Complex Analysis: In the Spirit of Lipman Bers, Second Edition [2nd Ed] (Instructor Solution Manual, Solutions) [2, 2nd ed, 2e] 1441973222, 9781441973221

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A solutions manual for Complex Analysis In the Spirit of Lipman Bers by Rub´ı E. Rodr´ıguez, Irwin Kra and Jane Gilman Vamsi P. Pingali

Preface About the manual This manual is intended to accompany the second edition of Complex Analysis in the spirit of Lipman Bers by R. Rodriguez, I. Kra and J. Gilman. It consists of solutions to most of the problems. Some problems involving filling in the details of a proof in the text and those involving the proof of a well known theorem that can be found in other texts have been omitted. Also, in some cases, wherein the procedure is clear and mechanical, an outline of the solution is given (as opposed to a complete one). In order to gain most out of the problems, readers are strongly encouraged to try solving the problems themselves before referring to the manual. Acknowledgements I thank Irwin Kra for delegating this task to me and for helping me out with it. I also thank R. Rodriguez and J. Gilman for their suggestions regarding the problems and otherwise. My appreciation also extends to my adviser Leon Takhtajan, my friends and my family. Contact The author would be most grateful if mistakes (whether typographical or mathematical) are pointed out or if suggestions are given. To do so, please feel free to email the author at [email protected]. The author’s current webpage (where errors shall be posted) is www.math.sunysb.edu/ ∼ vpingli Vamsi Pritham Pingali, Department of Mathematics, Stony Brook university.

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Chapter 2 1. (a) If we fix m, then |zn −zm | < ǫ ∀ n > m−1+ 1ǫ . Hence zn converges to zm as n → ∞. But this is true for all m. Thus, zm is a constant sequence (and can be any constant sequence. We don’t have enough information to discern the value of the constant). (b) Let C be such that |wn | ≤ C. Thus, if n > N , |zn | < Cǫ and hence |wn zn | < ǫ. Thus proved. 2. (a) Noting that |b|2 = b¯b, |¯ cz − 1|2 − |z − c|2 = |cz|2 + 1 − c¯z − z¯c − |z|2 − |c|2 + c¯z + z¯c = |cz|2 + 1 − |z|2 − |c|2 = (1 − |z|2 )(1 − |c|2 ) (b) If |z − c| < |¯ cz − 1|, then, by (a), (1 − |z|2 )(1 − |c|2 ) > 0, and hence, |z| < 1 (because |c| < 1). Conversely, if |z| < 1, then |z − c| < |¯ cz − 1|. The remaining parts of the problem are similar. 3. There is a t ∈ [0, 1] so that b = ta + (1 − t)c. In order to show that a complex number c is real, all one needs to do is to prove that c = c¯. Indeed, a−b a (a − ta − (1 − t)c)(¯ c − t¯ a − (1 − t)¯ c) − (c − ta − (1 − t)c)(¯ a − t¯ a − (1 − t)¯ c) ¯ − ¯b = − 2 ¯ c−b |c − b| c¯ − b = 0 4. (a) Every z that lies in the triangle lies on a line joining 1 to the side containing z1 |α| ≥ 1. Hence, and z2 . Let −(α + 1) be a point on the line. Note that δ = |z−1| |α| |1 − z| = 1 − |z| δ − |α + δ| |tz1 + (1 − t)z2 − 1| = δ − |δ − 1 + tz1 + (1 − t)z2 | where 0 ≤ t ≤ 1. We see that the above expression is decreasing as a function of delta (quite easily by computing the derivative). Hence, it is maximum when δ = 1; i.e., when z is on the line joining z1 and z2 . Maximising the resulting expression gives us 2

the K. This expression has a maximum because it is a continuous function of t (on a compact set [0, 1]). . This is because |1 − z| is largest for this (b) The maximum is attained when z = 1+i 2 1 z, and |z| is also largest for this z. Hence, K = √2−1 . 5. f (z) = z 3 = (x + ıy)3 = x3 − 3y 2 x + ı(−y 3 + 3x2 y). Hence, ux = 3x2 − 3y 2 = vy and uy = −6xy = −vx . 6. Let z = t(a + ıb) be a straight line connecting a + ıb to 0 (where a, b are real). lim

f (z) − f (0) t4 ab2 (a + ıb) = lim t→0 t(a + ıb)(t2 a2 + t4 b4 ) z tab2 = lim 2 t→0 a + t2 b4 = 0

Plugging in x = y 2 in f , lim

f (z) − f (0) y 4 (y 2 + ıy) = lim 2 y→0 (y + ıy)2y 4 z 1 = 2

′

Hence f (0) does not exist (the limit is different along different paths). 7. By the chain rule, (note that x2 + y 2 = r2 and

y x

= tan(θ) when x 6= 0)

∂r ∂θ + uθ ∂x ∂x sin(θ)uθ cos(θ)ur − r v r ry + v θ θ y cos(θ)vθ vr sin(θ) + r cos(θ)uθ ur sin(θ) + r sin(θ)vθ cos(θ)vr − r

ux = ur = vy = = uy = vx =

By imposing the usual CR equations (i.e. ux = vy and uy = −vx ) and solving for ur and uθ in terms of vr and vθ , we get the result for x 6= 0. One may prove this for x = 0 by considering cot(θ) = xy and going through the same argument. 8. Since f is C 1 , being holomorphic is equivalent to the Cauchy-Riemann equations. ∂ ∂ + ı ∂y )(u + ıv) = (ux − vy ) + ı(uy + vx ) which equals 0 if and only if Hence, fz¯ = ( ∂x ′ ′ the CR equations are satisfied. Since (in this case) f = ux + ıvx , f = fz . 3

9. By the previous problem, since f is C 1 , f is holomorphic if and only if fz¯ = 0, which is the same as Rz¯eıΦ + ıReıΦ Φz¯ = 0 (notice that the usual product rule carries over ∂ ). The last statement is equivalent to Rz¯ + ıRΦz¯ = 0. trivially to ∂∂z¯ and ∂z 10. Let f = u+ıv. We shall prove this for (g ◦f )z using the usual Chain rule for derivatives (the other case is similar) : (g ◦ f )z = (g ◦ f )x − ı(g ◦ f )y = gx ux + gy vx − ı(gx uy + gy vy ) 1 = (gx (fx + f¯x ) − ıgy (fx − f¯x ) − ı(gx (fy + f¯y ) − ıgy (fy − f¯y ))) 2 = gw fz + gw¯ f¯z 11. p is an entire function if and only if pz¯ = 0. Thus,

X

Pj (z)j z¯j−1 = 0 for all z¯.

j≥1

Comparing coefficients, we see that the condition is equivalent to Pj = 0 for all j ≥ 1. N X If indeed p is an entire function, then p(z) = P0 (z) = ck z k . This implies that k≥0

p(z) =

µ=k N X X k≥0 µ=0

k! ck xk ık−µ y k−µ . Hence, interchanging the order of summation (k − µ)!µ!

we see that aµν = ıν cµ+ν (µ+ν)! when 0 ≤ ν ≤ N − µ and 0 ≤ µ ≤ N (otherwise µ!ν! aµν = 0). 12. f is anti-holomorphic if f is holomorphic. Hence, if f = u + ıv, (by replacing v by −v in the usual CR equations), ux = −vy and uy = vx . In polar coordinates, if f = U +ıV , the equations are rUr = −Vθ and rVr = Uθ . If f is C 1 , the equation for f¯ to be holomorphic is f¯z¯ = 0. But it is easy to see from the definitions that f¯z¯ = (fz ), and therefore in this case the condition f is anti-holomorphic is equivalent to fz = 0. z )z = 0; the last statement follows by applying the Chain rule. 13. gz¯(z) = (f¯(¯ z ))z¯ = f (¯ Similarly, hz = f (¯ z )z = 0 (by the same argument as before). Hence, g is holomorphic and h is anti-holomorphic. If h is also holomorphic, then hz¯ = 0. Thus, hx = hy = 0. Hence, h = f (¯ z ) is a constant which implies f is too. 14. If f is holomorphic and anti-holomorphic, then fz = fz¯ = 0 and hence fx = fy = 0. Since D is connected, f is a constant (if it weren’t connected, f could be different constants on the connected components). 15. (a) No. If f = u + ıv is holomorphic and C 2 , then uxx + uyy = vxy − vyx = 0. But, clearly u = ex is C 2 and does not satisfy the condition and hence can’t be the real part of a holomorphic function. 4

(b) Yes. The functions f = ex (x cos(y) − y sin(y)) + ıex (x sin(y) + y cos(y)) + ıb, with b in R, are holomorphic (the real and imaginary parts are smooth functions that satisfy the Cauchy-Riemann equations) and their real part is the given function. In order to derive the formula for these functions, we shall solve the Cauchy-Riemann equations: ux = ex (x cos(y) − y sin(y) + cos(y)) v y = ux Z v = ux dy + g(x) Z x = e (x sin(y) + sin(y)) + ex y sin(y)dy + g(x)

= ex (x sin(y) + sin(y)) − ex (−y cos(y) + sin(y)) + g(x) = ex (x sin(y) + y cos(y)) + g(x)

One may see (quite easily) that the second Cauchy-Riemann equation (i.e. uy = −vx ) implies that g ′ (x) = 0; i.e., g(x) is a real constant. 16. We shall follow the hints in the text : (a) Firstly, note that |p(z)| ≥ ||z n |−|an−1 z n−1 +. . .+a0 ||. Now, |z k | ≤ |z n | ∀ 0 ≤ k ≤ n when |z| ≥ 1. Thus, |an−1 z n−1 + . . . + a0 | ≤ |z n−1 |(|an−1 | + . . . + |a0 |) = C|z n−1 |. So, |p(z)| ≥ C|z n−1 | > M when |z| > R = max(2C, 1, ( M )1/(n−1) ). C (b) Since S = (z; |z| ≤ R) is compact, |p(z)| (being continuous) attains a minimum over it. Since |p(z)| ≥ M for z ∈ C − S, the minimum of |p(z)| over C is attained at some point z0 . (c) Changing variables we define g(z) = p(z + z0 ), clearly g(0) = 0 if and only if p(z0 ) = 0. ˜ + 1 is a polynomial such that h(0) ˜ (d) Write g(z) = α + βz m h(z) (where h = h =0 and β 6= 0). Note that |g(0)| = α is the minimum of |g|. But, if α 6= 0, then defining γ m = − αβ , |g(γz)| = |α||1 − z m h(γz)| < |α| for some small, positive, real z. Indeed, ˜ ˜ |1 − z m (h(γz) + 1)| ≤ 1 − z m + z m |h(γz)| < 1 for a small positive, real z, because ˜ |h(γz)| is small. This is a contradiction and hence g(0) = 0. 17. We shall prove this by induction: The base case is trivial. If the statement is true for k = n, then, if p(z) is a degree n+1 polynomial, by the fundamental theorem of algebra, it has a root and hence p(z) = (z −z0 )h(z) where h(z) is a degree n polynomial. By the induction hypothesis, h(z) = (z − z1 ) . . . (z − zn ) and hence p(z) = (z − z0 )(z − z1 ) . . .. This implies that the number of roots with multiplicities) is indeed n. P (counted Given a real polynomial p(x) = an xn , we see that p(¯ z ) = p(z). Hence, the nonreal roots (if any) occur in conjugate pairs. Since every nonconstant degree d real polynomial can be factored into complex linear factors p(x) = ad (x − z1 )(x − z¯1 ) . . . = ad (x2 − (z1 + z¯1 )x + |z1 |2 ) . . ., it can be factored into products of real linear or quadratic factors.

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18. Once again we shall follow the hints in the text : (a) Consider θ ∈ F − R. The set {1, θ, θ2 , . . . θn−1 , θn } must be linearly dependent over R (because the dimension of F is assumed to be n). Hence there exists a non-constant real polynomial p(x) = xk + ak−1 xk−1 + . . ., so that p(θ) = 0. (b) Every real polynomial can be factored into a product of quadratics and linear factors (by the previous problem). Hence, there exist real numbers β and γ so that θ2 − 2βθ + γ = 0. (c) By (b) (θ − β)2 = β 2 − γ. If the right hand side is positive, then we could’ve factored the quadratic into real linear factors, but θ is not real. Hence, β 2 − γ = −δ 2 for some positive real number δ. Therefore, σ = θ−β satisfies σ 2 = −1. δ (d) Clearly, the field G in the text is isomorphic to C by sending σ → ı. Since very element of F satisfies a complex (in fact real) polynomial by (b), it follows from the fundamental theorem of algebra that F = G. 19. (a) Every automorphism σ takes 1 to 1. Hence, it is the identity on Q. If x > 0, then √ 2 σ(x) = (σ( x)) > 0. Thus, σ preserves the order relation on the reals. If we assume that σ(x) < x for some real x (without loss of generality), then σ(x) < a ≤ x for some a ∈ Q. But, if a ≤ x, then σ(a) = a ≤ σ(x) (it preserves the order). A contradiction. Hence, σ is the identity. (b) An automorphism σ fixing the reals takes a root of the polynomial x2 + 1 = 0 to another root. Hence, it takes ı to ı or to −ı. This means it maps bı to σ(b)σ(ı) = bı or to −bı for all b in R. Thus, σ(a + ıb) = a + ıb or a − ıb for all a in R. The former corresponds to the identity map and the latter to conjugation. (c) Note that σ(1) = 1 and hence σ restricted to the rationals is the identity. For every real r, there exists a sequence of rationals an → r. Since σ is continuous, an = σ(an ) → σ(r) and hence σ maps reals to reals. By (a), σ restricted to the reals is the identity. By (b), we are done. 20. Yes. By general topology (see Munkres for instance), every connected, locally path connected space is path connected. Every open set in C is obviously locally path connected (there is an open disc around every point). Hence connectedness and pathconnectedness are the same for open sets in C.

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Chapter 3 an where R is the radius of convergence. 1. We shall use the ratio test: R = limn→∞ an+1 We see (quite easily) that the radii of convergence are ∞, 1, 1 and 0 (i.e the last series does not converge anywhere except at z = 0) respectively. P zn P 1 (a) The series is absolutely convergent even when |z| = 1 (because is). n2 n2 P zn On the boundary, the series converges when z 6= 1 and diverges when z = 1 n (by the integral test). One may prove this by observing that this series equals ln(1 − z). To prove the statement without using this observation, the reader is referred to Rudin’s book “Principles of Mathematical Analysis” (the idea is to use summation by parts to prove the assertion). P zn does not converge at infinity because, as z → ∞ via the (b) The series ez = n! z negative reals e → 0, and along the positive reals it goes to ∞. 1 2. The given series is z2 +z−1 as is easily seen by multiplying it with z 2 + z − 1. Hence, it converges when |z| is less than the roots of z 2 + z − 1 (by theorem 3.55). Thus, the √ . radius equals 5−1 4

3. Since |an | ≤ M , limn→∞ (|an |)1/n ≤ 1 and hence ρ ≥ 1 (by Hadamard’s theorem). 4. (a) If one of the sequences is unbounded from above (and the other does not converge to 0), then the right hand side is ∞ and the result is trivial. If both of them are bounded from above, the lim sups are finite; supn≥p an bn ≤ supn≥p an supn≥p bn . Hence, inf p supn≥p an bn ≤ inf p (supn≥p an supn≥p bn ) = limp (supn≥p an supn≥p bn ) where the last equality holds because (supn≥p an supn≥p bn ) is decreasing. So, inf p supn≥p an bn ≤ limp supn≥p an limp supn≥p bn and the result follows. If an = 2 + (−1)n and bn = 5 + (−1)n+1 , the left hand side is 12 whereas the right hand side is 18. (b) If bn is not bounded from above, then supn≥p an bn = ∞ for every p (an gets close to the limit and bn has a subsequence going to ∞). Hence the result holds in this case. If bn is bounded from above, then there exists a subsequence bnk going to supn≥p bn for a given p. Hence, limk→∞ (ank bnk ) = lim an supn≥p bn i.e. limk→∞ (ank bnk ) ≥ lim an lim sup bn . By problem 18, the lim sup is the supremum of all subsequential limits. Hence the resut follows. 5. If lim an+1 = L, then (L − ǫ)an ≤ an+1 ≤ (L + ǫ)an ∀n ≥ N . Hence, (L − ǫ)n−N +1 aN ≤ an P an+1 ≤ (L + ǫ)n−N +1 aN . Using the geometric series, this implies that an converges 7

when L + ǫ < 1 for an ǫ > 0 and diverges when L − ǫ ≥ 1 for an ǫ > 0 i.e. the series converges when L < 1 and diverges when L > 1. P P P n 6. If an z n and b z converge (which they do when |z| < ρ), then, (an + bn )z n n P P converges to an z n + b n z n : |

n=N X n=0

n

(an + bn )z − (

∞ X n=0

n

an z +

∞ X n=0

n

bn z )| = | ≤ |

∞ X

n=N ∞ X

n=N

n

an z +

∞ X

n=N ∞ X

an z n | + |

ǫ ǫ + ≤ 2 2

bn z n |

n=N

bn z n |

PP P P P n for sufficiently large N . Also, cn z n converges to ( an z n )( bn z n ) i ai bn−i z = by lemma 3.17. Hence, the radii of convergence of these two series are ≥ ρ. 7. Changing variables to w = z1 and apply corollary 3.60 to f ( w1 ) to conclude that, f ( w1 ) = wn g. Hence, there exists a ball B centred at z = 0 such that outside the ball, f (z) = z1n g(z) where g is a holomorphic function on C − B and is a unit. 8. We shall prove that e−z ez = 1. By theorem 3.18, we are allowed to multiply the power P m P Pn=k (−1)n zk P n n = series (like polynomials). Hence, e−z ez = ( (−1)n! z )( zm! ) = k n=0 n!(k−n)! P zk k k k! (1 − 1) = 1. √ 9. (a) Certainly, a branch of f (z) = z is an algebraic function (which is clearly neither a polynomial nor a rational function). Indeed, (f (z))2 − z = 0. (b) otherwise. Then, there exist polynomials with rational coefficients so that P Assume pn (z)enz = 0. Substitute z = 1 in this equation to deduce that e is algebraic, a contradiction. 10. No. Log(eı2π ) = ln |e2πı | + ıArg(e2πı ) = 0 6= 2πı. 11. If exp(exp(z)) = 1, then exp(z) = 2nπı where n is a non-zero integer. Hence, z = ln(2|n|π) + ı((−1)sgn(n) π2 + 2kπ) where k ∈ Z and sgn(n) = 0 if n is positive and 1 otherwise. 12. (a) Let the period of cos(z) be p: eı(z+p) + e−ı(z+p) = eız + e−ız i.e. eız (eıp − 1) = e−ız (1 − e−ıp ). Hence, eıp = 1 and so p = 2π. The case of sin(z) is similar. (b) We see that sin(z) = 0 if and only if eız = e−ız i.e. e2ız = 1. Hence, z = nπ for some n ∈ Z. (c) cos(z) = 0 if and only if e2ız = −1 i.e. 2z = π + 2nπ.

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13. (a) eız + e−ız 2 1 y = ((e + e−y ) cos(x) + ı(e−y − ey ) sin(x)) 2 1 −y ((e + ey ) sin(x) + ı(ey − e−y ) cos(x)) sin(z) = 2

cos(z) =

(b) 1 2y (e + e−2y + 2 sin2 (x) − 2 cos2 (x)) 4 1 2y (e + e−2y + 2 sin2 (x) − 2 + 2 sin2 (x)) = 4 = sin2 (x) + sinh2 (y)

| sin(z)|2 =

Similarly for | cos(z)|2 . (c) Note that sinh(z)+cosh(z) = ez and cosh(z)−sinh(z) = e−z . Writing cosh(a+b) = 1 a b (e e + e−a e−b ) (similarly for sinh(a + b)) and using these identities, we arrive at 2 cosh(a + b) = cosh(a) cosh(b) + sinh(a) sinh(b) and sinh(a + b) = sinh(a) cosh(b) + cosh(a) sinh(b). z −z z −z = sinh(z) and that of sinh(z) = e +e = (d) The derivative of cosh(z) = e −e 2 2 2 cosh(z). Notice that cosh(ız) = cos(z) and sinh(ız) = ı sin(z). Hence, cosh (ız) − sinh2 (ız) = cos2 (z) + sin2 (z) = 1. √ ız −ız 14. By definition, e +e = 2 and√hence e2ız −4eız +1 = 0. Solving this, we get eız = 2± 3. 2 If z = x + ıy, e−y eıx = 4( 12 ± 23 ). Hence, y = − ln(4) and x = ± π3 + 2kπ where k is an integer. π

π

15. (a) ıı = eıln(ı) = eı(ı(2kπ+ 2 )) = e−(2kπ+ 2 ) ∀k ∈ Z. (b) Let b = x + ıy. ab = ebln(a) = ebln(|a|)+ıb(Arg(a)+2πk) = exln(|a|)−y(Arg(a)+2πk)+ı(yln(|a|)+x(Arg(a)+2πk)) Hence, infinitely many distinct values appear if and only if either y 6= 0 (so that the magnitudes are different), or xk is not an integer for infinitely many k, i.e. when x is irrational (so that the arguments differ by values other than 2πn where n ∈ Z). (c) By (b), firstly y = 0. Secondly, xn = p where p and n are coprime. n| n| 16. (a) This means the positive and negative parts of an i.e. pn = an +|a and qn = −an +|a 2 2 P P don’t converge (since (pnP +qn ) = an converges, if one of them did converge, then so P would the other have; But (pn − qn ) = |an | diverges). Since pn and qn are positive

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sequences (without loss of generality, we may throw out all the zeroes), this means their sums go to ∞. Given an a, there exists a sequence an converging to it. Let m1 and k1 be the smallest integers so that p1 +. . .+pm1 > a1 and p1 +. . .+pm1 −q1 −. . .−qk1 < a1 (First choose pi so that their sum is > a1 . Then choose qi so that the second condition is satisfied. This is possible because we may make the sums as large as we like owing to their smallest integers P Pk1divergence). Pm2 Similarly, choose Pm1 m2 , k2Ptok1be the P Pk2 so that m1 m2 p − q + p > a and p − q + p − 2 j=1 j j=1 j j=1 j j=1 j j=1 j j=1 j j=1 qj < a2 . Continue along these lines. Since pj and qj → 0 as j → ∞, and |sn − an | ≤ pmn (where sn is the nth partial sum of the series having the last term as pmn ), we have produced a rearrangement that converges to a. (b) If every rearrangement converges to the same sum, then, the series converges absolutely (by part (a)). Now assume that the series converges absolutely. Let akn be a rearrangement with partial sums Sn (let the original partial P sums be sn ). Given an ǫ > 0, there exists an N so that if m ≥ n ≥ N , then m j=n |ai | ≤ ǫ. Now choose p to be so large that (1, 2 . . . , N ) is contained in (k1 , . . . kp ). Then, if n > p, clearly |Sn − sn | < ǫ. Hence, the rearrangement converges and does so to the same sum as the original series. 17. We shall prove this for lim sup (the case for lim inf follows by replacing an by −an ): If there is a subsequence bn → ∞, then supn≥p bn = ∞∀p. Hence the result holds in this case. If bn is bounded from above, then, given an ǫ > 0, there exists a p so that supn≥p bn − lim sup bn < ǫ. There exists a convergent subsequence bnk going to supn≥p bn . This means, for every ǫ, there is a convergent subsequence whose limit α is such that α − lim sup bn < ǫ. Hence, sup α ≥ lim sup bn . Conversely, if bnk → α, then choosing p, k (sufficiently large) supm≥p bm − lim sup bm < 2ǫ and |bnk − α| < 2ǫ . This means α − ǫ < lim sup bm for every ǫ. Thus, α ≤ lim sup bm . The result follows. = an 6= 0). Now, the 18. (a) It is clear that p(∞) = ∞ (the reason is that limz→∞ p(z) zn rest of this statement maybe deduced from the fundamental theorem of algebra (every polynomial has a complex root): p(z) − α has a root. Alternatively, suppose there is an 1 has a convergent α ∈ C such that Im(p(z)) does not contain α. Then, h(z) = p(z)−α power series expansion (via Theorem 3.47 in the text for instance). But, as z → ∞, h(z) → 0, hence (changing w = z1 ), we see that h(z) = z1l r(z). For h to be a power series, it then has to be identically zero, a contradiction. (b) Define mp (z) = νz (p(u) − α) if α 6= ∞ and the order of the pole of p(z) if p(z) = ∞. Pk=n

a

wk

At ∞, p(w = 1/z) = k=0wnn−k . Thus, the sum of multiplicities at ∞ is n (because p(z) 6= ∞ if z is finite). The statement for α ∈ C follows via induction: The base case of a degree one polynomial is trivial. If the statement is true for polynomials of p(u)−α degree ≤ n − 1, then, if p(z) = α, then h(u) = (u−z) mp (z) is a polynomial of degree P n − mp (z). Hence by the induction hypothesis, u∈C;h(u)=0 mh (u) = n − mp (z) and ˆ P thus z∈C;p(z)=α mp (z) = n (noticing that the order of a zero of h(u) and p(u) − α are ˆ same away from z). 10

ˆ then by the previous problem there is a z ∈ C ˆ so that the polynomial 19. (a) If α ∈ C, P (z) − αQ(z) = 0. Hence p is surjective (note that we are using the fact that P and Q don’t have any common roots). (b) Define mp (z) = νz (P (u) − αQ(u)) if α 6= ∞ and the order of the pole of p(z) if p(z) = ∞; The topological degree is defined as max(deg(P ), deg(Q)). In what follows we shall assume deg(P ) ≥ deg(Q) (the other case being P similar). When α = ∞, mp (∞) = deg(P ) − deg(Q) and mp (z) = νz (Q). Since Q(z)=0 mp (z) = deg(Q) (by P the previous problem), we have p(z)=∞ mp (z) = deg(P ). When α ∈ C, by applying the previous problem to the polynomial P (z) − αQ(z) we get the result. 20. The stereographic map f is clearly smooth (being a restriction of a smooth map on R3 − (ζ = 1) to S 2 − (0, 0, 1)). If α ∈ C, we may solve f (x, y, z) = α : x + ıy = (1 − z)α (1 − z) |α|2 + z 2 = 1 (1 − z)2 |α|2 = (1 − z)(1 + z) 2

Hence, if z 6= 1, then the above equation has a unique solution. Thus, f is a bijection. We shall use the inverse function theorem to prove it is a diffeomorphism i.e. we shall show that the kernel of the derivative consists of the vectors not tangent to the sphere : The derivative matrix (of the smooth function on R3 − (ζ = 1) evaluated on points on the sphere) is ! 1 x 0 2 1−z (1−z) y 1 0 1−z (1−z)2 The kernel is easily seen to consist of non-tangential vectors (the dot product of a basis vector in the kernel with (x, y, z) is 1 − z for a point (x, y, z) on the sphere). In coordinates w around ∞, 1−z x + ıy (1 − z)(x − ıy) = x2 + y 2 (1 − z)(x − ıy) = 1 − z2 x − ıy = 1+z

w =

The last equation implies that the map extends smoothly (it is smooth when z 6= −1), mapping the north pole to w = 0. 21. A plane through (0, 0, 1) is of the form ax + by + c(z − 1) = 0. So, f (x, y, z) = u + ıv 11

x + ıy 1−z c(x + ıy) = ax + by au + bv = c =

Hence, maximal circles are mapped onto straight lines. Given a straight line au + bv = c, tracing the above equations backwards, we see that the unique maximal circle corresponding to it is the intersection of ax + by + c(z − 1) = 0 with the sphere. A non-maximal circle is the intersection of a plane ax + b(y − β) + c(z − γ) = 0 (where γ 6= 1 and β 2 + γ 2 = 1) with the sphere. The image of such an object is x + ıy 1−z = u + ıv x2 + y 2 + z 2 = 1 = (u2 + v 2 )(1 − z)2 + z 2 (u2 + v 2 )(1 − z) = 1 + z au + bv − bβ − cγ z = au + bv − c 2 2 (u + v )(c(γ − 1) + bβ) = 2(au + bv) − c(γ + 1) − bβ p a b Hence (u, v) lie on a circle with centre ( c(γ−1)+bβ , c(γ−1)+bβ ) and radius a2 + (bγ + cβ)2 . Given such a circle we may trace the argument backwards to prove that there is a unique non-maximal circle that maps to this. f (x, y, z) =

22. If A is a linear map between Rn and Rn such that kAxk2 = c2 kxk2 where c is a non-zero constant, then 4(Ax, Ay) = = = (Ax, Ay) = kAxkkAyk

kAx + Ayk2 − kAx − Ayk2 c2 kx + yk2 − c2 kx − yk2 4c2 (x, y) (x, y) kxkkyk

Hence such an A preserves angles. Now, we shall prove the stereographic projection f considered as a map from R3 − (z = 1) (as in the previous problems) takes vectors tangent to the sphere to ones whose norms are scaled by a constant (thus proving the result). A tangent vector at (x, y, z) is of the form (a, b, c) where ax + by + cz = 0. A basis for these is (y, −x, 0) and (z, 0, −x) (valid when all three coordinates are nonzero. The cases where they are zero maybe handled in a similar manner). Now, the 1 derivative of f (as computed in problem 18) maps these vectors to 1−z (y − ıx) and 12

1 (y 2 (1−z)2

+ z − 1 − ıxy) respectively. Their norms (squared) are

1 (x2 (1−z)2

+ y 2 ) and

1 1 ((y 2 + z − 1)2 + x2 y 2 ) = ((y 2 + z − 1)2 + (1 − y 2 − z 2 )y 2 ) 4 4 (1 − z) (1 − z) 1 ((z − 1)2 − y 2 (z − 1)2 ) = (1 − z)4 1 = (x2 + z 2 ) (1 − z)2 respectively. Hence, the result follows. ′

23. Note that sin(z) and cos(z) don’t have any common zeroes and that, since (cos(z)) = − sin(z), the roots of cos(z) are simple. Hence, tan(z) has simple poles at (2k + 1) π2 . Given, α ∈ C, we may solve tan(z) = α (thus proving surjectivity) by writing it as eız −e−ız = α. Hence, ı(eız +e−ız ) e2ız − 1 = ıα(e2ız + 1) 1 1 + ıα z = ln( ) 2ı 1 − ıα (a) The last equation implies there are infinitely many solutions for every α; z and η are solutions if and only if ζ − z = 2i1 (2πık) = πk where k is an integer (this is because the logarithm is well-defined upto 2πık). (b) The map is obviously holomorphic in the given region −π < R(z) < π2 (it has no 2 poles). It is one-one because the width of the region is less than π (and using (a)). The image would miss only those α which are tan( ±π + ıy) i.e. α = −ı coth(y) = 2 e2y +1 −ı e2y −1 . The imaginary part of the latter expression is a decreasing function of y (by computing the derivative). It has horizontal asymptotes at 1 and −1. Thus, its image is C − i((−∞, −1) ∪ (1, ∞)). From this it also follows that by including R(z) = π2 we get all of C. It is still one-one because the interval is semi-open. By changing to coordinates at ∞, w = cot(z) is clearly well-defined and holomorphic at z = π2 . ′ (c) This is an easy consequence of the quotient rule and that (sin(z)) = cos(z) and ′ (cos(z)) = − sin(z). 24. (a)The radius of convergence is easily computed to be 1. Hence, it defines a holomorphic function on |z| < 1. (b)The given function f (z) also has radius of convergence equal to 1 (by the ratio test). Hence it is holomorphic on |z| < 1. By theorem 3.19, we may differentiate f ′ 1 term-by-term and thus f (z) = 1+z 2. (c) and (d) These two follow from the previous exercise. Note that the local injectivity of tan(z) follows from the inverse function theorem. ′ ′ 1 2 (e) g (z) = f (tan(z)) sec2 (z) = 1+tan 2 (z) sec (z) = 1 whenever the derivatives make sense, i.e. when | tan(z)| < 1 and z 6= (2n + 1) π2 . Using a result of problem 14, | sin(z)|2 < | cos(z)|2 sin2 (x) + sinh2 (y) < cos2 (x) + sinh2 (y) 13

hence D = (− π4 + nπ < R(z) < π4 + nπ) where n is a positive integer. ′ (f) When |z| < π4 , then indeed z ∈ D. Hence, g (z) = 1 and g(z) = z. This means f (tan(z)) = z here. But, when |z| < 1, the inverse of tan is unique (because π2 > 1). Hence, f (z) = Arc tan(z) (by the Identity theorem). (g) This is because the derivative of f (z) has a pole at z = ı. Pn=M (−1)n (h) The given series converges. This is because, | n=N | < 2N1+1 . Hence, by 2n+1 Abel’s theorem (and the fact that Arc tan is continuous at 1), we are allowed to plugin z = 1 in the series to get Arc tan(1). P n 25. We may assume that f (z) = z α ( ∞ n=0 an z ) (where an 6= 0 and α ≥ 1) and g(z) = ′ P f (0) f n α−1 a0 b z ) (where b = 6 0). Hence, lim = z( ∞ = lim z . Now, ′ n z→0 g z→0 n=0 n b0 g (0) αz α−1 a0 . b0 ′ f (0) . g ′ (0)

limz→0 a0 b0

=

′

If α > 1, then f = 0 and so is limz→0 fg . If α = 1, then limz→0

f g

=

26. Yes. Define an = αn − βn . We shall prove that an = 0. Assume By the P otherwise. n definition of convergence of a doubly infinite series, f (z) = a z and g(z) = n≥0 n P −n are convergent and f (z) = −g(z) on 2 < |z| < 3. We shall show that n>0 a−n z f (z) actually converges on 0 < |z| ≤ 2 : Since f (z) converges when 2 < |z| < 3, by z n+1 | 1 ≤ 1. This means, 2 < |z| ≤ the ratio test, lim sup |an+1 |an+1 | i.e. the radius |an z n | lim sup

|an |

of convergence is atleast 2. Similarly, g(z) converges when |z| > 2. This implies that h(z) = f (z) when |z| < 3 and h(z) = −g(z) when |z| > 2 defines an entire function. We shall see (in a later chapter) that bounded, non-constant entire functions cannot be exist. Thus proved. P

i

bi (k+1) | 27. (a) Indeed, the radius of convergence is R = limk→∞ | | P = 1 (for sufficiently bi k i | n n large |z|, C1 |z| ≤ |p(z)| ≤ C2 |z| ). (b)We prove this by induction on n. For n =P0, a0 = b0 . Assume the induction hypoth−1 esis for n < N . For n = N , p(z) − bN z N = i=N ai z(i) by the induction hypothesis. i=0 P −1 Also notice that, zz(i−1) = z(i) + (i − 1)z(i−1) and that, z N −1 = j=N cj z(j) . Thus, j=0 Pj=N Pi=N N N −1 z = zz = j=0 cj (z(j+1) + jz(j) ) and hence, p(z) = i=0 a ˜i z(i) . (c) Note that, k(n) = k(n−1) (k − (nP − 1)) = k(n−2) (k − (n − 2))(k − (n − 1)) = ∞ 1 k! k . Also, since 1−z = . . . = (k−(n−2))! k=0 z , differentiating n times on both sides, P n! k−n = ∞ . k=n k(n) z (1−z)n+1 (d) All the series involved are absolutely convergent and hence, we may rearrange the order of summation. This observation yields the result easily.

14

Chapter 4 1. Let C be γ(t) R1 R (1) γ(t) = (2t − 1)ı. Hence, C |z|dz = 0 |2t − 1|2ıdt = ı R1 R (2) γ(t) = − sin(πt) − ı cos(πt). Hence, C |z|dz = 0 (−π cos(πt) + ıπ sin(πt))dt = 2ı. R R1 (3) γ(t) = sin(πt) − ı cos(πt). Hence, C |z|dz = 0 (π cos(πt) + ıπ sin(πt))dt = 2ı. 2. C = tı + t. Thus,

R

C

xdz =

R

tdt(1 + ı) =

1+ı . 2

R1 R 3. C = z0 + r(cos(2πıt) + ı sin(2πt)). Hence, C (z − z0 )m dz = 2πı 0 rm+1 e2πı(m+1)t dt (1) When m ≥ 0, the integral is 0. (2) When m 6= −1 and m < 0, the integral is 0. When m = −1, the integral is 2πı. 4. Z

3

z dz = γ1

Z

1

(1 + ıt)3 ıdt 0

5 4 Z Z z 3 dz = −πı = −

γ2

= 0 Z Z z 3 dz = πı γ3

Z

γ4

1

e−3πıt e−πıt dt 0

1

e3πıt eπıt dt 0

= 0Z 1 (1 + t2 + ıt)3 (ı + 2t)dt z 3 dz = 0

= 6ı

The other integrals are evaluated similarly. The answers are Z 1 1 π dz = ln(2) + ı z 2 4 Zγ1 1 +ı z¯dz = 2 γ1 15

Z

Zγ2

γ2

Z

Zγ3

1 dz = −πı z z¯dz = −πı 1 dz = πı z z¯dz = πı

γ3

1 1 1 dz = ln(5) + ıArctan( ) z 2 2 Zγ4 2 z¯dz = 2 + ı 3 γ4

Z

5. The case of the union is actually a special case of the Seifert-Van-Kampen theorem of algebraic topology. But we shan’t prove it that way (because it is cumbersome and deviates from complex analysis). We shall assume this (fairly believable) fact : The winding number of all loops in a domain D is zero about all points outside D if and only if D is simply connected. One direction has been proved in the text. The other is much harder (although intuitively plausible) and requires tools from topology (the idea is to first prove that this implies that the complement of D is connected and then prove that the latter is equivalent to simple-connectedness using a version of the Jordan curve theorem; See Ahlfors’ Complex Analysis book and Munkres’ Topology book). Let γ(t) be a loop in D1 ∩ D2 . Let p be a point outside D1 ∩ D2 . Then, if p is in D1 , consider the loop to lie within D2 so that the winding number about p is then zero (because it is contractible within D2 ). Similarly, if p is in D2 , the winding number is zero. If p is outside D1 and D2 , due to the same reason, the winding number is still zero. Hence D1 ∩ D2 is simply connected. If γ(t) is a loop in D1 ∪D2 , then, we shall “break” this loop up into pieces that lie within D1 and D2 (and thus, due to the same argument as above, the winding number about points outside vanishes and thus the loop is contractible). Let the Lebesgue number for the cover γ −1 (D1 )∪γ −1 (D2 ) be δ. Choosing a partition 0 < t1 . . . < tn < 1 finer than δ, we may assume (without loss of generality) that γ1 = γ[0, t1 ] ∈ D1 , γ2 = γ[t1 , t2 ] ∈ D2 etc. So, γ(t) = γn ∗ γn−1 ∗ . . . where the star denotes composition of paths. Hence, R the R dz dz the winding number of γ(t) about a point a equals γ1 z−a + γ2 z−a +. . .. We may make the portions of the loop γ that lie within D1 into a closed loop by joining γ(t1 ), γ(t3 ) (and other such pairs of points) by (piecewise) smooth paths. Travelling in the opposite direction along these paths will “complete” the portion of γ that lies within D2 into closed loops. The winding number does not change during the process (because the integrals around the newly introduced paths will cancel out). By the same argument as in the case of the intersection, the winding number about any point outside D1 ∪ D2 will vanish for all loops and hence D1 ∪ D2 is simply connected. 16

6. The map is f (t) = a + t(b − a). This map is obviously differentiable, 1 − 1 and onto. 7. If 0 < a1 < a2 < . . . < an < 1, and 0 < b1 < b2 . . . < bm < 1 are two partitions, then we may choose a common partition by subdividing each of these partitions. Hence, if we prove that the integral is unchanged under subdivision of partitions, we will have proved If 0 < Ra ˜1 < . . . < a ˜ k1 < a1 < a ˜k1 +1 . . . is a subdivision, then, notice R R the theorem. that γ[0,a˜ ] ω + γ[a˜ ,a˜ ] ω = γ[0,a˜ ] ω (by the addition property of integrals which we are 2 1 2 1 allowed to use because γ is actually differentiable on [0, a˜2 ]). If γ is differentiable, then choose the partition 0 < 1 to get back the usual definition for the integral over the path. 8. We need to verify that, at every point 0 ≤ t0 ≤ 1, there exists a neighbourhood Nt0 of f (t0 ) such that f (t) agrees with the restriction of a primitive Ft0 defined in that neighbourhood. Indeed, consider a neighbourhood Nt0 such that it does not contain the origin. A holomorphic branch of the logarithm ln(z) maybe defined there. Let Ft0 = ln(z) + C where the constant C is chosen so that Ft0 (exp(2πıt)) = 2πıt near t0 . Hence proved. 9. The form is closed if and only if d(P dz + Qd¯ z ) = 0. This means Pz¯ = Qz . 10. (1) Let Γ : [0, 1] × [0, 1] → D be the homotopy. By Green’s theorem (and the fact that Γ(0, s) = Γ(1, s)), Z dω = 0 Γ Z Z Z Z = ω− ω− ω+ ω ′ γ γ Γ(0,s) Γ(1,s) Z Z ω = ω− γ

γ

′

R R (2) This follows from : R∂a ω = a dω = 0 (by Stokes’ theorem). (3) This follows from : γ df = f (γ(1)) − f (γ(0)) = 0 (by the fundamental theorem of calculus). 11. (1) We shall prove this by induction (with the base case being n = 0). Let the minimum distance of the disc (t||t − z| < ǫ) from the range of γ be l (for a fixed small ǫ). If the statement is true for all k ≤ n − 1, then, R φ(u) |g (n−1) (z + h) − g (n−1) (z) − n! γ (u−z) n+1 du| = Lh |h| Lh

1 (|(n − 1)! = |h|

Z

(φ(u)

Z

(z,z+h)

n dtdu) (u − t)n+1 17

φ(u) du|) (u − z)n+1 Z Z φ(u) (u − z)n+1 − (u − t)n+1 n! | ( ( dt)du| = |h| (u − z)n+1 (z,z+h) (u − t)n+1 Z Z φ(u) (u − z)n (u − t) + . . . + (u − t)n (u − z) ≤ n! |( |( | ||dt|)|du| (u − z)n+1 (z,z+h) (u − t)n+1 M ≤ n+1 |h| l − n!

Z

where M is some constant. The last expression goes to 0 as h → 0. Hence proved. ′ (2) RLet I(w) be the desired integral. Choose φ(u) = 1 in (1). Then, I (w) = 1 1 dw which is zero by an explicit calculation. Hence I is a constant on each 2πi (z−w)2 connected component. Clearly I(w) = 0 on the unbounded component (take w → ∞). Taking w = z0 , we get I(w) = 1 again by an explicit calculation. Hence proved.

18

Chapter 5 1. The equivalence of (1) and (2) is a topological statement. To prove this, we first refer to Ahlfors’ book wherein (2) is proved to be equivalent to all curves in D being homologous to 0. Obviously, (1) implies that all the curves in D are homologous to 0 (the winding number is invariant under homotopy) and hence (1) implies (2). To prove the converse, since every cycle being homologous to zero implies (3), (2) implies (3) and we shall see that this is enough to prove that (2) implies (1). ′

(z) (1) ⇒ (3) : By Cauchy’s theorem, ff (z) (which is holomorphic because f 6= 0), has a primitive g(z). Choose g(z) such that g(a) = ln(f (a)) for some a ∈ D (and some ′ ′ branch of the logarithm). But, h (z) = ff , h(a) = b has a unique solution. This is ′ because, if h1 , h2 are two solutions, then f (h1 − h2 ) = 0. Since the logarithm exists locally, g(z) is the logarithm and hence it satisfies eg = f . 1 (3) ⇒ (4) : Define h = e n ln(f ) . This satisfies hn = f . (4) ⇒ (1) : This follows from the proof of the Riemann mapping theorem (wherein we just use this property to prove that every simply connected domain is equivalent to the unit disc. Refer to Ahlfors’ book for instance). If hn1 = hn2 , then (h1 /h2 )n = 1 and hence h2 = ζh1 where ζ is an nth root of unity. Similarly, if eg1 = eg2 , then g1 = g2 + 2kπı where k is an integer.

R R ′ (z)dz dw 2. I(β, ζ) = f ◦γ w−ζ = γ ff (z)−ζ where the last equality follows by applying the Chain rule. But, the last integral is zero by Cauchy’s theorem because D is simply connected ′ (z) is holomorphic on D (f (z) 6= ζ∀z ∈ D). and ff(z)−ζ 3. Since homotopy does not change the winding number, I(γ, z) = I(˜ γ , z) = 0 where γ˜ is the constant path. Hence the cycle is homologous to 0 in U. The converse is false. Let U be the disc minus p, q. Consider a closed path that winds once around p and once q (but in the opposite direction). Such a cycle is homologous to zero in U. But it is certainly not homotopic to the constant path in U . f (z1 )−z z1 −z 4. (1) Let m1 (z) = 1− and m2 (z) = 1− ¯ 1 )z where z1 is any fixed number in the disc. z¯1 z f (z −1 Notice that g = m2 ◦ f ◦ m1 maps 0 to 0 and hence satisfies the conditions of the ′ Schwarz lemma. Thus, |g (0)| ≤ 1. This gives the desired result (when expanded out). (2) Equality holds if and only if g(z) = αz where |α| = 1, i.e. when f (z) = α(m−1 2 ◦

19

m1 )(z). ′ (3) The inequality implies |f ( 21 )| ≤

32 . 27

(n)

5. (1) Let g(z) = f (z(R+ǫ)) where β = n!M . Notice that (by Cauchy’s estimates) g(z) β+ǫ Rn satisfies the conditions for the Schwarz lemma (for every ǫ > 0). Hence, |g(z)| ≤ |z|. |z|n+1 By the complex version of Taylor’s theorem, |f (z)| ≤ (M + ǫ) (R+ǫ) n+1 for every ǫ > 0. Hence, the desired inequality holds . The other inequality holds by Cauchy’s estimates. Equality holds if and only if g(z) = αz and hence when f (z) is of the given form. ′ (2) The Schwarz lemma implies that g(z) = αz if |g(ζ)| = |ζ| for some ζ or if |g (0)| = 1. This translates (as in (1)) into the desired form for f . 6. If M > 0, then f (0) = 0 and is hence a removable singularity. If M = 0, then |f (z)| ≤ C. Hence limz→0 zf (z) = 0. This means h(z) = z 2 (f (z)) when z 6= 0 and h(z) = 0 is holomorphic at 0 (with derivative equal to 0). By a power series argument, this implies that f (z) maybe extended holomorphically to 0 and hence it is a removable singularity. If M < 0, then |z −M f (z)| ≤ C. By the argument above, f (z) = h(z) where zM h is holomorphic at 0. Hence, the order of the pole at 0 is at most −M (in the previous cases, the order of the zero is atleast M ). 7. Every half-plane may (via a rotation) be mapped biholomorphically onto the upper half plane. The map z → z−ı maps this biholomorphically onto the unit disc (this is z+ı easy to show). Every disc maybe rotated and translated to coincide with the unit disc. Hence, we need to look only at self-maps of the unit disc D. Given a map f : D → D, we may compose it with fractional linear transformations to get a map g : D → D such that g(0) = 0. Hence, by Schwarz’s lemma, g(z) = αz for some |α| = 1. Since a composition of fractional linear transformations is also fractional linear, we are done. The region E = 1 < |z| ≤ ∞ maybe mapped onto D by means of z → z1 . Any self map of E, h : E → E can be composed with fractional linear transformations to map ∞ to ∞. Then we reduce this case to the previous problem. Hence, all such maps are fractional linear transformations. 8. By Cauchy’s estimates |f (n) (z)| ≤ n!M where M = max |f | on a disc of radius R centred Rn n!(a(R+|z|)b +c) (n) at z. Hence, |f (z)| ≤ . If n > b, then, taking R → ∞, we find that Rn (n) |f (z)| = 0∀z. Hence, f is a degree n polynomial, where n ≤ b. 9. Let f = u + ıv. Define h = ef . Then |h| = eu → 1 as z → ∞. Hence, h is a bounded entire function. By Liouville’s theorem, it is a constant. This, means h(0) = 1 = h(z). Hence, f (z) is identically 0. 1 is holomorphic on D and satisfies |h| = 1 on ∂D. 10. If f (z) 6= 0∀z ∈ D, then h(z) = f (z) Hence, by the maximum modulus principle, |h(z)| ≤ 1 (and hence |f (z)| ≥ 1). But, so is |f (z)| ≤ 1. This means, |f (z)| = 1 identically on D. By the maximum principle, this means f is a constant on D. This is a contradiction.

20

11. We shall prove the maximum principle (the minimum principle follows by applying the maximum principle to the negative of the function). If u(x0 ) ≥ u(x)∀x ∈ D, where R 2π 1 x0 is in D, then u(x0 ) ≥ 2π 0 u(x0 + reıθ )dθ. But, the last expression equals u(x0 ) by the mean value property. Hence, u is a constant = u(x0 ) on a small disc around x0 . Let A be the set of all x ∈ D such that u(x) = u(x0 ). By continuity, A is closed. But, we have proved that A is open as well. Since A is non-empty and D is connected, A = D. The mean value property implies that max(f ) ≤ max(f |∂D ) ≤ M and similarly for min(f ). 12. Let r < u < R. The supremum of |f | is always attained on the boundary of the disc. Hence, supB(0,r) |f (z)| = M (r) ≤ supB(0,u) |f (z)| = M (u). This means, M (r) is strictly increasing.

21

Chapter 6 1. Let p(z) = an z n + an−1 z n−1 + . . .. Let f (z) = an z n and g(z) = p(z) − f (z). Notice that n−2 |+... ) = R. |g(z)| ≤ (|an−1 | + |an−2 | + . . .)|z|n−1 ≤ |f (z)| when |z| ≥ max(1, |an−1 |+|a |an | Hence, if γ is the circle with radius R and centre 0, by Rouche’s theorem p = f + g has the same number of zeroes as f does, within γ. This implies the fundamental theorem of algebra. 2. (1) By Rouche’s theorem, z − tg(z) and z have the same number of roots in |z| < 1. This means that z = tg(z) has a unique solution for every |t| < 1. (2) Clearly s(0) = 0. One may prove a version of the Implicit Function Theorem for holomorphic functions (by replicating the proof for smooth functions). Hence, if ′ ′ ∂(z−tg(z)) = 1 − tg (z) 6= 0, we are done. If not, there exist t0 , z0 such that t0 g (z0 ) = 1 ∂z |g(z0 )| and z0 = t0 g(z0 ). Then, |g(z0 )| > |z0 | and by the Schwarz-Pick lemma, 1−|g(z 2 ≤ 0 )| |z0 | . 1−|z0 |2

x However, the function x → 1−x 2 is increasing and hence, the last inequality implies that |g(z0 )| ≤ |z0 |, thus producing a contradiction. ′ ′ (3) Differentiating on both sides and simplifying, s (t)(1 − tg (s(t))) = g(s(t)). Hence, ′ s (t) = 0 if and only if g(0) = 0. (4) If s(t1 ) = s(t2 ) = z, then, t1 g(z) = t2 g(z). If g(z) 6= 0, then, t1 = t2 . If g(z) = 0, then, z = t1 g(z) = 0. Hence, if g(0) = 0, s(t) = 0 for every value of t. If not, s is injective. P P n−1 3. If f (z) − c = g(z) = n≥−M an (z − ζ)n , then F = Panan(z−ζ) n . If M = 0 and the n (z−ζ) k−1

+... = first non-zero coefficient is ak (i.e. ν(f (z) − c, ζ) = k), then F = kaakk(z−ζ) (z−ζ)k +... kak + g where g is holomorphic at z = ζ. Hence, the residue of F at ζ is k. If (z−ζ)(ak +...)

M 6= 0, (ν(f (z) − c, ζ) = −M ), then F = Hence, the residue is −M . Thus proved.

−M a−M (z−ζ)−1 +h a−M +a−M +1 (z−ζ)+...

where h is holomorphic.

4. Notice that 0 is certainly not an essential singularity (by the Casorati-Weierstrass 2 theorem). We see that h(z) = f (z)−1 is such that |h(z)| ≤ 1. Clearly h(z) (being bounded) extends holomorphically to zero. If 0 is a pole for f (z), then h(0) = 0. By the Schwarz lemma, this means, |h(z)| ≤ |z|. If the pole is of order 1, then the residue is atleast 2. 5. (a) Writing 2 cos(θ) = eıθ + e−ıθ , on the unit circle we have 2 cos(θ) = z + z1 . Hence the 22

given integral is (let C be the curved part of the semicircle in the upper half-plane) Z Z −2ıdz −2ıdz √ √ = 1 2 C z + 2 5z + 1 C z(2 5 + z + z ) √ √ . Close C The integrand has poles at z = − 5 ± 2. The residue at − 5 + 2 is −ı 2 √ √ with (−1, − 5 + 2 − ǫ) ∪ √C1 ∪ (− 5 + 2 + ǫ, 1) where C1 is a semicircle in the lower half-plane with centre − 5 + 2 and radius ǫ. Z Z 2π −2ıdz 2dθ √ = 2 4 + ǫeıθ C1 z + 2 5z + 1 π π → 2 as ǫ → 0. Also, lim( ǫ→0

Z

√ − 5+2−ǫ −1

+

Z

1

)

√

− 5+2+ǫ

x2

−2ıdx √ = 0 + 2 5x + 1

Hence, the given integral is π2 . 1 1 ). (there is a pole at z = 71/7 (b) The integral is 2πı 49 (c) The poles of the integrand are at kπ where 1 ≤ k ≤ 199. Hence, the integral is (by P 2 the residue theorem) 2πı kπ = 2π ı × 19900. (d) The only pole is at z = π2 . Hence the integral is −π 2 i. 6. (a) Consider the semicircular contour C with centre at 0, radius R and situated R dz in the upper half-plane. Denote the curved part of it by E. Notice that C 1+z6 = R dz R R dx R dz R π Rdθ → 0 as R → ∞. Hence, by the residue + . But, | | ≤ 6 6 6 E 1+z −R 1+x 0 R6 −1 E 1+z R∞ 1 −ı5π/6 dx −ı25π/6 +e + e−ıπ/2 ). theorem, −∞ 1+x6 = 2πı 6 (e R Rπ (b) Use the same contour as above. Once again, | E (1+z)dz | ≤ 0 (1+R)Rdθ → 0. 1+z 4 R4 −1 ıπ/4

ı3π/4

+ 1+e ). Similar to above, the integral is (by the residue theorem), 2πı( 1+e 4eı3π/4 4eı9π/4 1 (c) Similar to the above. The answer is 2πı( 4ı ). (d) Consider the contour CR given by CR = [0, R] ∪ C1 ∪ L where C1 is the portion of 2πı Reı theta where 0 ≤ θ ≤ 2π and L is the line joining Re 3 to the origin. The only pole 3 ıπ R ıπ 1 2πıe 3 . By the Residue theorem, . F (z)dz = − this contains is e 3 . Let F (z) = 1+z 3 3 CR R 2πı R R Also, L F (z)dz = e 3 0 F (r)dr. We have the following estimate |

Hence, as R → ∞, − √ . is 32π 3

ıπ 2πıe 3

3

Z

C1

F (z)dz| ≤

= (1 − e

2πı 3

)

R∞ 0

Z

2π 3

0

Rdθ R3 − 1

F (x)dx. This means, the desired integral

(e) This integral is very similar to example 4 in the text and shall be omitted. 23

7. Since f is entire, f (z) = a0 + a1 z + a2 z 2 + . . .. Since |f (z)| = 1 when |z| = 1, on the circle we have, (a0 + a1 z + . . .)(a¯0 +

a¯1 a¯2 + 2 + . . .) = 1 z z

Comparing the coefficients of the powers of z (which we are allowed to do because of the uniqueness of Fourier series i.e. just divide by the appropriate power of z and integrate over the circle), X |ai |2 = 1 a0 a¯1 + a1 a¯2 + . . . = 0 a0 a¯2 + a1 a¯3 + . . . = 0 and so on. Now, if a0 6= 0, define λi = as,

ai . a0

Hence, the above equations maybe written

X

|ai |2 = 1

λ¯1 + λ1 λ¯2 + . . . = 0 λ¯2 + λ1 λ¯3 + . . . = 0 and so on. Notice that all but the first equation just imply that the vector (λ¯1 , λ¯2 , . . .) is in the kernel of the matrix that takes (1, 0, . . .) to (1, 0, . . .), (0, 1, 0, . . .) to (λ1 , 1, 0 . . .) and so on. This matrix is easily seen to be invertible (we may produce an explicit inverse via recurrence relations). Hence, λi = 0. The same proof shows that exactly one of the ai is non-zero. Hence, f (z) = az k where |a| = 1. This means, f (0) = a where either a = 0 or |a| = 1. In the former case, f (17) = a and in the latter, f (17) = a17k . (z− 12 )2 f (z) (z+ 31 )3 h(z)

8. Notice that equal to e

is a non-vanishing holomorphic function on the disc. Hence, it is

where h is holomorphic. Since, |f (z)| = 1 on the circle, 1 1 ℜ(h(eıθ )) = 2 ln(|(eıθ − )|) − 3 ln(|(eıθ + )|) 2 3

We may expand the right hand side as a power series in z = eıθ thereby obtaining a Fourier series for ℜ(h) restricted to the unit circle. From this, we may recover on Ph(z)ınθ ıθ the unit disc and thus the most general form for f (z). Indeed, ifP ℜ(h(e )) = an e , P then, we claim that ℜ(h(z)) = an z n and that ℑh(z) = c − ıan nz n where c is a real constant. This is most easily proved by using harmonic functions (treated in a later chapter). The real part of h is harmonic. A harmonic function on the disc is uniquely determined by its value on the unit circle. Indeed, the function h constructed above is harmonic and restricts to the right value on the circle. Given, the real part, the imaginary part is easily constructed by solving the Cauchy-Riemann equations in the polar form. 24

9. If we try to extend f to ∞, then we see that ∞ is an essential singularity (it can’t be a pole because it does not have a well-defined order). Hence, by the Picard theorem, f (z) assumes all but one value infinitely often. This means g(z) has infinitely many roots. 10. By the argument principle, this is the number of zeroes minus the number of poles of f (z) − 1 within the circle |z| = 3. There are no poles. On |z| = 3, |2 − 2z + z 2 | = z3 |1 + (z − 1)2 | ≥ 3 ≥ | 81 |. Hence, by Rouche’s theorem, f (z) and 2 − 2z + z 2 have the same number of zeroes within the circle. This means the answer is 2. 11. Taking n to ∞, we see (by the continuity of f ) that f (0) = 0. Hence, f (z) = z k g(z) k (where g(0) 6= 0). Notice that g( n1 ) = 7n . By taking n to ∞, it is clear that the only n3 ′′′ 3 way g(0) 6= 0 is if k = 3. Hence, f (z) = z g(z). This means, f (0) = 6g(0) = 42. R f (z)dz 7! . 12. Let C be a circle centred at 0 with radius R. Notice that f (8) (10.001) = 2πi C (z−10.001)9 So, |f (8) (10.001)| ≤

7! R23/3 2πR 2πı (R−10.001)9

which goes to 0 as R → ∞. Hence the answer is 0. R ∞ ıx 13. The integrals are the real and imaginary parts of −∞ e1+xdx2 . As usual, we choose the semicircular contour of radius R centred at 0. The integral over the curved part is bounded by Z π −R sin(θ) e Rdθ 2πR ≤ 2 R −1 R2 − 1 0 → 0 −1

Hence, the integral (by the Residue theorem) is 2πı e2ı (whose imaginary part is 0). 14. (a) On r < |z| (where r > 1), we have z2 z 3 (1/z − 1)2 (1/z + 1) X (−1)k X 1 = ( )( l( )l ) k z z k=0 l=0 X n − (n − 1) + (n − 2) . . . = zn n=0 X −(n + 1) = 2z 2n+1 n=0

f (z) =

On |z| < 1, we have, f (z) =

1 3z − 1 + 2 4(z − 1) 4(1 + z)

Each of the above maybe easily expanded to give a Taylor series about z = 0.

25

(b) There are four Laurent series for the given function, In |z| < 1, 1 1 1 − + 2(z − 1) z − 2 2(z − 3) X1 1 1 (−1 + k − k+1 )z k = 2 2 3 k=0

f (z) =

In 1 < |z| < 2, 1 1 1 z − 1 + 2z(1 − z ) 2(1 − 2 ) 6(1 − z3 ) ∞ ∞ X 1 X 1 1 1 = ( + ( k − k+1 ))z k k+1 2 k=0 z 2 3 k=0

f (z) =

In 2 < |z| < 3, 1 1 1 1 − 2 − 2z(1 − z ) z(1 − z ) 6(1 − z3 ) ∞ ∞ X 1 1 X zn 1 k = ( − 2 ) k+1 − 2 z 6 n=0 3n k=0

f (z) =

In 3 < |z|, 1 1 1 1 − 2 + 2z(1 − z ) z(1 − z ) 2z(1 − z3 ) ∞ X 3k 1 1 = ( − 2k + ) k+1 2 2 z k=0

f (z) =

(c) There are three Laurent series for the given P k function. In 0 < |z| < 1, the function 1 is (we shall use the geometric series 1−a = a when |a| < 1) 1 1 2 − ) f (z) = ( − z)( z 1−z 2−z ∞ X 2 1 = ( − z)( (1 − k+1 )z k ) z 2 k=0 = −

∞

X 3 1 7 − + z n−1 (1 − n ) 3 2 z z 2 n=0

In 1 < |z| < 2, f (z) =

z2 − 2 2z 2 (1 − z1 )(1 − z2 ) 26

n=∞ 1 1 X zn = ( − 2 )( ) 2 z n=−∞ 2n−1 ∞ X 3z n−2 = 2n−1 n=−∞

Similarly, in 2 < |z|, 2 − z2 z 3 (1 − z1 )(1 − z2 ) ∞ 1 X 2n+1 2 ) = ( 3 − )( z z n=0 z n

f (z) =

∞ X 1 2 2m = − − 2− z z z m+1 m=−∞ ′′

15. Since f (z) is entire, by example 6.11 and Picard’s theorem, it can miss at most 1 point ′′ ′′ if it is non-constant. But, |f − 3| ≥ 0.001 implies that f is indeed constant. Hence, f is a quadratic polynomial. From the given data, it is clear that f (z) = z(3z − 1). Hence, f (ı) = ı(3ı − 1). 16. Let h(z) = ef (z) . The given condition on f implies that |h(z)| > e−2 . But, by example 6.11 in the text along with Picard’s theorem, every non-constant entire function can miss at most 1 point. This implies that h(z) is a constant. Hence, f (−ı) = ı + 2. 17. Since f (z) is unbounded near 0, it certainly does not have a removable singularity. If it had a pole of order k, then g(z) = z k f (z) is well-defined at 0 (and is non-vanishing). 2 3 But, limn→∞ g( n1 ) = limn→∞ nnk whereas limn→∞ g( −1 ) = limn→∞ nnk which aren’t equal n (unless g(0) = 0 which isn’t allowed). Hence, it has an essential singularity at 0. 18. Firstly, f (z) = 0 somewhere in D. If not, f1 is holomorphic in D. By the maximum modulus principle, | f1 | ≤ 1c which implies |f | ≥ c which is a contradiction (by the maximum modulus principle once again). Secondly, on ∂D, since |f (z)| = c > |a|, by Rouche’s theorem f (z) − a has as many zeroes inside D as f (z) itself. This implies that f (z) = a somewhere in D. 1 . z

This implies Z that the equation holds trivially Z in this case. τ f (τ ) f (τ ) 1 1 If |z| < 1, notice that zf (z) = 2πı dτ and that f (z) = 2πı dτ . |τ |=1 τ − z |τ |=1 τ − z Hence, multiplying the first equation by z¯ and adding, we get the desired equality. zτ | = 1 (by an If |z| = 1, then, the estimate holds trivially. If |z| < 1, notice that | 1−¯ τ −z easy computation). Hence, the desired inequality follows.

19. If |z| = 1, then z¯ =

27

20. We shall compute f (11) (z) and show that it is 0. Let CR be a circle of radius R centred at 0. Z (A + BR10 )Rdθ 10! (11) |f (z)| ≤ 2π CR (R − |z|)12 → 0 as R → ∞. Hence f is a polynomial (of degree ≤ 10). 21. Notice that f (z) has finitely many poles and zeroes in the unit disc (if it had infinitely many of them, then they would’ve had a limit point and that is not allowed unless f is identically ∞ or 0). Let the poles be pi with orders ki and the zeroes be ai with orders li . Hence, in the same vein as problem 8, there exists a holomorphic function h(z) so k k 1 ) 1 (z−p2 ) 2 ... that eh(z) = f (z)(z−p . Then, the constraint |f (z)| = 1 on |z| = 1, implies (z−a1 )l1 (z−a2 )l2 ... (exactly as in problem 8), that we may recover ℜ(h(z)) and from that h(z) itself. ′

22. Note that f (z) and f (z) are holomorphic. Hence, by Goursat’s theorem, the first two integrals are 0. The third integral is (by the argument principle) 2π times the number 2 . Hence, by of zeroes inside C. On C, | exp(z 21 )(z − 21 )2 (z − 13 )2 | ≥ 1e × ( 21 )2 × ( 23 )3 ≥ 81 Rouche’s theorem, the number of zeroes inside C is 5. Thus, the third integral is 10πı. 23. Clearly f (z) has a non-removable singularity at 0. If f has a pole at 0, then f (z) = g(z) zk where g(z) is holomorphic and non-zero at 0. Hence, n2 = nk g(1/n). The constraint on g implies that k = 2. This means, g(1/n) = 1. But, g(−1/n) = 2. Thus limz→0 g(z) does not exist. This is a contradiction. Hence, f has an essential singularity at 0. 24. By Picard (and example 6.11), since an entire function can miss at most one point, f (z) is a constant. 25. Notice that 0 is certainly not a pole. We shall prove that it isn’t a removable singularity thus implying that it is an essential singularity. Indeed, if it were a removable one, then f (0) = 0. This means, f (z) = z k g(z) where g is holomorphic and non-zero at 0. Noting that, g( n1 ) = nk−2 , we see that k = 2. Hence, g(− n1 ) = 2. This implies that limz→0 g(z) does not exist. A contradiction. By Casorati-Weierstrass, the image near 0 is dense. Hence, lim inf z→0 |f (z) − 2| = 0. R∞ R∞ 2 4ıx 4ız 26. Notice that 0 sinx(2x) dx = ℜ( 14 −∞ 1−e dx). Let F (z) = 1−e . It has a pole at 0 2 x2 z2 with residue −4ı. If C is the contour (−R, −ǫ)∪C1 ∪(ǫ, R)∪C2 where C1 is a semicircle in the lower half plane centred at 0 with radius ǫ and C2 is one in the upper half plane centred at 0 with radius R. Z Z π 1 + e−4R sin(θ) | F (z)dz| ≤ dθ R C2 0 2π ≤ R → 0 28

as R → ∞. Also |

Z

C1

F (z)dz| = |

Z

2π

(4ıeıθ + π

(4ıǫeıθ )2 + . . .)e−ıθ dθ ǫ

→ 4π

as ǫ → 0. Hence, the integral is (by the Residue theorem), π. 27. If we prove that |f (0)| ≤ A, we will have solved the problem. This is because, there exists a Mobius transformation φ that takes the unit circle to itself and sends 0 to any given point a within the disc. But, Z 2π 1 |f (0)| = f (reıθ )dθ 2π 0 Since f is bounded, R 2π usingıθ the dominated convergence theorem (by taking limr→1 ), we 1 have |f (0)| ≤ 2π 0 |f (e )|dθ ≤ A. R n (w) 28. Firstly f is holomorphic. This is because 2πıf (z) = 2πı lim fn (z) = lim fw−z dw = R R f (w)dw fn (w) lim w−z dw = , where the interchange of limits and integrals is justified using w−z ′ uniform convergence. Also, by Cauchy’s estimates, fn → f uniformly on compact subsets. If f is not identically 0, then the argument principle gives us the number of times it vanishes on Dr where r < 1. Indeed, Z Z ′ ′ fn (z) f (z) dz = dz lim fn (z) ∂Dr ∂Dr f (z) Z ′ fn = lim dz ∂Dr fn ≤ 7 Hence the result. ′

z 1 ) = (1−z) 29. Notice that f (z) = z(1 + 2z + 3z 2 + . . .) = z( 1−z 2 . If f (z1 ) = f (z2 ), then 2 2 z1 z2 + z1 = z2 + z1 z2 . Hence, z1 z2 (z2 − z1 ) = z2 − z1 . Hence, if z2 6= z1 , and |z1 | < 1, then |z2 | > 1. So, f is injective on D. 1 If |z| = 1 and z 6= 1, f (z) = z¯+z−2 . Hence it gets mapped onto (−∞, −1 ). If α = f (z), 4 √ √ 4α+1 2α+1± then z 2 α − z(2α + 1) + α = 0. This implies z = . If 4α + 1 = w, then, 2α w+1 z = w−1 or z = w−1 . So, the image consists of all complex numbers except the ones w+1 1 w+1 for which | w−1 | = 1 i.e. C − (−∞, − 4 ).

30. By the Riemann mapping theorem (proved in a later chapter), the triangle maybe mapped biholomorphically via a map g onto the disc with the map extending continuously to the boundary. Such a map would take 21 (1 + ı) to a point b on the exterior 1 of the unit disc and 10 (1 + ı) to a on the interior. Notice that g ◦ f assumes a (by exercise 6.17). It certainly does not assume b (by the maximum principle). 29

R 2π 1 31. By Cauchy’s integral formula, f (0) = 2π f (erıθ )dθ for ever r. Hence, as r → 1, 0 by the continuity of f , the formula holds for r = 1. But this means f (0) = 0. By Schwarz’s lemma, |f (z)| ≤ |z| on |z| < 1. By continuity, |f (z)| ≤ |z| on |z| ≤ 1. But, |f (ı)| = 2. Hence, there is no such function. 32. (1) By Fubini’s theorem, if we evaluate a multi-variable integral in one order and the result is finite, the integral is well-defined and can be evaluated in any order. Z ∞Z ∞ 2 2 2 e−(x +y ) dxdy I = Z0 ∞ Z0 2π 2 e−r dθrdr = = π

0

0

√ Hence, I = π. (2) This is actually a very nontrivial problem. The reader is referred to the article “On evaluating eax(x−2b) by Contour integration round a parallelogram ” by D.Desbrow in Am. Math. Monthly, ’98. ˆ 33. The Residue theorem maybe stated as : Let f be holomorphic in a domain D ∈ C except for isolated singularities at z1 , z2 , . . . ∈ D. Let γ be a positively oriented Jordan curve homotopic P to a point in D such that zj ∈ i(γ). If ∞ is not on γ or in i(γ), R then γ f (z)dz = 2πı Reszj . Otherwise, we may find a fractional linear transformaR ′ tion φ taking γ ∪ i(γ) → φ(γ) ∪ φ(i((γ))) ∈ C. Then, φ◦γ) f (φ−1 (z))(φ−1 ) (z)dz = R R P ′ 2π Res((f ◦ φ−1 (z))φ (z))φ(zj ) . Define γ f (z)dz = φ◦γ f (φ(z))dz. Notice that this is well-defined Z 1 Z ′ ′ ′ −1 −1 ˜ ˜ ˜ f (φ (z))φ )(z)dz = f (γ(t))(φ(γ)) (t)(φ˜−1 ) (z)dt ˜ φ◦γ Z0 a ′ ′ ˜ f (γ(t))(φ(γ)) (t)(φ˜−1 ) (z)dt = Z0 1 ′ ′ ˜ + f (γ(t))(φ(γ)) (t)(φ˜−1 ) (z) a

where γ(a) might be ∞. Changing variables, we see that indeed the last integral equals the corresponding one for φ. With this definition, the same thing can be done with the argument principle. The same statement as given in the text holds provided we define the integral as above. 34. Near ∞, let w = z1 . Near w = 0, (a) g(w) = exp(w) has no (i.e. a removable) singularity. Hence, so does f at ∞. (b) g(w) = w1n has a pole of order n at w = 0 and hence so does f at ∞. (c) g(w) = w sin( w1 ) has a Laurent series about 0 with infinitely many negative powers. Hence, ∞ is an essential singularity for f . 30

35. We shall prove that infinity is either a pole or a removable singularity, thus implying that f is a polynomial. This would mean (by the fundamental theorem of algebra) that f is of degree one. Indeed, if infinity is an essential singularity, then, A = f (|z| > 1) is dense. Since f has an inverse, this means that A is an open set and so is B = f (|z| < 1). But, the injectivity of f implies that, A ∩ B is empty. This is a contradiction and hence infinity is either a pole or a removable singularity.

31

Chapter 7 1

1. (a) Let fn (x) = |x| n +1 . Clearly fn (x) is differentiable everywhere (including at 0). Now, if K is a compact set, let maxK (|x|) = C. Then, 1

1

||x| n +1 − |x|| = |x|(|x| n − 1) 1

≤ C(C n − 1) → 0

as n → ∞. Hence, fn (x) uniformly approximates the absolute value function. 1 (b) Let D be the open unit disc. The functions fn (z) = |z| n are continuous on D and converge to 1 on D − (0) and to 0 at 0. n+1

2. Let fn (x) = xn+1 . As n → ∞, fn (x) converges uniformly to 0. This is because, ′ 1 |fn (x)| ≤ n+1 . But, fn (x) = xn which does not converge uniformly (the pointwise limit is 0 when x < 1 and 1 when x = 1; This isn’t continuous). 3. Let D be the open unit disc. If fn (z) = z + 25, then, fn (z) → z + 25 which does not vanish anywhere on D. If fn (z) = z n , then, fn (z) → 0 uniformly on compact subsets of D. 4. Notice that X 1 1 π cot(πz) − 2 = 2 z z z − n2 P 1 P 1 π cot(πz) −1 1 Hence, = lim − . This is true because converges uniformly z→0 2 2 n 2 z z z−n2 on small compact sets containing 0. −1 π cot(πz) πz cot(πz) − 1 1 −1 lim − 2 = lim 2 z→0 z z 2 z→0 z2 πz cos(πz) − sin(πz) −1 lim = 2 z→0 z 2 sin(πz) 2

3

+ . . .) − (πz − (πz) + . . .) πz(1 − (πz) −1 2! 3! = lim 3 (πz) 2 z→0 z 2 ((πz − 3! + . . .)) π2 = 6 32

5. If z = x + ıy, X

−x + n y 1 = +ı 2 2 n−z (x − n) + y (x − n)2 + y 2

For sufficiently large n the real is > 0. Hence we may use the integral test to R ∞part−x+t R ∞ −x+t decide convergence. Clearly, 1 (x−t)2 +y2 dt > 1 2(x−t)2 dt = ∞. Hence, the series diverges. 6. If A is strongly bounded, then choosing ǫ = 1, for a given compact set K, there exists a λK so that kgkK ≤ λK ∀g ∈ A. Conversely, if there exists an M (K) so that g kgkK ≤ M (K)∀g ∈ A, then given an ǫ and a K, let λK,ǫ = M (K) . Then λK,ǫ = f where ǫ kf k ≤ ǫ. 7. (1) If fk ∈ B(K, S) → f uniformly on K, then f is continuous on K. Let gnk → fk be a sequence of rational functions converging to fk uniformly on K. Let us choose n1 , n2 . . . so that kgnk k − fk kK ≤ 21k . We claim that gnk k → f uniformly. Indeed kgnk k − f kK

≤ kgnk k − fk kK + kfk − f kK 1 + kfk − f kK ≤ 2k → 0

as k → ∞. Hence B(K, S) is closed under uniform limits. (2) The usual C 0 norm kf k = supK |f | works. Clearly this is a norm. The algebra operations are obviously continuous. The completeness of this norm is a standard fact about C0 norms. (3) One may refer to Rudin’s book (Real and Complex Analysis) for a functional analytic proof using the Hahn-Banach theorem. ′

′

′

8. If fn → f in the H(D), then, fn → f and hence, f (0) = 0 and f (0) = 1. By Hurwitz’s ′ theorem, f is injective (it can’t be a constant because f (0) = 1) and hence the set S is closed.

33

Chapter 8 1. tA(z) =

taz+tb tcz+td

=

az+b cz+d

= A(z). (z

−z

)(z

−z

)

π(3) π(2) π(4) where π is a permutation. It is easy to 2. (zπ(1) , zπ(2) , zπ(3) , zπ(4) ) = (zπ(1) π(1) −zπ(4) )(zπ(2) −zπ(3) ) see that the number of permutations fixing the cross-ratio is 4. Hence, the number of orbits under the action of the permutation group is 6. They are (let (z1 , z2 , z3 , z4 ) = C).

(z1 , z3 , z4 , z2 ) = (z1 , z4 , z3 , z2 ) = (z1 , z2 , z4 , z3 ) = (z1 , z3 , z2 , z4 ) = (z1 , z4 , z2 , z3 ) =

1 1−C C C −1 1 C 1−C C −1 C

3. If we prove that the circum-radii R1 and R2 of the two triangles z3 , z4 , z1 and z3 , z4 , z2 (respectively) are equal, then, their circumcentres coincide and we will be done (because the circumcentre is obtained by finding the intersection of two arcs drawn with z3 , z4 as centres). By the sine rule of triangles, a = 2R sin(A) where a is the length of the side opposite to the angle A. Let a be |z3 − z4 |. Then, R1 = R2 if and only if sin(A1 ) = sin(A2 ) where A1 and A2 are the angles subtended by the line z3 , z4 at z1 and z2 respectively. Hence, A1 = A2 if and only if R1 = R2 . 4. (1) We showed in an earlier chapter that Mobius transformations map straight lines to ˆ (we actually showed that rotations of the Riemann sphere correspond to circles in C Mobius transforms; Rotations take circles to circles). (2) The cross ratio (z1 , z2 , z3 , z4 ) is the image of z1 under a particular Mobius transformation (taking the other three points to points on R ∪ ∞). Hence, by part (1), the image is on R if and only if z1 , z2 , z3 , z4 lie on a circle. 5. If f (z) is of the given form, then, when |z| = 1,

1 + |a|2 − a ¯z − a¯ z 2 1 + |a| − a ¯z − a¯ z = 1

|f (z)|2 =

34

By the maximum modulus principle, |f (z)| ≤ 1∀|z| ≤ 1. It is easy to see that f (z) is ye−ıθ +a ). bijective (by solving y = f (z) we see that z = 1+¯ aye−ıθ such that |u|2 −|v|2 = 1. If f is an automorphism of D, then, by theorem 8.18, f = vuz+v ¯z+¯ u v z−a u ¯ −ıθ Let a = − u . Then, f (z) = v¯ z+ u¯ . Let e = u . These substitutions force f to be of u u the desired form (notice that f (a) = 0 and hence a ∈ D). 6. Omitted. ˆ containing ∞, firstly, D misses atleast 1 7. Given a simply connected domain D in C ′ point a. Consider a Mobius map φ that maps a to ∞, φ (ζ) to 1 and ζ to b. Then, φ(D) ∈ C. The usual Riemann mapping theorem implies that there is a unique map ′ g : φ(D) → D such that g(b) = 0 and g (b) > 0. Then, g ◦ φ is the desired Riemann map. 8. We note that a Riemann map always takes the boundary to the boundary (an easy observation that follows from the fact that conformal maps preserve angles between tangent vectors). Also, three points uniquely determine a Mobius transformation. Besides, a Mobius map always maps circles to straight lines or circles. (1) The map ı(z+1) maps the curved part of this region onto the negative real axis. 1−z It maps the straight line part of this region to the imaginary axis. Hence, it maps the semi-disc onto the second quadrant. Multiplying by −ı rotates this to the first z−ı quadrant. The map z 2 maps this to the upper half-plane. The map z+ı maps the upper half-plane onto the disc. The inverse of the composition of these maps gives the answer. . Note that f (0) = ∞, f (2) = 0 and f (−2) = 2ı. Indeed f (2) Consider f (z) = ı(z−2) z maps |z + 1| = 1 to a straight line. It maps a point on the exterior of the circles (ı) to a point in the strip ı − 2. Also, it is forced to map the circle |z − 1| < 1 to a straight line (the image contains ∞). Since f (−2) = 2ı, it is mapped to the line ℑ(z) = 2. (3) First map it to the strip as in (2). Then, compose it with the map exp( π2 z). This maps it onto the half-plane. (4) The map nLn(z) takes the sector to the upper half-plane. Then, z−ı maps the z+ı upper half-plane onto the unit disc. (5) First map the strip to the unit disc in any way (as above). Then, using a Mobius transformation, √ one may find the desired map. (6) The map z maps it onto the semi-disc which maybe mapped onto the disc using the inverse of the map in 1. z−ı . Then, the inverse of (4) maps (7) Map the upper half-plane to the unit disc by z+ı 2 the unit disc to the semi-disc. The map z takes the semi-disc to the desired region. 9. Riemann maps are unique. So we just need to find one map that does the job. The map ız 50 maps the domain to the upper half-plane. Then, z−ı maps the upper halfz+ı 50 plane to the disc. The composition g of these maps is g(z) = zz50 −1 . This satisfies the +1 250 −1 desired properties. Hence, g(2) = f (2) = 250 +1 . 35

10. The map z 2 maps the quadrant onto the upper half-plane which is then mapped onto 2 −ı z−ı . Hence, the composition is f (z) = zz2 +ı . By composing with an the unit disc by z+ı appropriate Mobius transformation, we get g(z) (because Riemann maps are unique). √ ′ 4 2 z 2 −ı Hence, g(z) = i z2 +ı . Thus, |g (1 + ı)| = 9 . 11. By the theorem in the text on Blaschke products, we see that |f (z)| ≤ |B0 (z)B 1 (z)|. 2 This implies the inequality. 12. The sequence gn = ıfn have their images in the upper half-plane. Hence, by composing z−ı with z+ı , we get a sequence of holomorphic functions hn which map into the unit disc. Since |hn (z)| < 1, this sequence is uniformly bounded on compact sets. Hence (by the theorems in the preceding chapter), it contains a subsequence that converges uniformly on compact sets. This implies fn also has such a subsequence. 13. We just need to prove that ρ is a metric. The fact that it is invariant under the Mobius group holds because ρ = ln(|(z ∗ , w∗ , w, z)|) and cross-ratios are preserved by the Mobius group. Indeed, for any domain D, ρ is clearly symmetric and ρD (x, y) ≥ 0. Also, ρD (x, z) = inf γx,z l(γx,z ) ≤ inf γx,y l(γx,y )+inf γy,z l(γy,z ) where γx,y is a path joining x and y. If ρD (x, y) = 0, then, using the Riemann map π, ρD (π −1 (x), π −1 (y)) = 0. Hence, if we prove that ρD (x, y) = 0 implies x = y, then we will have proved that ρD is a metric. Indeed, Z 2 l(γx,y ) = |dz| 2 γx,y 1 − |z| Z ≥ 2 |dz| γx,y

Hence, if inf l(γx,y ) = 0, then, inf

R

γx,y

|dz| = 0. But this implies x = y.

14. This follows from the previous and the next problems. 15. We may locate the centre of the semicircle by finding the intersection of the perpen+ tı(w − z) with the real axis. Then, finding z ∗ and w∗ dicular bisector of z, w i.e. z+w 2 is easy t = z∗ = = w∗ = =

ℑ(z + w) 2ℜ(z − w) z+w ℑ(z + w) w−z ℑ(z + w) + ı(w − z) − | + ı(w − z)| 2 2ℜ(z − w) 2 2ℜ(z − w) |z|2 − |w|2 − |w − z||z − w| ¯ 2ℜ(z − w) z+w ℑ(z + w) w−z ℑ(z + w) + ı(w − z) + | + ı(w − z)| 2 2ℜ(z − w) 2 2ℜ(z − w) |z|2 − |w|2 + |w − z||z − w| ¯ 2ℜ(z − w) 36

Using these, we may find the cross-ratio and prove that ρH (z, w) = | ln(z ∗ , w∗ , w, z)| = |z−w|+|z−w| ¯ . This gives the desired result. ln |z− w|−|z−w| ¯ 16. Consider the map T (z) =

z−ı z+ı

that maps H to D. Indeed,

ρD (0, z) = ρH (T −1 (0), T −1 (z)) = ln(

|−ı−

ı(z+1) | 1−z ı(z+1) | 1−z

|−ı− 1 + |z| = ln( ) 1 − |z|

+ |ı −

− |ı −

ı(z+1) | 1−z ı(z+1) | 1−z

17.

ρ(z, w) = = = = =

√

a2 − 1) |z − w| ln(1 + 2 ) |z − w| ¯ − |z − w| |z − w|(|z − w| + |z − w|) ¯ ) ln(1 + 2 2 2 |z − w| ¯ − |z − w| |z − w|(|z − w| + |z − w|) ¯ ln(1 + ) 2y1 y2 |z − w|2 (|z − w||z − w|) ¯ ln(1 + + ) 2y1 y2 2y1 y2 |z − w|2 ) arccosh(1 + 2y1 y2

arccosh(a) = ln(a +

where the last equality follows from a trivial calculation. 18. We shall assume that a geodesic is a curve that extremises the length between two points. This means that γ is a geodesic between a and b if any perturbation of this curve (fixing the endpoints) does not change the length to the first order (in the sense of calculus of variations) i.e. Z p ((γx + sǫx )′ )2 + ((γy + sǫy )′ )2 d ( )(s = 0) = 0 ds γy + sǫy (t) This implies the following differential equation for γ ( All of this is standard in Riemannian geometry. Do Carmo’s book is a good reference) ′

γx

( γy q

(γx′ )2 + (γy′ )2 γy2

+( γy 37

q q

′

)

′

′

= 0

(γx )2 + (γy )2 ′

γy ′

(γx

)2

′

+ (γy

) )2

′

= 0

(1) We easily see that γ(t) = it satisfies the differential equation and is hence a geodesic. (2) If the length of the segment equals ρ(a, b), then, the curve is an extremiser and is hence a geodesic by definition. Conversely, a geodesic is a minimiser (and hence the length of the segment is ρ(c, d)) because the second variation is positive. This (easy) calculation is done in Do Carmo’s book (stability of geodesics in the case of negative sectional curvature). (3) Such semicircles are certainly geodesics. Indeed, upon plugging in γ(t) = a + r cos(t) + ır sin(t) (a semicircle with centre (a, 0) and radius r) into the geodesic equation, we see that it is satisfied. Besides, given any point and a direction at that point, there is a unique semicircle passing through that point tangent to the given direction (just draw a line perpendicular to the tangent and see where it intersects the real axis; that gives the centre). By part (5), this implies that the only geodesics are such semicircles. (4) Given two points a and b, the centre of a circle containing these points lies on the perpendicular bisector of the line joining them. Since we established that the geodesics are semicircles perpendicular to the real axis, the centre lies on the intersection of the perpendicular bisector with the real axis. Hence, they lie on a unique geodesic. (5) Given a point and a direction vector at that point, the geodesic equation being a second order ODE has a unique solution. 19. (1) If A(z) =

az+b cz+d

(ad − bc = 1) fixes ı, then, aı + b cı + d b a 2 a + b2

′

= ı = −c = d = 1

1 2ab 2 2 2 (2) A (ı) = (a−ıb) 2 = (a + ıb) = a − b + 2ıab. Hence, tan(θ) = a2 −b2 . From this and a2 + b2 = 1, upon squaring the first equation and simplifying, 4a4 − 4a2 + sin2 (θ) = 0. ; This means a = ± cos( 2θ ) (the other choice does not satisfy the Hence, a2 = 1±cos(θ) 2 tan(θ) equation). The two signs give the same matrix in PSL(2,R). Thus, there is a unique matrix given θ. Notice that, by choosing a fixed map that takes i to a, the same result holds for any a ∈ H. (3) There exists a M¨obius map (in P SL(2, R)) A0 taking z to w (Indeed one may find such a map from the disc to itself and then convert that into one from H to itself. To find one on the disc, use the explicit formula from theorem 8.18 to solve for a in terms of b). Every map A taking z to w is A0 ◦ B where B(z) = z. By (2), such a map B is ′ ′ uniquely specified by arg B (z) and hence so is A by arg A (z). (4) If there exists such an A, then, clearly ρ(a, b) = ρ(z, w). Conversely, if the latter holds, choose some fixed map B that takes a to z. Now, every map that takes a to z is the composition of B with some map C that fixes z. Clearly, ρ(z, B(b)) = ρ(z, w). ′ ′ ′ Notice that, if γ(t) is a curve, then arg(f ◦ γ(t)) = arg f + arg γ (t) i.e. a hyperboliclength-preserving holomorphic map f rotates (in the usual Euclidean metric) a tangent

38

′

vector at z by arg f (z). If we can find a map C that fixes z and takes the tangent vector (at z) of the (unique) geodesic joining z and B(w) to the one joining z and w, then, the desired map A is certainly A = C ◦B (because a geodesic is determined by its initial tangent vector). Indeed, let the angle that the tangent vector has to be rotated ′ be θ. By (2), there exists a unique map C such that C(z) = z and arg(C (z)) = θ. 20. By the theorem in the text on Blaschke products, |f (z)| ≤ |B 2ı (z)B −ı (z)| = |g(z)|. 2

(z) is a bounded holomorphic function on the disc (it is bounded by 1 Hence, G(z) = fg(z) everywhere and hence extends holomorphically to ±ı ). 2

21. Without loss of generality, one may assume that the disc is the unit disc (otherwise, M¨obius-transform it to the unit disc). Then, this follows from theorem 8.18. 22. Omitted.

39

Chapter 9 1. This is an easy exercise based on the Chain rule and we omit it. ¯z ) (and vice-versa). Analytic functions are harmonic. 2. If f is analytic, then, so is f (¯ 3. ω = (f + g)dx + ı(f − g)dy. From this, the result follows easily. 4. d is complex linear. If u1 = ıv (where v is real) then v is harmonic if and only if u is, and d(u∗1 du2 − u∗2 du1 ) = ıd(v ∗ du2 − u∗2 dv) = 0 (from the proof for real valued harmonic functions). Similarly, when u2 = ıv, the form is closed. The form is linear in u1 and u2 and hence the form is closed for complex valued harmonic functions. 5. We shall prove the maximum principle. The minimum one follows by replacing u with −u (where u is our harmonic function). (1) If u attains a local maximum at x0 , Z 2π 1 u(x0 + reıθ )dθ u(x0 ) = 2π 0 ≤ u(x0 ) Equality implies that u is locally constant. The set of x such that u(x) = u(x0 ) is closed and the preceding argument implies that it is open as well. Since it is non-empty and connected, u is a constant throughout. (2) Let us assume (without loss of generality) that u attains a local maximum at 0. This means, 0 ≤ g(0) − g(z) when |z| < r where r is sufficiently small. Then, using g(0) − g(z) as the positive harmonic function in Harnack’s inequality, we obtain, 0 ≤ g(0) − g(z) ≤ 0. Hence, g(z) = g(0) in that small disc. By the same argument as in (1), g(z) is a constant. 6. Cover K by small discs Ui (with centres ak ) lying entirely in D such that the centre of one of these discs U1 is z2 and the centre of another Um is z1 . Harnack’s inequality 2) 2) implies that, on U1 , u(z ≤ u(z) ≤ A1 u(z2 ). If Uk overlaps with U1 , then, Au(z ≤ A1 1 Ak u(ak ) ≤ u(z2 )A1 Ak . If Ul overlaps with any of the Uk that overlapped with U1 , then 2) (again by the Harnack inequality), Au(z ≤ u(al ) ≤ u(z2 )A1 Ak Al . Continuing this 1 Ak Al way (i.e. using induction), we see that (since by connectedness, there is a finite chain 1) ≤ c where c depends on the of overlapping sets that lead to Um ), we see that, 1c ≤ u(z u(z2 ) 40

U and K. We may choose the maximum of all products ΠAk to be c so as to make it independent of z1 and z2 . 7. The only fact that needs to be justified is done so as follows : un+1 − un is harmonic. Also, since the family is non-decreasing, un+1 − un ≥ 0. Hence, by the Harnack inequality, sup(un+1 (z)−un (z)) ≤ C inf(un+1 (z)−un (z)) ≤ C(un+1 (z0 )−un (z0 )). Both possibilities listed in the conclusion occur: Consider the domain to be the entire plane. Then, un (x + ıy) = x is a constant sequence and the limit is harmonic; un (x + ıy) = 2n goes to infinity everywhere. 8. Choose a disc of radius r and centre ζ in D. Let v be the harmonic function on the disc that agrees with u on the boundary. Now, F = u − v + ǫ|z − ζ|2 satisfies the maximum principle. Indeed, if F attains a local maximum at z0 , then, since u is twice differentiable separately in x and y, Fxx ≤ 0 and Fyy ≤ 0. But, Fxx + Fyy = 2ǫ > 0. A contradiction. Hence, F can’t attain a local maximum. This means, F ≤ 0 on the disc. Hence, u(z) ≤ v(z) + ǫ|z − ζ|2 for every ǫ. Hence, as ǫ → 0, u(z) ≤ v(z). The same argument maybe applied by reversing the roles of f and v. Thus, u = v. This means u is harmonic. 9. Yes, it does. Clearly harmonic functions satisfy the area mean value property. Using the same notation as above, F (z) = f − v satisfies the maximum principle because it satisfies the area mean value property (the proof is exactly the same as in the case of the usual mean value property). Hence, f ≤ v. Applying the argument to −F , we see that f = v. 10. By the Poisson formula, we get ı 3 u( ) = 2 8π 1 = − 4

Z

2π 0

cos(2θ) + sin(2θ) dθ |eıθ − 2ı |2

11. This follows trivially from the Harnack inequality. 12. a) Indeed f is holomorphic for all values of α (it can be expanded as a power series in z using the geometric series). b) The real and imaginary parts of such an f (being harmonic) are determined by φ. Indeed, if fα = uα + ivα , uα = u1 and vα = αv1 . For α = 1, f (z) = z is the desired function. Hence, fα = x + iαy which is holomorphic if and only if α = 1. c) It is true for all values of α (the condition doesn’t even depend on the imaginary part of φ). 13. By the formulae series interpretation), Pin the text P (in the section 1onR the Fourier P [g](z) = a0 + an z n + bn z¯n where bn = 2π g(ζ)ζ n dζ. Such an f exists if and only if bn = 0 (in that case f = P [g]). Hence, the result. 41

14. RNotice that ζ + ζ −1 = ζ + ζ¯ on the unit circle. The previous exercise tells us that, since g(ζ)dζ = 2πı 6= 0, there is no such f . 2

2

∂f ∂ ∂ f 2 = 2f ∂z∂ f + 2 ∂f . Hence, if f 2 and f are harmonic, f is holo15. Notice that ∂z∂ z¯ z¯ ∂z ∂ z¯ morphic or anti-holomorphic. If f is holomorphic or anti-holomorphic, obviously f is harmonic. The above identity then shows that so is f 2 .

16. It is easy to see that the harmonicity of h implies that of the functions mentioned (be∂2 cause the laplacian is proportional to ∂z∂ . We may assume (without loss of generality) z¯ that h is real-valued. It is then the real part of a holomorphic function g. We shall ′ prove that |g | satisfies or violates the desired inequalities (thus implying the same for the derivatives of h). R ′ 1 Notice that, |g (0)| ≤ 2π | g(z) dz| ≤ M . z2 R ′ Consider the sequence of functions gn (z) = z n on the unit disc. Here M = 1. |gn (z)| can be made arbitrarily close n by choosing a point close enough to the boundary. Hence, c > n for every n. This means, there is no such universal constant c that works for z 6= 0. and p(w) = g ◦ φ−1 . Consider a disc D with centre z and radius r. Let φ(w) = r2 r2z−w −¯ zw We see that, φ is an automorphism of D sending z to 0 and by the first part of this ′ ′ r M problem, |p (0)| ≤ Mr . Hence, |g (z)| ≤ r2M ≤ r−|z| . Sending r → R, we get the −|z|2 desired inequality. 17. Assume h is real (we may repeat the same argument for the imaginary part if there is any). A harmonic function on a disc is the real part of a holomorphic function f . Now |ef | = eh which is bounded. Hence, the holomorphic function has a removable singularity at z0 . Thus, h = ln(|ef |) extends as a harmonic function to the entire disc. 18. If the set of critical points has a limit point p in D, then choose a small disc U in D about p. The harmonic function u restricted to U is the real part of a holomorphic function f . Wherever du = 0, dv = 0 where v is the imaginary part of f . Hence, fz = 0 on a set having a limit point. By the identity theorem fz = 0. But u is a non-constant function. This is a contradiction. 19. u is superharmonic if and only if −u is subharmonic. The rest of the problem follows from the properties of subharmonic functions. 20. Z Z

u(z)dA =

Z rZ 0

|z−ζ|≤r

≤

Z

2π

u(ζ + reıθ )dθtdt 0

r

2πu(ζ)tdt 0

≤ πu(ζ)r2 42

21. (a) Let u be a non-constant, bounded subharmonic function on the plane. Also, let M = supC u, m be the maximum of u on the unit circle, and v = u − ln(|z|). By the maximum principle, on the unit disc, u ≤ m. Since v is subharmonic away from the origin, it satisfies the maximum principle there. Since v → −∞ as |z| → ∞ and v = u on the unit circle, by the maximum principle, v ≤ m away from the unit disc. Taking c → 0, we see that, u ≤ m throughout C. But this is a contradiction because m < M . (b) If u is a nonconstant subharmonic function on D = C − (p1 , p2 , . . .), then, we shall construct a bounded, non-constant function that is subharmonic on the entire plane. We then derive a contradiction by using part (a). Indeed, choose sufficiently small (so that the function we end up defining is not a constant) discs around the punctures pi . Harmonise u on these discs i.e. Let v = u outside these discs and equal to the unique harmonic function on the discs agreeing with u on the boundary. 22. (a) It is obviously positive and harmonic on 0 < |z| < 1. Also, − log(|z|) + log(|z|) = 0 is harmonic throughout. If g is another function satisfying the two properties, then, g + log(|z|) attains its minimum (taken over |z| ≤ r < 1) on |z| = r. Hence, on that set g + log(|z|) ≥ min|z|=r g + log(r) ≥ log(r) → 0 as r → 1. Thus, g + log(|z|) ≥ 0. This means − log(|z|) is indeed the Green’s function. z−a (b) It is − log(| 1+¯ |). This function is certainly positive. It is harmonic on |z| < 1−(a) az z−a ). If g is another because it is the real part of the holomorphic function − log( 1+¯ az z−a function satisfying those two properties, then (let φ(z) = 1+¯az ) g ◦ φ−1 is harmonic and positive on 0 < |z| < 1 and hence (by part (a)) is greater than − log(|z|). Thus, z−a g ≥ − log(| 1+¯ |). az 23. (a) Notice that gD2 (z, a) is harmonic and positive on D1 − a, and gD2 (z, a) + log(|z − a|) extends to D1 . Hence, by definition of gD1 , gD1 (z, a) ≤ gD2 (z, a). Now, g = gD2 − gD1 is harmonic and non-negative on D1 . If g(z0 ) = 0 for z0 ∈ D1 , then z0 is a point of local minima for g and thus g is a constant. A contradiction, Thus, gD1 < gD2 . (b) From the next problem we see that log(r) − log(|z|) is the Green’s function for a disc of radius r with the singularity at 0. Hence, by part (a), we are done. 24. Clearly, − log(|f (z)|) > 0 and harmonic on D − (a) (because it is the real part of ′ − log(f (z))). Since f (a) = 0, f (z) = g(z)(z − a). Also, since f (a) > 0, g(a) 6= 0. This implies that − log(|f (z)|) + log(|z − a|) = − log(|g(z)|) which extends to a harmonic function on D (because it is the real part of a holomorphic function on D). Notice that this may not be true if the normalisation condition was dropped. If g˜ is another function satisfying these properties, then g˜ ◦ f −1 is positive and harmonic on D − (0). Besides, |˜ g ◦ f −1 (z)| + log(|z|) = |˜ g (y)| + log(|f (y)|) = |˜ g (y)| + log(|y − a|) + log(|g(y)|) which extends to a harmonic function on D. Hence, g˜ ◦ f −1 ≥ − log(|z|). This implies that − log(|f |) is the Green’s function for D. 25. This problem is very similar to the previous one (the conformality of the map replaces ′ the condition that f (a) > 0). 43

26. Any proper domain that contains ∞ can be Mobius-transformed by z → z1 into one that is in C. Since Green’s functions are conformal invariants (in the sense of the previous problem), one may use results on the usual Green’s functions and translate them back (via the Mobius map) into the domain containing ∞.

44

Chapter 10 1. This problem is somewhat hard. The reader is referred to Problem 9 in chapter 3 of Knopp’s problem book, Volume 2, and to the Wolfram Alpha article on infinite products. P 2. (1) If the product converges absolutely, then | ln(1 + an )| converges. But, | ln(1 + |xn | an )| ≥ ln(|1 + an |) ≥ ln(1 + xn ) ≥ 2 (notice that ln(1 + x) ≥ |x| when x is small). 2 P yn | ≥ |y2n | . Hence P |xn | converges. Now, | ln(1 + an )| ≥ | arg(1 + an )| = | arctan 1+x n Hence |yn | converges and so the sum converges absolutely. yn If the sum converges, then since arg(1 + an ) = arctan( 1+x ) ≤ arctan(yn ) ≤ yn . Hence, n N X n=1

| ln(1 + an )| ≤ ≤ ≤

N X

n=1 N X n=1

X

ln(|1 + an |) + |an | ln(1 + |an |) + |an | +

X

|an |

X

|an |

Thus the product converges absolutely. (2) This follows from the result for sums (proved in the exercises of the first chapter). (3) Insufficiency: Omitted. 1 . Then Non-necessity: Let an = − n+1 converges to 0.

P

an diverges whereas, the product 21 . 32 . 43 . . .

√ 2 3. By Legendre’s duplication formula, πΓ( 31 ) = 2− 3 Γ( 61 )Γ( 23 ). By Γ(a)Γ(1 − a) = 2π π csc(πa), we have Γ( 32 )Γ( 13 ) = √ . From these two, we get the desired result. 3 4. Note that Γ(a)Γ(1 − a) = π csc(πa). Hence, the residue at z = −n is lim (a + n)Γ(a) =

a→−n

=

lim (a + n)π

a→−n

1 sin(πa)Γ(1 − a)

1 (−1)n Γ(1 + n) 45

=

(−1)n n!

5. The gcd of D1 = Πziνi , D2 = Πaµi i is defined as g = Πbαi i such that D1 and D2 are divisible (as formal products) by g and if any divisor divides both, it divides g too. The lcm is defined as l = D1gD2 where the empty set is treated as 1. The formulae for g and l are Πz min(νz ,µz ) and Πz max(νz ,µz ) respectively. It is easy to verify this. 6. Note that f and g have the same zeroes (with multiplicity) if and only if fg and fg exist as non-vanishing meromorphic functions. They have the same poles (with multiplicity) if and only if and fg exist as holomorphic functions. Hence, they define the same divisor if and only if h = fg and fg are non-vanishing holomorphic functions. In fact the first condition is enough (because if h is non-vanishing, then h1 is holomorphic and nonvanishing). 7. Obvious. 8. If f = 0, then h = g is the gcd. If both are not identically zero, then consider the gcd a of the divisors (f ), (g). By theorem 10.10, we may construct an h so that (h) = a. Such an h is obviously the gcd of f , g. 9. We have eC

′

+C( 21 +ıy)

′ 1 1 1 1 1 π ( + ıy)ıy eJ( 2 +ıy) eC +C( 2 −ıy) ( − ıy)−ıy eJ( 2 −ıy) = 2 2 cosh(πy) 1 ′ ( + ıy)ıy J( 1 +ıy) J( 1 −ıy) π e 2 = e 2 e2C −1 12 ıy cosh(πy) ( 2 − ıy)

As y → ∞, J → 0. Hence, e

′

2C −1

( 21 − ıy)ıy 2π = lim πy y→∞ e + e−πy ( 12 + ıy)ıy

Noting that z a = eaLn(z) and e−y → 0 as y → ∞, we use L’Hospital’s rule to get ′ . C = ln(2π) 2 10. Let the Weierstrass factors be denoted by Epn (z). The function f (z) = ΠEpn ( azn ) has zeroes at ai . If we find a meromorphic function g with specified poles (with their residues), we will be done because f g (for the appropriate residues at the poles) will then be our desired function. This is the statement of the classical Mittag-Leffler theorem. The proof of this is standard and maybe found in Ahlfors’ book. 11. This boils down to showing that every non-empty set of integral divisors has a greatest common divisor (once we have this, we may use Theorem 10.10 to construct the holomorphic function). Indeed if Dα = Πz ναz is a family of divisors, then g = Πz minα ναz 46

is definitely well-defined (it has support on a discrete set. If not, then there exists a member of that family having support on a non-discrete set). Indeed g is the gcd (as is easily verified). P 12. If h 6= 1, divide by h throughout in the equation fk gk = h to reduce the problem to the case where h = 1. If h = 1 and n = 2, we ought to prove that there exist g1 and g2 so that f1 g1 + f2 g2 = 1. Since f1 , f2 have 1 as their gcd, they don’t have common roots. We see that g1 = 1−ff12 g2 . Hence, we must choose g2 so that if f1 (a) = 0, then g2 (a) = f21(a) with the same multiplicity. This maybe accomplished through exercise 10 in this chapter. For n > 2 and h = 1, we shall use induction. The base case has been proven. Assuming the induction hypothesis for n = Pk, for n = k + 1, assume the gcd of f1 , . . . fk is u. Then there exist v1 , . . . vk so that fj vj = u. The gcd of u and fk+1 is 1. Hence, by the base case there exist p1 , p2 so that p1 u + p2 fk+1 = 1. Thus, gi = p1 vi for i ≤ k and gk+1 = p2 . 13. Let h be the gcd of f1 , . . . fk . By the previous exercise, h is in I. In fact, by definition of gcd, I is generated by h. 14. Let a1 , a2 , . . . ∈ D. Consider a maximal ideal I containing the ideal generated by (z − a1 )(z − a2 ), (z − a3 )(z − a4 )(z − a5 ), . . .. It is not finitely generated. If it were, it would have been a principal ideal (f ). The function f must vanish somewhere. If it vanishes at a, then (f ) = (z − a). But, if (z − a1 )(z − a2 ) = (z − a)g(z), then g(z) must have a pole at a unless a = a1 or a = a2 . But the same argument produces a contradiction when applied to (z − a3 )(z − a4 )(z − a5 ). Hence I is a non-principal maximal ideal. 15. If φ is not onto, then φ = 0. If φ 6= 0, then φ is onto. This implies that the kernel is a maximal ideal (C is a field). Define a = φ(z). Define I to be the ideal of functions that vanish at a. It is (z − a) and is a maximal ideal. Indeed, if a maximal ideal contained it, then it contains a function that does not vanish at a. If it does not vanish anywhere, the ideal is H(D); If it vanishes at b, then z − b is in the ideal and hence so is 1. Since I contains (z − a), it is the kernel. This means that φ(g) = φ(g − g(a) + g(a)) = g(a). 16. If I = (f ) is a principal maximal ideal, then f has to vanish somewhere. If it vanishes at a ∈ D, then (z −a) contains I. Since I is maximal, I = (z −a). A maximal principal ideal is a principal ideal which no non-trivial principal ideal contains. If (f ) is such an ideal, by the same argument as above, (f ) = (z − a). Hence, in this case there is no difference between these two concepts. 17. Omitted.

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