z z −1 ρ C⊕ (1) ! Z Z 1 1 1 − dz dz − . 2 z z C⊖ (ρ) C⊕ (ρ) Z

Both I and the integral over C⊕ (1) are already known. Let us calculate the integrals over C⊕ (ρ) and C⊖ (ρ), IC⊕ (ρ) =

Z

= ı

IC⊖ (ρ)

π

dθ ız 0 Z

1 z

π

dθ 0

= ıπ, Z 2π 1 dθ ız = z π Z π dθ = ı 0

= ıπ,

so that these two integrals cancel off in the expression for I, and we have therefore

269

CALCULATION OF INTEGRALS BY RESIDUES I =

Z

1 + z

dz C⊕ (1)

= ıπ +

Z

−ρ

dx

−1

= ıπ,

Z

−ρ −1

1 + z

dx

1 + z

1

Z

dx

ρ

1 z

Z

1

dx ρ

1 z

from which it follows that Z

−ρ

dx

−1

1 + z

Z

1

dx ρ

1 = 0. z

The ρ → 0 limit, which defines the Cauchy principal value in the complex plane, corresponds therefore to the limit lim

ρ→0

Z

−ρ

−1

1 dx + z

Z

ρ

1

1 dx z



= 0,

which is the corresponding definition of the Cauchy principal value in the real case. Problem 11. Consider the complex function w(z) = 1/z, and the real axis in the complex z plane. Calculate the integral of the function on the real axis according to the criterion of the Cauchy principal value, that is, according to the definition Z ∞ dz w(z) I = −∞

=

lim

Z

R

R→∞ −R

dz w(z),

with respect to the asymptotic part, and the Cauchy criterion discussed in the text for the integration through the singularity at the origin. Do this in the two ways listed below and interpret the meaning of the Cauchy principal value of all the integrals that appear. (a) Do the calculation directly by elementary means, that is, do the real integral using the fundamental theorem of the calculus, for finite R, and then take the limit R → ∞.

270

SOLUTIONS 15

(b) Close the contour in the complex plane by means of a circular arc, with finite radius R, consider the integral over this arc of circle, use the residue theorem in order to calculate the integral, and then take the limit R → ∞. Answer: I = 0. Complete Solution: (a) Let us calculate directly the integral of w(z) on the real axis. Using the criterion of the Cauchy principal value for both the asymptotic part and the singular region around zero, we have, with z = x + ıy and y = 0, Z ∞ 1 dx I = x −∞  Z −ǫ Z R 1 1 dx = lim lim . dx + R→∞ ǫ→0 x x ǫ −R Making the transformation of variables x′ = −x in the first integral we obtain  Z ǫ Z R 1 ′ 1 dx I = lim lim dx ′ + R→∞ ǫ→0 x x ǫ R   Z R Z R 1 1 dx = lim lim − dx′ ′ + R→∞ ǫ→0 x x ǫ ǫ Z R  Z R 1 1 = lim lim dx − dx R→∞ ǫ→0 x x ǫ ǫ = 0, independently of either R or ǫ. Note that it is not even necessary to do the integration, which in any case can easily be done, R

 Z R 1 1 I = lim lim dx − dx R→∞ ǫ→0 x x ǫ   ǫ  R R − ln = lim lim ln R→∞ ǫ→0 ǫ ǫ = lim lim ln(1) Z

R→∞ ǫ→0

= 0.

CALCULATION OF INTEGRALS BY RESIDUES

271

It follows, of course, that we can simply take the two limits involved and therefore that the answer is I = 0. (b) Let us calculate the open integral of w(z) on the upper semicircle C⊕ (R), of radius R, of the circle of radius R, that is, the integral over the circle of radius R, in the positive direction, between the points (R, 0) and (−R, 0). Using z = R exp(ıθ), with θ going from 0 to π, we have that dz = ız dθ, and therefore that IC⊕ (R) =

Z

= ı

π

dθ ız

0

Z

1 z

π

dθ

0

= ıπ.

We can now use this semicircle in order to close the integral IR of w(z) on the real axis from −R to R, and so we have a closed contour integral, whose value is the sum of the integral IR we want to calculate with the integral IC⊕ (R) we have just calculated, that is, IR + IC⊕ (R) = IR + ıπ. We now use the residue theorem in order to calculate the integral of w(z) on this closed contour. Following the criterion of the Cauchy principal value when the integral passes over the pole at z = 0, we have that the value of the residue is 1, and therefore that the closed-contour integral has the value 2ıπ(1/2) = ıπ, that is, we have that IR +ıπ = ıπ, from which it follows that IR = 0. This is independent of R, and thus holds for all R, so that, making R → ∞, we have the answer I = 0. Problem 12.

Calculate by residues the following integral, Z ∞ 1 dx. −∞ cosh(x/2)

Consider the following steps. (a) Write cosh(x/2) in terms of the real exponentials exp(±x). (b) Extend the variable x to the complex plane of z = x + ıy, and verify to what the integral reduces over two infinite straight lines: the real axis y = 0 and the line y = 2π. (c) Consider how to close a contour which comprises these two straight lines, in order to enable the calculation of the integral.

272

SOLUTIONS 15

(d) Locate the singularities of the function being integrated and determine which ones are relevant. (e) Show that the integrals over the two vertical segments which are included in order to close the contour vanish when these additional segments are taken to x → ±∞. (f) Assume that the relevant pole is of the first order and calculate the residue by means of the limit that applies to simple poles. Use the residue theorem in order to find the value of the integral. Answer: 2π. (g) (Challenge Item) Find the Laurent series of the function being integrated, around the relevant pole, including a sufficient number of terms to determine the order of the pole and the value of the residue. Complete Solution: We must calculate the following asymptotic integral, Z ∞ 1 dx. I= −∞ cosh(x/2) (a) Writing the hyperbolic cosine in terms of real exponentials we have I = = =

Z

∞

−∞ ∞

Z

−∞ ∞

Z

−∞

2 dx ex/2 + e−x/2 2 ex/2 dx ex + 1 √ 2 ex dx. ex + 1

(b) Extending the integral to the complex plane of z = x + ıy we have Iz = 2

Z

C

ez/2 dz, ez + 1

on some contour C yet to be specified. On the real axis y = 0 the integral is reduced, of course, to that from which we began,

CALCULATION OF INTEGRALS BY RESIDUES I = I0 Z = 2

∞

273

ex/2 dx. ex + 1

−∞

However, on the axis y = 2π we have for the integral, I2π = 2

Z

∞

ex/2 eıπ dx ex e2ıπ + 1

−∞ Z ∞

= −2

= −I0 ,

−∞

ex/2 dx ex + 1

since exp(2ıπ) = 1 and exp(ıπ) = −1. (c) If we close the contour formed by the straight lines y = 0 and y = 2π by two vertical segments of finite length 2π, which are taken to infinity, one in each direction, and we traverse the resulting contour in the positive direction, the bottom line y = 0 will contribute I0 to the integral, and the top line y = 2π will contribute −I2π = I0 , that is, the same value, because this line is being traversed in its negative direction. If the two finite segments did not contribute to the integral in the limit in which they are taken to infinity, then we would be able to calculate the original integral using this closed contour C, that is, I ez/2 1 2 dz I = z 2 C e +1 I ez/2 = dz. z C e +1 (d) The singularities of the integrand, considering that the exponential function is analytic on the whole complex plane, and that the integrand is given by ez/2 , ez + 1 are given by the zeros in the denominator of this ratio. The zeros are therefore at the points defined by exp(z) = −1, that is, at the coordinates x and y given by

274

SOLUTIONS 15 ex eıy =

ex [cos(y) + ı sin(y)]

= −1

x

e cos(y) = −1,

⇒

ex sin(y) = 0.

From the second relation, since the real exponential never vanishes, we conclude that it is necessary to have y = kπ, for some integer value of k. This implies that the cosine is equal to (−1)k , and therefore, from the first relation, that ex (−1)k = −1. Since the real exponential is never negative, it follows that k must be odd, and hence it follows that x = 0, so that this equation can be satisfied. In conclusion, the singularities are on the imaginary axis, at the coordinates x = 0 and y = (2k + 1)π, with integer k. Of all these infinitely many singularities only one is within the contour C described above, the one with k = 0, located at (0, π). (e) The additional integrals over the segments that go to infinity, located at ±x0 , are given by 2π

ez/2 ıdy ez + 1 0 Z 2π 1 dy = ı z/2 e + e−z/2 0 Z 2π 1 = ı dy. x /2 ıy/2 0 e e + e−x0 /2 e−ıy/2 0

Ix 0 =

Z

The exponentials with imaginary arguments in y are limited functions, and when we make x0 → ±∞ the exponentials of x0 go to zero or to infinity, depending on the signs involved. From the two terms in the denominator, one always goes to zero, and can be disregarded when calculating the limit. The other term always goes to infinity, and the bounded oscillating function that multiplies it is never zero. Thus we see that in both cases the limit of the above integral is zero, lim Ix0 = 0.

x0 →±∞

CALCULATION OF INTEGRALS BY RESIDUES

275

(f) The residue ξ of the pole located at z = (0, π) can be calculated using the limit ξ = lim z→ıπ

(z − ıπ) ez/2 . ez + 1

Analyzing this limit, we see that we have a zero in the numerator and a zero in the denominator, while the exponential in the numerator has a finite and non-zero limit. There are several ways of calculating this limit, including the use of the l’Hˆopital rule (in fact, due to Bernoulli). What we will do here is to calculate the limit of the inverse ratio of the two factors that cancel each other out, using the Taylor series of the exponential around the point ıπ. It is not difficult to verify that this series is given by ez =

∞ X −1

n=0

n!

(z − ıπ)n ,

where we have that ∞ X 1 e +1=− (z − ıπ)n , n! n=1 z

and therefore that ez + 1 z − ıπ

= − = −

∞ X 1 (z − ıπ)n−1 n!

n=1 ∞ X

n=0

1 (z − ıπ)n . (n + 1)!

This makes it very easy to take the limit, because only the first term of the remaining series is non-zero in the limit, and we have immediately that ez + 1 = −1. lim z→ıπ z − ıπ From this it also follows that

276

SOLUTIONS 15 lim

z→ıπ

z − ıπ ez + 1

1 −1 = −1.

=

Finally, it follows that we have for the original limit (z − ıπ) ez/2 lim z→ıπ ez + 1 ıπ/2 = −e

ξ =

= −ı.

We can now calculate the integral, ez/2 dz z C e +1 = (2ıπ)ξ

I =

I

= (2ıπ)(−ı) = 2π. (g) In order to formalize and clarify what we just did, we calculate the initial part of the Laurent series of the integrand with respect to the point z0 = ıπ. In its complex form, the integrand can be written as f (z) = = = =

ez/2 ez + 1 1 ez/2 + e−z/2 1 2 cosh(z/2) ∞ X an (z − ıπ)n ,

n=−∞

where we wrote the general form of the representation in terms of the Laurent series. Recall now that, as shown above, the limit ez/2 = lim (z − ıπ)f (z) lim (z − ıπ) z z→ıπ e + 1 z→ıπ

277

CALCULATION OF INTEGRALS BY RESIDUES

exists, is finite and non-zero. It follows from this fact that the largest negative power present in the Laurent series of f (z) is n = −1, because if there were a negative power greater than this, the limit would not be finite. In other words, the existence of this limit shows that the pole is of the first order, that is, a simple pole. Thus, we can write for f (z) f (z) = = g(z) = = =

g(z) z − ıπ ∞ X an (z − ıπ)n

n=−1 ∞ X

n=−1 ∞ X

n=0 ∞ X

n=0

⇒

an (z − ıπ)n+1

an−1 (z − ıπ)n

bn (z − ıπ)n ,

where bn = an−1 are now the coefficients of the power series around z0 = ıπ which represents g(z), that we now realize is a Taylor series, since g(z) is an analytic function at z0 , if we define it at that point by continuity. We have, moreover, that g(z) = (z − ıπ)f (z) (z − ıπ) = , 2 cosh(z/2) so that we can construct the Taylor series of g(z), calculating its derivatives at the point z0 , which in each case will involve the calculation of a limit for z → z0 . However, before doing this it is interesting to determine the parity properties of this function, with respect to the variable z ′ = z−ıπ. Let us start by determining the parity of cosh(z/2). Making z = z ′ + ıπ we have for this function 2 cosh(z ′ /2 + ıπ/2) = =

′

′

′

′

ez /2+ıπ/2 + e−z /2−ıπ/2 ez /2 eıπ/2 + e−z /2 e−ıπ/2 ′

′

= ı ez /2 − ı e−z /2

= 2ı sinh(z ′ /2).

278

SOLUTIONS 15 Thus we see that cosh(z/2) is an odd function with respect to the variable z ′ , and therefore so is f (z). Since we have that g(z) = z ′ f (z), it follows that g(z) is an even function of z ′ , so that the Taylor series of g(z) around z0 = ıπ only has the even powers, g(z) =

∞ X k=0

b2k (z − ıπ)2k .

Writing the function and its series in terms of z ′ , in order to facilitate the derivation of the coefficients, we have g z′




z′ 2ı sinh(z ′ /2) ∞ X
2k . b2k z ′ =

=

k=0

The first coefficient is simply b0 = a−1 , that is the residue of the pole and that, as we have already seen, can be obtained by the limit lim g z′ ′

z →0




z′ z →0 2ı sinh(z ′ /2) 1 = lim z ′ →0 ı cosh(z ′ /2) 1 = ı = −ı, =

lim ′

where we used the l’Hˆopital rule. Calculating the first derivative of g(z ′ ) we have g′ z ′




= =

1 z ′ cosh(z ′ /2) − 2ı sinh(z ′ /2) 4ı sinh2 (z ′ /2) 2 sinh(z ′ /2) − z ′ cosh(z ′ /2) . 4ı sinh2 (z ′ /2)

If we now calculate the z ′ → 0 limit of the derivative, which is once more indeterminate, we have

CALCULATION OF INTEGRALS BY RESIDUES lim g′ z ′ ′

z →0




= = = = =

279

2 sinh(z ′ /2) − z ′ cosh(z ′ /2) z →0 4ı sinh2 (z ′ /2) cosh(z ′ /2) − cosh(z ′ /2) − z ′ sinh(z ′ /2)/2 lim z ′ →0 4ı sinh(z ′ /2) cosh(z ′ /2) ′ −z sinh(z ′ /2)/2 lim z ′ →0 4ı sinh(z ′ /2) cosh(z ′ /2) −z ′ lim z ′ →0 8ı cosh(z ′ /2) 0, lim ′

as predicted, and where we used once again the l’Hˆopital rule. Calculating the second derivative of g(z) we have g′′ z ′




cosh(z ′ /2) − cosh(z ′ /2) − z ′ sinh(z ′ /2)/2 + 4ı sinh2 (z ′ /2) 2 sinh(z ′ /2) − z ′ cosh(z ′ /2) cosh(z ′ /2) −2 2 4ı sinh3 (z ′ /2)

=

−z ′ sinh2 (z ′ /2) 4 sinh(z ′ /2) cosh(z ′ /2) − 2z ′ cosh2 (z ′ /2) − 8ı sinh3 (z ′ /2) 8ı sinh3 (z ′ /2) −z ′ sinh2 (z ′ /2) − 4 sinh(z ′ /2) cosh(z ′ /2) + 2z ′ cosh2 (z ′ /2) . 8ı sinh3 (z ′ /2)

= =

If we now calculate the z ′ → 0 limit of the second derivative, which is once more indeterminate, we have lim g′′ z ′ ′

z →0




−z ′ sinh2 (z ′ /2) − 4 sinh(z ′ /2) cosh(z ′ /2) + 2z ′ cosh2 (z ′ /2) z →0 8ı sinh3 (z ′ /2)   − sinh2 (z ′ /2) −z ′ sinh(z ′ /2) cosh(z ′ /2) +  −2 cosh2 (z ′ /2) −2 sinh2 (z ′ /2) +  2 ′ ′ ′ ′ +2 cosh (z /2) +2z cosh(z /2) sinh(z /2) + = lim z ′ →0 12ı sinh2 (z ′ /2) cosh(z ′ /2) − sinh(z ′ /2) − z ′ cosh(z ′ /2) − 2 sinh(z ′ /2) + 2z ′ cosh(z ′ /2) = lim z ′ →0 12ı sinh(z ′ /2) cosh(z ′ /2)   − cosh(z ′ /2)/2 − cosh(z ′ /2) −z ′ sinh(z ′ /2)/2 + − cosh(z ′ /2) +2 cosh(z ′ /2) +z ′ sinh(z ′ /2) + = lim 2 2 z ′ →0 6ı cosh (z ′ /2) + 6ı sinh (z ′ /2) =

lim ′

280

SOLUTIONS 15 − cosh(z ′ /2)/2 + z ′ sinh(z ′ /2)/2 z →0 6ı cosh2 (z ′ /2) + 6ı sinh2 (z ′ /2) −1/2 = 6ı ı , = 12 =

lim ′

where we used the l’Hˆopital rule twice. Therefore we have the first two non-zero terms of the Taylor series of g(z) around z0 = ıπ, g(z) = −ı +

ı (z − ıπ)2 + . . . , 24

and it finally follows that we have the first two non-zero terms of the Laurent series of f (z) around z0 = ıπ, f (z) = −

ı ı + (z − ıπ) + . . . , z − ıπ 24

where the residue ξ = −ı is now in clear view.

Solutions 16

Residues on Riemann Surfaces We present here complete and commented solutions to all problems proposed in Chapter 16 of the text. For reference, the propositions of the problems are repeated here. The problems are discussed in the order in which they were proposed within the problem set of that chapter. Problem 1. text,

Consider the first complex integral that was discussed in the

I¯ =

I

C

dz

√

z , z2 + 1

on the closed contour C defined in the text. (a) Show with all rigor that the part of the integral over the segment Cε of the contour is zero in the limit ε → 0. (b) Show with all rigor that the part of the integral over the segment CR of the contour is zero in the limit R → ∞. Hint: take absolute values and use the triangle inequalities. Complete Solution: The complex integral to be discussed is √ I z . dz 2 I¯ = z +1 C 281

282

SOLUTIONS 16

(a) On the segment Cε of the contour we have I¯Cε =

Z

√

2π

dz 0

z , +1

z2

where z = ε exp(ıθ) and thus dz = ız dθ. Taking absolute values and using the triangle inequalities we have 2π

Z

I¯Cε ≤

|dz|

0

√ | z| , |z 2 + 1|

√ √ where |dz| = εdθ and | z| = ε. For the denominator we can write, using once again the triangle inequalities, 1 = (z 2 + 1) − z 2 ⇒ 1 ≤ z 2 + 1 + z 2 ⇒ 2 z + 1 ≥ 1 − ε2 .

We can therefore upper-bound the integrand and write I¯Cε ≤

Z

2π

dθ ε 0

√

ε . 1 − ε2

Taking now the ε → 0 limit we obtain lim I¯Cε ≤

ε→0

Z

2π

dθ 0



ε3/2 . ε→0 1 − ε2 lim

Since the integral that remains is limited and the limit shown is zero, we have that lim I¯Cε = 0,

ε→0

and therefore that

lim I¯Cε = 0,

ε→0

which is the required result.

283

RESIDUES ON RIEMANN SURFACES (b) On the segment CR of the contour we have I¯CR =

Z

√ z , dz 2 z +1

2π 0

where z = R exp(ıθ) and thus dz = ız dθ. Taking absolute values and using the triangle inequalities we have I¯C ≤ R

Z

2π 0

√ | z| |dz| 2 , |z + 1|

√ √ where |dz| = Rdθ and | z| = R. For the denominator we can write, using once again the triangle inequalities, z 2 = (z 2 + 1) − 1 2 z ≤ z 2 + 1 + 1 2 z + 1 ≥ R2 − 1.

⇒

⇒

We can therefore upper-bound the integrand and write I¯C ≤ R

Z

2π

dθ R 0

√

R . R2 − 1

Taking now the R → ∞ limit we obtain lim I¯CR ≤

R→∞

Z

2π

dθ 0



R3/2 . R→∞ R2 − 1 lim

Since the integral that remains is limited and the limit shown is zero, we have that lim I¯CR = 0,

R→∞

and therefore that

lim I¯CR = 0,

R→∞

which is the required result.

284

SOLUTIONS 16

Problem 2. the text,

Consider the second complex integral that was discussed in

I¯ =

I

dz C

ln(z) , 1 + z + z2

on the closed contour C defined in the text. (a) Show with all rigor that the part of the integral over the segment Cε of the contour is zero in the limit ε → 0. (b) Show with all rigor that the part of the integral over the segment CR of the contour is zero in the limit R → ∞. Hint: take absolute values and use the triangle inequalities. Complete Solution: The complex integral to be discussed is I ln(z) dz I¯ = . 1 + z + z2 C (a) On the segment Cε of the contour we have I¯Cε =

Z

2π

dz 0

ln(z) , 1 + z + z2

where z = ε exp(ıθ) and thus dz = ız dθ. Taking absolute values and using the triangle inequalities we have I¯Cε ≤

Z

2π 0

|dz|

|ln(z)| , |1 + z + z 2 |

where |dz| = εdθ. For the logarithm we can write, using once again the triangle inequalities, ln(z) = ln(ε) + ıθ |ln(z)| ≤ |ln(ε)| + θ.

⇒

For the denominator we can write, using yet again the triangle inequalities,

285

RESIDUES ON RIEMANN SURFACES 1 = (1 + z + z 2 ) − z − z 2 ⇒ 1 ≤ 1 + z + z 2 + |z| + z 2 ⇒ 1 + z + z 2 ≥ 1 − ε − ε2 .

We can therefore upper-bound the integrand and write 2π

|ln(ε)| + θ 1 − ε − ε2 0 Z 2π Z 2π ε ε |ln(ε)| dθ θ + . dθ = 2 1−ε−ε 1 − ε − ε2 0 0

I¯Cε ≤

Z

dθ ε

Taking now the ε → 0 limit we obtain lim I¯Cε ≤

ε→0

Z

2π

0



ε |ln(ε)| dθ lim + ε→0 1 − ε − ε2

Z

2π

dθ θ 0



ε . ε→0 1 − ε − ε2 lim

Since the integrals that remain are limited and the limits shown are zero, we have that lim I¯Cε = 0,

ε→0

and therefore that

lim I¯Cε = 0,

ε→0

which is the required result. (b) On the segment CR of the contour we have I¯CR =

Z

2π

dz

0

ln(z) , 1 + z + z2

where z = R exp(ıθ) and thus dz = ız dθ. Taking absolute values and using the triangle inequalities we have I¯C ≤ R

Z

0

2π

|dz|

|ln(z)| , |1 + z + z 2 |

286

SOLUTIONS 16 where |dz| = Rdθ. For the logarithm we can write, using once again the triangle inequalities, ln(z) = ln(R) + ıθ |ln(z)| ≤ ln(R) + θ.

⇒

For the denominator we can write, using yet again the triangle inequalities, z 2 = (1 + z + z 2 ) − z − 1 ⇒ 2 z ≤ 1 + z + z 2 + |z| + 1 ⇒ 1 + z + z 2 ≥ R2 − R − 1.

We can therefore upper-bound the integrand and write 2π

ln(R) + θ R2 − R − 1 0 Z 2π Z 2π R ln(R) R dθ 2 = dθ θ 2 + . R −R−1 R −R−1 0 0

I¯C ≤ R

Z

dθ R

Taking now the R → ∞ limit we obtain lim I¯CR ≤

R→∞

Z

2π



R ln(R) + lim 2 R→∞ R − R − 1 0  Z 2π R dθ θ lim 2 + . R→∞ R − R − 1 0 dθ

Since the integrals that remain are limited and the limits shown are zero, we have that lim I¯CR = 0,

R→∞

and therefore that

lim I¯CR = 0,

R→∞

which is the required result.

287

RESIDUES ON RIEMANN SURFACES Problem 3. text,

Consider the third complex integral that was discussed in the

I¯ =

I

dz

C

zα , +1

z2

with 0 < α < 1, on the closed contour C defined in the text. (a) Show with all rigor that the part of the integral over the segment Cε of the contour is zero in the limit ε → 0. (b) Show with all rigor that the part of the integral over the segment CR of the contour is zero in the limit R → ∞. Hint: take absolute values and use the triangle inequalities. Complete Solution: The complex integral to be discussed is I zα dz 2 I¯ = , z +1 C where 0 < α < 1. (a) On the segment Cε of the contour we have I¯Cε =

Z

2π

dz 0

zα , z2 + 1

where z = ε exp(ıθ) and thus dz = ız dθ. Taking absolute values and using the triangle inequalities we have I¯Cε ≤

Z

0

2π

|dz|

|z α | , + 1|

|z 2

where |dz| = εdθ and |z α | = εα . For the denominator we can write, using once again the triangle inequalities, 1 = (z 2 + 1) − z 2 ⇒ 1 ≤ z 2 + 1 + z 2 ⇒ 2 z + 1 ≥ 1 − ε2 .

288

SOLUTIONS 16 We can therefore upper-bound the integrand and write I¯Cε ≤

Z

2π

dθ ε 0

εα . 1 − ε2

Taking now the ε → 0 limit we obtain lim I¯Cε ≤

ε→0

Z

2π



dθ 0

ε1+α . ε→0 1 − ε2 lim

Since the integral that remains is limited and the limit shown is zero, we have that lim I¯Cε = 0,

ε→0

and therefore that

lim I¯Cε = 0,

ε→0

which is the required result. (b) On the segment CR of the contour we have I¯CR =

Z

2π

dz 0

zα , z2 + 1

where z = R exp(ıθ) and thus dz = ız dθ. Taking absolute values and using the triangle inequalities we have I¯C ≤ R

Z

2π 0

|dz|

|z α | , |z 2 + 1|

where |dz| = Rdθ and |z α | = Rα . For the denominator we can write, using once again the triangle inequalities, z 2 = (z 2 + 1) − 1 2 z ≤ z 2 + 1 + 1 2 z + 1 ≥ R2 − 1.

⇒

⇒

289

RESIDUES ON RIEMANN SURFACES We can therefore upper-bound the integrand and write I¯C ≤ R

Z

2π

dθ R 0

Rα . −1

R2

Taking now the R → ∞ limit we obtain Z

lim I¯CR ≤

R→∞

2π

dθ 0



R1+α . R→∞ R2 − 1 lim

Since the integral that remains is limited and the limit shown is zero, given that α < 1, we have that lim I¯CR = 0,

R→∞

and therefore that

lim I¯CR = 0,

R→∞

which is the required result. Problem 4.

Calculate by residues the real asymptotic integral given by Z ∞ α + βx dx I= , 1 + x + x2 + x3 0

for arbitrary real α and β. Justify explicitly every step of the procedure. Answer: π(α + β)/4. Complete Solution: We must calculate the integral Z ∞ dx I= 0

α + βx . 1 + x + x2 + x3

The denominator can be easily factored, for example by using the formula for the sum of a geometric progression, resulting in
1 + x + x2 + x3 = (x + 1) x2 + 1 = (x + 1)(x − ı)(x + ı).

290

SOLUTIONS 16

Since there is in the integrand no singularity of the branch-point type, we consider the complex integral (α + βz) ln(z) 1 + z + z2 + z3 IC (α + βz) ln(z) dz , = (z + 1)(z − ı)(z + ı) C

I¯ =

I

dz

on the closed contour C described in the text. Let us now discuss the two arcs. On the segment Cε we have z = ε exp(ıθ) and thus dz = ıε exp(ıθ)dθ, and therefore we obtain I¯Cε

= ≈

Z

Z

2π

dz 0

(α + βz) ln(z) 1 + z + z2 + z3

2π

dz (α + βz) ln(z) 0

Z

2π

  dθ ε eıθ α + βε eıθ [ln(ε) + ıθ] 0 Z 2π  Z 2π  dθ eıθ ε ln(ε) − α = ıα dθ θ eıθ ε + 0 0  Z 2π  Z 2π 2ıθ 2ıθ 2 dθ e dθ θ e ε ln(ε) − β ε2 . +ıβ

= ı

0

0

Taking now the ε → 0 limit we obtain lim I¯Cε

ε→0

2π

Z 2π  dθ θ eıθ lim ε + lim ε ln(ε) − α ε→0 ε→0 0 0 Z 2π  Z 2π  2ıθ 2ıθ 2 dθ e dθ θ e +ıβ lim ε ln(ε) − β lim ε2 .

= ıα

Z

dθ eıθ

0



ε→0

0

ε→0

The remaining integrals are all limited, and the limits shown are zero, so that we have lim I¯Cε = 0.

ε→0

On the segment CR we have z = R exp(ıθ) and thus dz = ıR exp(ıθ)dθ, and thus we obtain

RESIDUES ON RIEMANN SURFACES I¯CR

291

2π

(α + βz) ln(z) 1 + z + z2 + z3 0 Z 2π (α + βz) ln(z) dz ≈ z3 0
Z 2π ıθ [ln(R) + ıθ] ıθ α + βR e dθ R e = ı R3 e3ıθ 0 Z 2π  Z 2π  1 −2ıθ −2ıθ ln(R) dθ θ e −α + dθ e = ıα 2 2 R R 0 0 Z 2π Z 2π   ln(R) 1 dθ e−ıθ dθ θ e−ıθ +ıβ −β . R R 0 0 =

Z

dz

Taking now the R → ∞ limit we obtain lim I¯CR

R→∞

Z 2π  1 ln(R) −2ıθ dθ θ e −α lim 2 + dθ e = ıα lim 2 R→∞ R R→∞ R 0 0 Z 2π  Z 2π  ln(R) 1 dθ e−ıθ lim dθ θ e−ıθ lim +ıβ −β . R→∞ R→∞ R R 0 0 Z

2π

−2ıθ



The remaining integrals are all limited, and the limits shown are zero, so that we have lim I¯CR = 0.

R→∞

The discussion of the two straight segments of the contour is exactly the same as the one that was made explicitly in the text in a case similar to this one, from which it results that we have for the real integral I=

I¯ . −2πı

All that remains to be done here is the calculation of I¯ by means of the residue theorem. Since the relevant singularities are three simple poles, we obtain for each of them, by the method of the limit, (α + βz) ln(z) z→−1 (z + 1)(z − ı)(z + ı) (α + βz) ln(z) = lim z→−1 (z − ı)(z + ı)

b1⊖1 =

lim (z + 1)

292

SOLUTIONS 16 = = = b1⊕ı = = = = = = = b1⊖ı = = = = = = =

(α − β) ln(−1) (−1 − ı)(−1 + ı) (α − β) ln[exp(ıπ)] (1 + ı)(1 − ı) ıπ(α − β) , 2 (α + βz) ln(z) lim (z − ı) z→ı (z + 1)(z − ı)(z + ı) (α + βz) ln(z) lim z→ı (z + 1)(z + ı) (α + ıβ) ln(ı) (ı + 1)(ı + ı) (α + ıβ) ln[exp(ıπ/2)] (1 + ı)2ı ıπ(α + ıβ) 4ı(1 + ı) π(α + ıβ)(1 − ı) 8 π[(α + β) − ı(α − β)] , 8 (α + βz) ln(z) lim (z + ı) z→−ı (z + 1)(z − ı)(z + ı) (α + βz) ln(z) lim z→−ı (z + 1)(z − ı) (α − ıβ) ln(−ı) (−ı + 1)(−ı − ı) (α − ıβ) ln[exp(3ıπ/2)] − (1 − ı)2ı 3ıπ(α − ıβ) − 4ı(1 − ı) 3π(α − ıβ)(1 + ı) − 8 3π[(α + β) + ı(α − β)] . − 8

Thus we have for the integral I¯ I¯ = 2πı(b1⊖1 + b1⊕ı + b1⊖ı )

RESIDUES ON RIEMANN SURFACES

293



ıπ(α − β) + 2  π[(α + β) − ı(α − β)] 3π[(α + β) + ı(α − β)] + − 8 8      1 3 1 1 3 − − − + π(α + β) = 2πı ıπ(α − β) 2 8 8 8 8 α+β . = −π 2 ı 2 = 2πı

We have therefore for the original real integral I = = =

I¯ −2πı −π 2 ı α + β −2πı 2 π(α + β) . 4

As expected, this is a real result, which is positive for positive α and β, a fact that serves as a partial confirmation of the calculations. We can make another partial verification of the correctness of the result since in the case α = 1, β = 1 this integral reduces to one which was calculated in the text of the previous chapter, and that actually has the value π/2 as is the case here. Problem 5.

Calculate by residues the real asymptotic integral given by Z ∞ xα , dx I= (x + 1)2 0

for a real constant α in the open interval (0, 1). Justify explicitly every step of the procedure. Answer: πα/[sin(πα)]. Complete Solution: We must calculate the integral I

∞

xα 1 + 2x + x2 Z0 ∞ xα dx = . (x + 1)2 0 =

Z

dx

294

SOLUTIONS 16

Since the integrand has a branch point at z = 0, we consider the complex integral I zα dz I¯ = 1 + 2z + z 2 IC zα dz = (z + 1)2 C I eα ln(z) , dz = (z + 1)2 C

on the closed contour C described in the text. Note that here we have a double pole. Let us now discuss the two arcs. On the segment Cε we have z = ε exp(ıθ) and thus dz = ıε exp(ıθ)dθ, and therefore we obtain Z 2π zα dz I¯Cε = 1 + 2z + z 2 0 Z 2π dz z α ≈ 0 Z 2π dθ ε eıθ εα eıαθ = ı 0 Z  2π ı(1+α)θ dθ e = ı ε1+α . 0

Taking now the limit we obtain Z lim I¯Cε = ı ε→0

2π

dθ e

ı(1+α)θ

0



lim ε1+α .

ε→0

The remaining integral is limited and the limit shown is zero, so that we have lim I¯Cε = 0.

ε→0

On the segment CR we have z = R exp(ıθ) and thus dz = ıR exp(ıθ)dθ, and therefore we obtain Z 2π zα dz I¯CR = 1 + 2z + z 2 0 Z 2π zα ≈ dz 2 z 0 Z 2π Rα eıαθ dθ R eıθ 2 2ıθ = ı R e 0 Z  2π 1 −ı(1−α)θ dθ e . = ı 1−α R 0

295

RESIDUES ON RIEMANN SURFACES Taking now the limit we obtain Z ¯ lim ICR = ı R→∞

2π

−ı(1−α)θ

dθ e

0



lim

1

R→∞ R1−α

.

The remaining integral is limited and the limit shown is zero, since 0 < α < 1, so that we have lim I¯CR = 0.

R→∞

The discussion of the two straight segments of the contour is exactly the same as the one that was made explicitly in the text in a case similar to this one, from which it results that we have for the real integral I=

I¯ . 1 − e2ıπα

All that remains to be done here is the calculation of I¯ by means of the residue theorem. Since the relevant singularity is a pole of order two, we consider the function φ(z) given by φ(z) = (z + 1)2

zα (z + 1)2

= zα. We have therefore that the residue b1 is the Taylor coefficient of order 2−1 = 1 of this function, around the point z = −1, that is, we have that  dz α b1 = dz −1 Calculating the derivative we obtain b1 = αz

α−1



−1

(α−1) ln(z)

= αe



(α−1) ln(−1)

= αe

−1

= α e(α−1) ln[exp(ıπ)] = α e(α−1)ıπ = α e−ıπ eıπα = −α eıπα .

296

SOLUTIONS 16

Thus we have for the integral I¯ I¯ = −2ıπα eıπα . It follows therefore that we have for the original real integral I = = = =

I¯ 1 − e2ıπα −2ıπα eıπα 1 − e2ıπα 2ıπα ıπα e − e−ıπα πα . sin(πα)

As expected, this is a real, positive and finite result, since 0 < α < 1, which serves as a partial confirmation of the calculations. Furthermore, in the case α = 0 the result above exists and is equal to one, as we can see by taking the limit α → 0. In this case the integral can be done by elementary means, resulting in fact the value one, which once again confirms the correctness of the result. Problem 6.

Calculate by residues the real asymptotic integral given by Z ∞ xα , dx I= 1 + βx + x2 0

for real constants 0 < α < 1 and 0 ≤ β ≤ 2. Justify explicitly every step of the procedure. Discuss in particular the cases β = 0 and β = 2. Answer: sin[α(π − θ⊕ )] , I = π sin(πα) sin(θ⊕ ) p sin(θ⊕ ) = 1 − (β/2)2 ,

cos(θ⊕ ) = −β/2. Complete Solution: We must calculate the integral Z I=

0

∞

dx

xα . 1 + βx + x2

RESIDUES ON RIEMANN SURFACES

297

Since the integrand has a branch point at z = 0, we consider the complex integral I zα dz I¯ = 1 + βz + z 2 C I eα ln(z) dz = 1 + βz + z 2 C I eα ln(z) dz , = (z − z⊕ )(z − z⊖ ) C in the closed contour C described in the text. By the Baskara formula the two roots of the polynomial in the denominator are given by p −β ± β 2 − 4 z± = 2p −β ± ı 4 − β 2 ⇒ = 2s  2 β β , z⊕ = − + ı 1 − 2 2 s  2 β β z⊖ = − − ı 1 − , 2 2 since 0 ≤ β ≤ 2. Note that the two roots, and therefore the two singularities of the integrand, are located on the unit circle, with the same negative real parts and imaginary parts of opposite signs. Let us now discuss the two circular segments of the contour. On the segment Cε we have z = ε exp(ıθ) and thus dz = ıε exp(ıθ)dθ, and therefore we obtain Z 2π zα dz I¯Cε = 1 + βz + z 2 0 Z 2π dz z α ≈ 0 Z 2π dθ ε eıθ εα eıαθ = ı 0 Z 2π  ı(1+α)θ dθ e = ı ε1+α . 0

Taking now the ε → 0 limit we obtain

298

SOLUTIONS 16

lim I¯Cε = ı

ε→0

Z

2π

dθ e

ı(1+α)θ

0



lim ε1+α .

ε→0

The remaining integral is limited and the limit shown is zero, so that we have lim I¯Cε = 0.

ε→0

On the segment CR we have z = R exp(ıθ) and thus dz = ıR exp(ıθ)dθ, and thus we obtain Z 2π zα ¯ dz ICR = 1 + βz + z 2 0 Z 2π zα ≈ dz 2 z 0 Z 2π Rα eıαθ dθ R eıθ 2 2ıθ = ı R e 0 Z  2π 1 −ı(1−α)θ . dθ e = ı 1−α R 0 Taking now the R → ∞ limit we obtain Z 2π  dθ e−ı(1−α)θ lim lim I¯CR = ı R→∞

1

R→∞ R1−α

0

.

The remaining integral is limited and the limit shown is zero, since 0 < α < 1, so that we have lim I¯CR = 0.

R→∞

The discussion of the two straight segments of the contour is exactly the same as the one that was made explicitly in the text in a case similar to this one, from which it results that we have for the real integral I=

I¯ . 1 − e2ıπα

All that remains to be done here is the calculation of I¯ by means of the residue theorem. In the case β = 2 we have a double pole whose residue was already calculated in the previous problem, b1 = −α exp(ıπα). In any

299

RESIDUES ON RIEMANN SURFACES

other case we have two simple poles and we can calculate the residues by the method of the limit, thus obtaining b1⊕ = = = b1⊖ = = =

lim (z − z⊕ )

z→z⊕

lim

z→z⊕ α z⊕

zα (z − z⊖ )

z⊕ − z⊖

,

lim (z − z⊖ )

z→z⊖

lim

z→z⊖ α z⊖

zα (z − z⊕ )

z⊖ − z⊕

zα (z − z⊕ )(z − z⊖ )

zα (z − z⊕ )(z − z⊖ )

.

We have therefore for the sum of the two relevant residues for the calculation ¯ of the integral I, b1⊕ + b1⊖ = =

α − zα z⊕ ⊖ z⊕ − z⊖ eα ln(z⊕ ) − eα ln(z⊖ ) . z⊕ − z⊖

We now observe that we can write z⊕ as eıθ⊕ , where s  2 β , sin(θ⊕ ) = 1− 2 β cos(θ⊕ ) = − , 2 and that for z⊖ , which is the complex conjugate of z⊕ , we can write eıθ⊖ , where s  2 β sin(θ⊖ ) = − 1 − , 2 β cos(θ⊖ ) = − . 2 ∗ is equivalent to the relation θ = 2π − θ between The fact that z⊖ = z⊕ ⊖ ⊕ the angles, so that we can write for the sum of the residues

300

SOLUTIONS 16 b1⊕ + b1⊖ = = = = = =

eıαθ⊕ − eıαθ⊖ eıθ⊕ − eıθ⊖ eıαθ⊕ − eıα(2π−θ⊕ ) eıθ⊕ − eı(2π−θ⊕ ) ıαθ e ⊕ − e2ıπα e−ıαθ⊕ eıθ⊕ − e−ıθ⊕ e−ıπα eıαθ⊕ − eıπα e−ıαθ⊕ eıπα eıθ⊕ − e−ıθ⊕ ıα(θ −π) ⊕ e − e−ıα(θ⊕ −π) eıπα eıθ⊕ − e−ıθ⊕ sin[α(π − θ⊕ )] . − eıπα sin(θ⊕ )

Note that in the limit β → 2, which corresponds to θ⊕ → π, in which we have sin(θ⊕ ) ≈ π − θ⊕ ,

sin[α(π − θ⊕ )] ≈ α(π − θ⊕ ), we obtain for this sum of residues sin[α(π − θ⊕ )] β→2 sin(θ⊕ ) α(π − θ⊕ ) lim β→2 (π − θ⊕ )

lim (b1⊕ + b1⊖ ) = − eıπα lim

β→2

= − eıπα

= −α eıπα = b1 .

Thus we see that in this limit, in which the two simple poles are united at the position z = −1, we recover the residue of the double pole b1 calculated in the previous problem. Returning to the general case we have for the integral I¯ sin[α(π − θ⊕ )] . I¯ = −2ıπ eıπα sin(θ⊕ ) It follows therefore that we have for the original real integral I¯ 1 − e2ıπα eıπα sin[α(π − θ⊕ )] = −2ıπ 1 − e2ıπα sin(θ⊕ )

I =

RESIDUES ON RIEMANN SURFACES

301

sin[α(π − θ⊕ )] 2ı ıπα −e sin(θ⊕ ) sin[α(π − θ⊕ )] = π . sin(πα) sin(θ⊕ )

= −π

e−ıπα

As expected, this is a real and positive result and, since 0 < α < 1 and 0 ≤ β ≤ 2, implying that π/2 ≤ θ⊕ ≤ π. This serves as partial confirmation of the calculations. These inequalities also imply that the result is obviously finite in all cases except in the case β = 2, which corresponds to θ⊕ = π. In this particular case we consider again the limit in which β → 2 and thus θ⊕ → π, in which we have, as we have seen before, sin[α(π − θ⊕ )] β→2 sin(πα) sin(θ⊕ ) α(π − θ⊕ ) = π lim β→2 sin(πα)(π − θ⊕ ) πα = . sin(πα)

lim I = π lim

β→2

This is in fact the same result that was obtained in the previous problem for this case. The case in which β = 0 and thus θ⊕ = π/2 can be calculated directly, and we obtain in this case I

sin(απ/2) sin(πα) sin(απ/2) = π 2 sin(απ/2) cos(απ/2) π . = 2 cos(απ/2)

= π

This is in fact the same result that was obtained in the text for this case. Problem 7.

Calculate by residues the real asymptotic integral given by Z ∞ xα dx I= , 1 + 2x2 + x4 0

for a real constant α in the open interval (0, 3). Justify explicitly every step of the procedure. Answer: −π(α − 1)/[4 cos(πα/2)]. Complete Solution:

302

SOLUTIONS 16

We must calculate the integral Z I=

∞

dx

0

xα . 1 + 2x2 + x4

Since the integrand has a branch point at z = 0, we consider the complex integral I zα ¯ dz I = 1 + 2z 2 + z 4 C I eα ln(z) dz = (1 + z 2 )2 C I eα ln(z) , dz = (z − ı)2 (z + ı)2 C in the closed contour C described in the text, where we show the two double poles of the integrand. Let us now discuss the two circular segments of the contour. On the segment Cε we have z = ε exp(ıθ) and thus dz = ıε exp(ıθ)dθ, and thus we obtain Z 2π zα ¯ dz ICε = 1 + 2z 2 + z 4 0 Z 2π dz z α ≈ 0 Z 2π dθ ε eıθ εα eıαθ = ı 0 Z  2π ı(1+α)θ dθ e = ı ε1+α . 0

Taking now the ε → 0 limit we obtain Z 2π  ı(1+α)θ dθ e lim I¯Cε = ı lim ε1+α . ε→0

0

ε→0

The remaining integral is limited and the limit shown is zero, so that we have lim I¯Cε = 0.

ε→0

On the segment CR we have z = R exp(ıθ) and thus dz = ıR exp(ıθ)dθ, and thus we obtain

303

RESIDUES ON RIEMANN SURFACES I¯CR

2π

zα 1 + 2z 2 + z 4 0 Z 2π zα dz 4 ≈ z 0 Z 2π Rα eıαθ dθ R eıθ 4 4ıθ = ı R e 0 Z  2π 1 dθ e−ı(3−α)θ = ı . 3−α R 0

=

Z

dz

Taking now the R → ∞ limit we obtain Z 2π  dθ e−ı(3−α)θ lim lim I¯CR = ı R→∞

R→∞

0

1 . R3−α

The remaining integral is limited and the limit shown is zero, since 0 < α < 3, so that we have lim I¯CR = 0.

R→∞

The discussion of the two straight segments of the contour is exactly the same as the one that was made explicitly in the text in a case similar to this one, from which it results that we have for the real integral I=

I¯ . 1 − e2ıπα

All that remains to be done here is the calculation of I¯ by means of the residue theorem. Since the two relevant singularities are poles of order two, we consider in the case of the pole z⊕ = ı the function φ⊕ (z) given by φ⊕ (z) = (z − ı)2 =

zα (z − ı)2 (z + ı)2

zα . (z + ı)2

We have then that the residue b1⊕ is the Taylor coefficient of order 2 − 1 = 1 of this function, around the point z = ı, that is, we have that  zα d . b1⊕ = dz (z + ı)2 ı Calculating the derivative we obtain

304

SOLUTIONS 16 b1⊕ = = =



"

αz α−1 2z α − (z + ı)2 (z + ı)3 α e(α−1) ln(z) (z + ı)2

−



ı#

2 eα ln(z) (z + ı)3



ı

α e(α−1) ln[exp(ıπ/2)] 2 eα ln[exp(ıπ/2)] − (2ı)2 (2ı)3

eıπα/2 α eıπ(α−1)/2 −ı = − 4 4 ı ıπα/2 = (α − 1) e . 4 In the case of the pole z⊖ = −ı we consider the function φ⊖ (z) given by φ⊖ (z) = (z + ı)2 =

zα (z − ı)2 (z + ı)2

zα . (z − ı)2

We have then that the residue b1⊖ is the Taylor coefficient of order 2 − 1 = 1 of this function, around the point z = −ı, that is, we have that  d zα b1⊖ = . dz (z − ı)2 −ı Calculating the derivative we obtain b1⊖ = = =



"

2z α αz α−1 − (z − ı)2 (z − ı)3 α e(α−1) ln(z) (z − ı)2

−



−ı

2 eα ln(z) (z − ı)3

#

−ı

α e(α−1) ln[exp(3ıπ/2)] 2 eα ln[exp(3ıπ/2)] − 2 (−2ı) (−2ı)3

α e3ıπ(α−1)/2 e3ıπα/2 +ı 4 4 ı 3ıπα/2 . = − (α − 1) e 4

= −

We have therefore for the sum of the two residues that must be used to calculate the integral I¯

305

RESIDUES ON RIEMANN SURFACES i ıh (α − 1) eıπα/2 − (α − 1) e3ıπα/2 4 i ı(α − 1) eıπα h −ıπα/2 = e − eıπα/2 4  πα  (α − 1) eıπα = . sin 2 2 Thus we have for the integral I¯ b1⊕ + b1⊖ =

I¯ = 2ıπ(b1⊕ + b1⊖ ) = ıπ(α − 1) e

ıπα

sin

 πα  2

.

It follows therefore that we have for the original real integral I =

I¯ 1 − e2ıπα

 πα  eıπα sin 1 − e2ıπα 2  πα  1 sin ıπ(α − 1) −ıπα e  − eıπα 2 πα π(α − 1) sin 2 − 2 sin(πα)  πα  sin π(α − 1)  πα  2  πα  − 4 sin cos 2 2 π(α − 1)  πα  . − 4 cos 2

= ıπ(α − 1) = =

=

=

As expected, this is a real result. In addition, although not obvious at first glance, it can be verified that it is positive for 0 < α < 3. In order to do this, we separate this interval into two parts, one in which 0 < α < 1 and another in which 1 < α < 3. The point α = 1 needs to be considered separately, in which case both the numerator and the denominator vanish. For 0 < α < 1 the numerator is strictly negative and argument of the cosine varies from 0 to π/2, so that the cosine is strictly positive. It follows that the result is strictly positive in this case. For 1 < α < 3 the numerator is strictly positive, and the argument of the cosine varies from π/2 to 3π/2, so that the cosine is strictly negative. It follows once again that the result is strictly positive in this case. In order to examine the case α = 1 we take the limit α → 1, getting

306

SOLUTIONS 16  πα  sin π(α − 1) 2 lim I = − lim α→1 α→1 2 sin(πα) 1 π(α − 1) = − lim α→1 2 (π − πα) 1 . = 2

Again, we have a positive result. It follows that we have a strictly positive result for every value of α in (0, 3). This serves as a partial confirmation of the result, since the real original integral is clearly positive. Problem 8. (Challenge Problem) Calculate by residues the real asymptotic integral given by Z ∞ ln(x) dx 2 I= . x +1 0 Justify explicitly every step of the procedure. Answer: 0. Hint: consider a contour contained in the upper half-plane. Complete Solution: We must calculate the integral Z ∞ ln(x) dx 2 I = x +1 Z0 ∞ ln(x) . dx = (x − ı)(x + ı) 0 Since the integrand has a branch point at z = 0, we consider the complex integral I ln(z) dz 2 I¯ = z +1 IC ln(z) dz , = (z − ı)(z + ı) C in the closed contour C described in Figure 16.1, contained in the upper half-plane. Note that it is not necessary to consider in detail the cut, which should only be taken out of the way.

307

RESIDUES ON RIEMANN SURFACES y

CR

C ı Cε

C⊖

ε

C⊕

R x

−ı

Figure 16.1: The integration contour in the complex plane, showing also the branch cut. Let us now discuss the two semicircular segments. On the segment Cε we have z = ε exp(ıθ) and thus dz = ıε exp(ıθ)dθ, and thus we obtain Z π ln(z) dz 2 I¯Cε = z +1 Z0 π dz ln(z) ≈ 0 Z π dθ ε eıθ [ln(ε) + ıθ] = ı 0 Z π  Z π  ıθ ıθ dθ e dθ θ e = ı ε ln(ε) − ε. 0

0

Taking now the ε → 0 limit we obtain Z π  Z ıθ dθ e lim I¯Cε = ı lim ε ln(ε) − ε→0

0

ε→0

0

π

dθ θ e

ıθ



lim ε.

ε→0

The remaining integrals are limited, and the limits shown are zero, so that we have lim I¯Cε = 0.

ε→0

On the segment CR we have z = R exp(ıθ) and thus dz = ıR exp(ıθ)dθ, and thus we obtain

308

SOLUTIONS 16 I¯CR

π

ln(z) z2 + 1 Z0 π ln(z) dz 2 ≈ z 0 Z π ln(R) + ıθ = ı dθ R eıθ R2 e2ıθ 0Z π  Z π  1 −ıθ ln(R) −ıθ dθ e = ı dθ θ e − . R R 0 0

=

Z

dz

Taking now the R → ∞ limit we obtain Z π  Z π  1 ln(R) −ıθ −ıθ ¯ dθ θ e − . lim dθ e lim ICR = ı lim R→∞ R R→∞ R→∞ R 0 0 The remaining integrals are limited, and the limits shown are zero, so that we have lim I¯CR = 0.

R→∞

Let us now consider the segment C⊕ of the contour. In this case we have z = x and dz = dx, so that we obtain immediately I¯C⊕

R

ln(z) z2 + 1 ε Z R ln(x) dx 2 = . x +1 ε =

Z

dz

In the limit in which ε → 0 and R → ∞ this reduces to our original real integral. Let us now consider the segment C⊖ of the contour. In this case we also have z = x and dz = dx, so that we obtain Z −ε ln(z) ¯ dz 2 IC⊖ = z +1 −R Z −ε ln(x) , dx 2 = x +1 −R where the signs and the order of the extremes are exchanged in relation to our original real integral. Of course, all values of x are negative in this interval, which would be a problem for the real version of the logarithm, but is not a problem here because this is the complex version of this function. Let us make the transformation of variables x → −x, to obtain

309

RESIDUES ON RIEMANN SURFACES I¯C⊖

= −

Z

ε

dx

R R

ln(−x) (−x)2 + 1

ln[x exp(ıπ)] x2 + 1 ε Z R ln(x) + ıπ . = dx x2 + 1 ε

=

Z

dx

Adding all portions of the integral and taking the limit, we therefore have, since the integrals on the semicircles vanish, Z R Z R ln(x) ln(x) + ıπ ¯ dx 2 lim lim IC = dx + ε→0 R→∞ x + 1 x2 + 1 ε ε Z ∞ Z ∞ 1 ln(x) dx 2 + ıπ dx 2 = 2 x +1 x +1 0 0 2 π = 2I + ı , 2 where the value of the additional integral that appears here was calculated in the previous chapter, and turns out to be π/2. We can omit the limits on the left-hand side of the equation, with the understanding that the integral should be calculated by residues, including only the pole located at z = ı. We have therefore for our original real integral I, I=

I¯C π2 −ı . 2 4

All that remains to be done here is the calculation of I¯ by means of the residue theorem. Since the only relevant singularity is a simple pole, we can get the residue by the method of the limit, b1+ = = = = = =

lim (z − ı)

z→ı

ln(z) (z − ı)(z + ı)

ln(z) lim z→ı (z + ı) ln(ı) 2ı ln[exp(ıπ/2)] 2ı ıπ 4ı π . 4

310

SOLUTIONS 16

Thus we have for the integral I¯ π I¯ = 2ıπ 4 2 ıπ = . 2 It follows therefore that we have for the original real integral I¯C π2 −ı 2 4 π2 ıπ 2 −ı = 4 4 = 0.

I =

As expected, this is a real and finite result. There was no need for the result to be positive, since the logarithm changes sign within the integration interval. Here we see that the negative part of the integral in (0, 1) exactly cancels the positive part in (1, ∞).

Bibliography [1] J. L. deLyra, Complex Calculus, vol. 1 of Mathematical Methods for Physics and Engineering. Amazon, first ed., 2018. ISBN-13: 9781793012050.

311

Index Chapter 1, 1 Problem 1, 1 Problem 2, 1 Problem 3, 2 Problem 4, 3 Problem 5, 4 Problem 6, 8 Problem 7, 11 Problem 8, 13 Chapter 2, 15 Problem 1, 15 Problem 2, 16 Problem 3, 18 Problem 4, 22 Problem 5, 25 Problem 6, 26 Problem 7, 27 Problem 8, 28 Problem 9, 30 Problem 10, 32 Chapter 3, 35 Problem 1, 35 Problem 2, 37 Problem 3, 38 Problem 4, 40 Problem 5, 41 Problem 6, 43 Chapter 4, 47 Problem 1, 47 Problem 2, 48 Problem 3, 51 Problem 4, 54

Problem 5, Problem 6, Problem 7, Problem 8, Chapter 5, 63 Problem 1, Problem 2, Problem 3, Problem 4, Problem 5, Chapter 6, 75 Problem 1, Problem 2, Problem 3, Problem 4, Problem 5, Chapter 7, 95 Problem 1, Problem 2, Problem 3, Problem 4, Problem 5, Problem 6, Problem 7, Chapter 8, 109 Problem 1, Problem 2, Problem 3, Problem 4, Problem 5, Problem 6, Problem 7, 312

55 57 59 61 63 64 66 67 71 75 80 81 84 89 95 96 97 99 101 102 106 109 112 113 114 115 117 119

Problem 8, 122 Problem 9, 125 Problem 10, 127 Problem 11, 128 Problem 12, 130 Chapter 9, 137 Problem 1, 137 Problem 2, 138 Problem 3, 142 Problem 4, 143 Problem 5, 144 Problem 6, 146 Problem 7, 148 Chapter 10, 151 Problem 1, 151 Problem 2, 152 Problem 3, 154 Problem 4, 155 Problem 5, 157 Problem 6, 159 Chapter 11, 169 Problem 1, 169 Problem 2, 171 Problem 3, 173 Problem 4, 175 Problem 5, 178 Problem 6, 181 Problem 7, 183 Problem 8, 186 Chapter 12, 189 Problem 1, 189 Problem 2, 190 Problem 3, 192 Problem 4, 194 Problem 5, 196 Problem 6, 201 Problem 7, 205 Chapter 13, 213 Problem 1, 213

Problem 2, 216 Problem 3, 218 Problem 4, 221 Chapter 14, 223 Problem 1, 223 Problem 2, 225 Problem 3, 226 Problem 4, 228 Chapter 15, 231 Problem 1, 231 Problem 2, 234 Problem 3, 237 Problem 4, 240 Problem 5, 246 Problem 6, 251 Problem 7, 256 Problem 8, 261 Problem 9, 263 Problem 10, 266 Problem 11, 269 Problem 12, 271 Chapter 16, 281 Problem 1, 281 Problem 2, 284 Problem 3, 287 Problem 4, 289 Problem 5, 293 Problem 6, 296 Problem 7, 301 Problem 8, 306

313