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Concise Complex Analysis Solution of Exercise Problems Ai Shu Xue March 9, 2008

1

Contents 1 Calculus

3

2 Cauchy Integral Theorem and Cauchy Integral Formula

14

3 Theory of Series of Weierstrass

27

4 Riemann Mapping Theorem

52

5 Differential Geometry and Picard’s Theorem

53

6 A First Taste of Function Theory of Several Complex Variables

53

7 Elliptic Functions

53

8 The Riemann ζ-Function and The Prime Number Theory

53

2

This is a solution manual of selected exercise problems from Concise Complex Analysis, Second Edition, by Gong Sheng (World Scientific, 2007). This version solves the exercise problems in Chapter 1-3, except the following: Chapter 1 problem 37-42; Chapter 2 problem 47, 49; Chapter 3 problem 15 (xi).

1

Calculus

1. Proof. See, for example, Munkres [4], §38. 2. (1) √ 4 Proof. |2i| = 2, arg(2i) = π2 . |1 − i| = 2, arg(1 − i) = 74 π. |3 + 4i| = ¡5, arg(3 ¢ + 4i) = arctan 3 ≈ 0.9273 5 (Matlab command: atan(4/3)). | − 5 + 12i| = 13, arg(−5 + 12i) = arccos − 13 ≈ 1.9656 (Matlab command: acos(−5/13)). (2) Proof. (1 + 3i)3 = −26 − 18i.

10 4−3i

=

8 5

+ 56 i.

2−3i 4+i

=

5 17

−

14 17 i.

n

(1 + i)n + (1 − i)n = 2 2 +1 cos n4 π.

(3) ¯ ¯ √ ¯ ¯ Proof. | − 3i(2 − i)(3 + 2i)(1 + i)| = 3 130. ¯ (4−3i)(2−i) (1+i)(1+3i) ¯ =

√ √5· √5 2· 10

= 25 .

3. √ π √ 1 Proof. Let θ = arctan 15 , α = arctan 239 . Then 5 − i = 26e−iθ , 1 + i = 2ei 4 and (5 − i)4 (1 + i) = √ i( π −4θ) √ 676 2e 4 . Meanwhile (5 − i)4 (1 + i) = 956 − 4i = 676 2e−αi . So we must have π4 − 4θ = −α + 2kπ, π 1 1 k ∈ Z, i.e. 4 = 4 arctan 15 −arctan 239 +2kπ, k ∈ Z. Since 4 arctan 15 −arctan 239 +2π > 0− π2 +2π = 74 π > π4 1 1 π π and 4 arctan 5 − arctan 239 − 2π < 4 · 2 − 2π = 0 < 4 , we must have k = 0. Therefore π4 = 4 arctan 15 − 1 arctan 239 . 4. Proof. If z = x + yi, 3

2

3

1 z¯

=

x x2 +y 2

+

y x2 +y 2 i.

z 2 = x2 − y 2 + 2xyi.

1+z 1−z

=

1−x2 −y 2 (1−x)2 +y 2

+

2y (1−x)2 +y 2 i.

z z 2 +1

=

r2 z

=

2

x +xy +x+i(−y −x y+y) . (x2 −y 2 +1)2 +4x2 y 2

5. Proof. a = 1, b = α + iβ, c = γ + iδ. So ∆ = b2 − 4ac = (α2 − β 2 − 4γ) + i(2αβ − 4δ) and √ −(α + iβ) ± ∆ z= . 2

6. 2

Proof. Denote arg z by θ, then z + rz = reiθ + 2 reiθ − rer iθ = r(eiθ − e−iθ ) = 2ir sin θ = 2iImz.

r2 reiθ

7.

3

= r(eiθ + e−iθ ) = 2r cos θ = 2Rez, and z −

Proof. If a = r1 eiθ1 and b = r2 eiθ2 , then ¯ ¯ ¯ ¯ ¯ a − b ¯ ¯ r1 − r2 ei(θ2 −θ1 ) ¯ ¯ ¯=¯ ¯. ¯1 − a ¯b ¯ ¯ 1 − r1 r2 ei(θ2 −θ1 ) ¯ Denote θ2 − θ1 by θ, we can reduce the problem to comparing |r1 − r2 eiθ |2 and |1 − r1 r2 eiθ |2 . Note |r1 − r2 eiθ |2 = (r1 − r2 cos θ)2 + r22 sin2 θ = r12 − 2r1 r2 cos θ + r22 and

|1 − r1 r2 eiθ |2 = (1 − r1 r2 cos θ)2 + r12 r22 sin2 θ = 1 − 2r1 r2 cos θ + r12 r22 . ¯ ¯ ¯ a−b ¯ So |1 − r1 r2 eiθ |2 − |r1 − r2 eiθ |2 = (r12 − 1)(r22 − 1). This observation shows ¯ 1−¯ ab ¯ = 1 if and only if at least ¯ ¯ ¯ a−b ¯ one of a and b has modulus 1; ¯ 1−¯ ab ¯ < 1 if |a| < 1 and |b| < 1. 8. Proof. We prove by induction. The equation clearly holds for n = 1. Assume it holds for all n with n ≤ N . Then ¯N ¯2 Ã N ¯N +1 ¯2 !Ã N ! ¯X ¯ ¯X ¯ X X ¯ ¯ ¯ ¯ ai bi ¯ = ¯ ai bi + aN +1 bN +1 ¯ = ai bi + aN +1 bN +1 a ¯i¯bi + a ¯N +1¯bN +1 ¯ ¯ ¯ ¯ ¯ i=1 i=1 i=1 i=1 ¯ ¯2 N N N ¯X ¯ X X ¯ ¯ = ¯ ai bi ¯ + |aN +1 |2 |bN +1 |2 + a ¯N +1¯bN +1 ai bi + aN +1 bN +1 a ¯i¯bi ¯ ¯ i=1

=

N X

i=1

|ai |2

N X

i=1

|bi |2 −

i=1

+|aN +1 |2 X

−

X

X

i=1

|ai¯bj − aj ¯bi |2 + |aN +1 |2 |bN +1 |2 + |bN +1 |2

X

|ai |2

1≤i m, n X k=m+1

ak bk =

n X

(Sk − Sk−1 )bk = Sb bn−1 +

n X

Sk (bk − bk+1 ) − Sm bm+1 .

k=m+1

k=m+1

Denote by K a bound of the P sequence (Sn )∞ bm+1 | ≤ K|bm+1 |, and n=1 , then |Sn bn+1 | ≤ K|bn+1 |, |SmP Pn n ∞ | k=m+1 Sk (b − b )| ≤ K |b − b |. Since lim b = 0 and k+1 n→∞ n k=m+1 k n=1 |bn − bn+1 | < ∞, Pkn k+1 a b can be arbitrarily small if m is sufficiently large. So under conditions (1)-(3), we conclude k=m+1 k k P∞ a b is convergent. n=1 k k In the Dirichlet criterion, we require (1) (Sn )∞ n=1 is bounded; (2) limn→∞ bn = 0; (3’) (bn )∞ n=1 is monotone. Clearly, (2)+(3’) imply (3), so the Dirichlet criterion is a special case of the result in current problem. In the Abel criterion, we require (1”) (Sn )∞ n=1 is convergent; (2”) (bn )∞ n=1 is bounded; (3”) (bn )∞ n=1 is monotone. Define b0n = bn − b, where b = limn→∞ bn is a finite number. Then (Sn )∞ (1) and (b0n )∞ n=1 satisfies n=1 P∞ P P N N 0 0 satisfies (2) and (3’). By the Dirichlet criterion, a b converges. Note a b = a b +bS n n n n N. n n n=1 n=1 n=1 P∞ By condition (1”), we conclude n=1 an bn converges. Remark 1. The result in this problem is the so-called Abel-Dedekind-Dirichlet Theorem. 45. (1) 1

Proof. For any P ε ∈ (0, 1 − q), there exists N P ∈ N, such that for any n ≥ N , |an | n < q + ε < 1. By comparing P ∞ ∞ ∞ n (q + ε) , we conclude |a | and n=1 an is absolutely convergent. n=1 n=1 n (2) Proof. By definition of upper limit and the fact q >P 1, we can find ε > 0 and infinitely many n, such that 1 ∞ |an | n > q − ε > 1. Therefore limn→∞ |an | = ∞ and n=1 an is divergent. 46. | Proof. If q < 1, for any ε ∈ (0, 1 − q), we can find N ∈ N, such that for any n ≥ N , |a|an+1 < q + ε < 1. So n| P P ∞ ∞ k k |aN +k | ≤ |aN |(q + ε) , ∀k ≥ 0. Since k=1 (q + ε) is convergent, we conclude n=1 an is also convergent. If q > 1, we have two cases to consider. In the first case, the upper limit is assumed to be a limit. Then we can find ε > 0 such that q − ε > 1 and for n sufficiently large, |an+1 | > |an |(q − ε) > |an |. This implies limn→∞ an 6= 0, so the series is divergent. In the second case, the upper limit is assumed not to be a limit. Then we can manufacture counter examples where the series is convergent. Indeed, consider (k ≥ 0) ( 1 if n = 2k + 1 2, an = (k+1) 2 , if n = 2k + 2. (k+1)2

Then

P∞ n=1

¯ ¯ ¯ ¯ an is convergent and limn→∞ ¯ aan+1 ¯ = 2. n

47.

12

´ ³ Proof. We choose ε > 0, such that limn→∞ n | aan+1 | − 1 < −1 − 2ε. By the definition of upper limit, there n ³ ´ | 1 exists N1 ∈ N, such that for any n ≥ N1 , n | aan+1 | − 1 < −1 − 2ε, i.e. |a|an+1 < 1 − 1+2ε n . Define bn = n1+ε . n n| Then µ ¶1+ε µ ¶ nε 1 bn+1 1 1+ε = = 1 − = 1 − + O . bn (n + 1)1+ε n+1 n+1 (n + 1)2 So there exists N2 ∈ N, such that for n ≥ N2 , n ≥ N,

an+1 an


1−

¯ ¯ ¯ aN +2 ¯ bN +2 ¯ ¯ ¯ aN +1 ¯ < bN +1 , · · ·

Multiply these inequalities, we get |aN +k | ≤ P∞

n=1 bn

Define N = max{N1 , N2 }, then for any

In particular, we have ¯ ¯ ¯ aN +1 ¯ bN +1 ¯ ¯ ¯ aN ¯ < bN ,

Since

1+2ε n+1 .

is convergent, we conclude

P∞ n=1

¯ ¯ ¯ aN +k ¯ ¯ ¯ < bN +k . ,¯ aN +k−1 ¯ bN +k−1

|aN | bN +k . bN

an must also converge.

1

48. (1) Proof. Since (an )∞ n=1 monotonically decreases to 0, for n sufficiently large, ln an < 0 and limn→∞ 1 n

So limn→∞ an = e

limn→∞ lnnan

ln an n

≤ 0.

0

≤ e = 1. By Hadamard’s formula, R ≥ 1.

(2) Proof. The claims seems problematic. For a counter example, assume R > 1 and an = θ 6= 0, we have N X n=0

P∞

So

n=0

an z n =

1 Rn .

If z = Reiθ with

N N N X X X sin (N +1)θ sin (N +1)θ cos N2θ sin N2θ 1 n inθ 2 2 + i . · R e = cos nθ + i sin nθ = Rn sin θ2 sin θ2 n=0 n=0 n=0

an z n is not convergent.

49. Proof. By definition of uniform convergence, for any ε > 0, there exists N ∈ N, such that for P∞any m ≥ ∗N , P ∞ n ∗ ∗ |a ||z − z | < ε, ∀z ∈ D. In particular, for any z ∈ ∂D, by letting z → z , we get 0 n=m |an ||z − n=m n P∞ z0 |n ≤ ε. That is, for this given ε, we can pick up N , such that for any m ≥ N , n=m |an ||z − z0 |n ≤ ε, ¯ This is exactly the definition of uniform convergence on D. ¯ ∀z ∈ D. 50. Proof. We choose a sequence (rn )∞ n=1 such that rn ∈ (0, 1) and limn→∞ rn = 1. Define βn = f (rn z0 ). We show limn→∞ βn = 0. Indeed, we note ¯ ¯ ¯ ¯ n ∞ ∞ n ¯X ¯ ¯X ¯ X X ¯ ¯ k k¯ k k k¯ |βn | = ¯ ak z0 − ak (rn z0 ) ¯ = ¯ ak (1 − rn )z0 − ak (rn z0 ) ¯ ¯ ¯ ¯ ¯ k=0

≤ ≤

n X k=0 n X

k=0

k=n+1

∞ X k |ak |rnk |ak |(1 − rn )(1 + rn + · · · + rnk−1 ) + n k=n+1

|ak |(1 − rn )k +

k=0 1 It

k=0

εn+1 1 , n 1 − rn

¯ ¯ ¯ ¯a seems the condition limn→∞ ¯ n+1 ¯ = 1 is redundant. a n

13

Pn k=0

ak z0k −

P where εn = k≥n k|ak |. We choose rn in such a way that limn→∞ n(1 − rn ) = 1, e.g. rn = 1 − n1 . Then εn+1 limn→∞ n(1−r = 0 since limn→∞ nan = 0. By Problem 18 (2), we have n) n X

Pn |ak |(1 − rn )k = (1 − rn )n ·

k=0

|ak |k → 0, as n → ∞. n

k=0

Combined, we conclude limn→∞ βn = 0. Remark 2. The result in this problem is a special case of the so-called Landau’s Theorem, see, for example, Boos [1], §5.1 Boundary behavior of power series.

2

Cauchy Integral Theorem and Cauchy Integral Formula

1. Proof. (i) The integral is equal to 2πi · sin i = π(e−1 − e). (ii) The integral is equal to 2eπi. (iii) For sufficiently small ε > 0, the integral is equal to · ¸ Z Z cos π cos(−π) cos z cos z dz + dz = 2πi + = 0. 2π −2π |z−π|=ε (z + π)(z − π) |z+π|=ε (z + π)(z − π) ³ ´(n−1) ¯¯ 1 ¯ (iv) The integral is equal to 2πi z−3 = − (n−1)! = 2πi · −(n−1)! 2n 2n−1 πi. ¯ z=1

(v) For sufficiently small ε > 0he integral is equal to · ¸ Z Z dz 1 dz 1 1 1 · + · = 2πi · + = 0. 2 2 (−2i) · 3 2i · 3 |z−i|=ε (z + i)(z + 4) (z − i) |z+i|=ε (z − i)(z + 4) (z + i) (vi) By the Fundamental Theorem of Algebra, the equation z 5 − 1 = 0 has five solutions z1 , z2 , z3 , z4 , z5 . For any i ∈ {1, 2, 3, 4, 5}, we have (zi − z1 ) · · · (zi − zi−1 )(zi − zi+1 ) · · · (zi − z5 ) = lim

z→zi

5 z5 − 1 z 5 − zi5 = lim = 5zi4 = . z→zi z − zi z − zi zi

So for sufficiently small ε > 0, the integral is equal to 5 Z X i=1

= 2πi

|z−zi |=ε 5 X i=1

= 2πi

dz (z − z1 )(z − z2 )(z − z3 )(z − z4 )(z − z5 )

1 (zi − z1 ) · · · (zi − zi−1 )(zi − zi+1 ) · · · (zi − z5 )

6 X zi i=1

5

= 0, where the last equality comes from the fact that the expansion of (z−z1 ) · · · (z−z5 ) is z 5 −1 and −(z1 +· · ·+z5 ) is the coefficient of z 4 . 2πi If (vii) If |a|, |b| > R, the integral is equal to 0. If |a| > R > |b|, the integral is equal to (b−a) n. ³ ´(n−1) ¯¯ 1 ¯ |b| > R > |a|, the integral is equal to 2πi z−b = 2πi · (−1)n−1 (a − b)−n . If R > |a|, |b|, the ¯ z=a i h 1 1 n−1 integral is equal to 2πi (b−a) n + (−1) (a−b)n = 0. 14

(viii) Denote by z1 , z2 , z3 the three roots of the equation z 3 − 1 = 0. Then the integral is equal to ·

1 1 1 2πi + + + (z1 − z2 )(z1 − z3 )(z1 − 2)2 (z2 − z1 )(z2 − z3 )(z2 − 2)2 (z3 − z1 )(z3 − z2 )(z3 − 2)2 · ¸ z1 z2 z3 12 = 2πi + + − . 3(z1 − 2)2 3(z2 − 2)2 3(z3 − 2)2 49

µ

1 z3 − 1

2. Proof. 1 2πi

Z

ezζ dζ 1 = · ζn ζ 2πi

C

Z X Z ∞ ∞ X zk 1 1 1 zn dζ zk dζ = · = , · · n−k n−k k! ζ ζ k! 2πi C ζ ζ n! C k=0

k=0

where the last equality is due to Cauchy’s integral formula. We need to justify the exchange of integration and infinite series summation. Indeed, for m > n, we have ¯ ¯m Z Z X ∞ ¯X z k 1 1 1 zk 1 dζ dζ ¯¯ ¯ · · − · ¯ ¯ ¯ k! 2πi C ζ n−k ζ 2πi C k! ζ n−k ζ ¯ k=0 k=0 ¯ ¯ ∞ ¯ 1 Z X zk 1 dζ ¯¯ ¯ = ¯ · ¯ ¯ 2πi C k! ζ n−k ζ ¯ k=m+1 Z ∞ X 1 |zζ|k |dζ| ≤ 2π C k! |ζ|n+1 k=m+1 Z ∞ X 1 M k |dζ| ≤ , 2π C k! |ζ|n+1 k=m+1

P∞ P∞ k where M = |z| maxζ∈C |ζ|. Since k=m+1 Mk! is the remainder of eM , for m sufficiently large, k=m+1 can be smaller than any given positive number ε. So for m sufficiently large, ¯ ¯ Z Z X Z ∞ m ¯X 1 1 dζ 1 zk 1 dζ ¯¯ |dζ| 1 zk ¯ · · − · ¯≤ε . ¯ ¯ k! 2πi C ζ n−k ζ 2πi C k! ζ n−k ζ ¯ 2π C |ζ|n+1 k=0

k=0

This shows

P∞

k

z k=0 k!

·

Mk k!

1 2πi

R

1 C ζ n−k

·

dζ ζ

converges to

1 2πi

R P∞ C

zk 1 k=0 k! ζ n−k

·

dζ ζ .

3. R f (ζ) 1 . If |z| < 1, we apply Cauchy’s integral formula to 2πi dζ and apply |ζ|=1 ζ−z R g(ζ) g(ζ) 1 1 Cauchy’s integral theorem to 2πi |ζ|=1 ζ− 1 dζ (regarding z as a parameter and ζ− 1 an analytic function of z z R f (ζ) (ζ) 1 ζ). If |z| > 1, we apply Cauchy’s integral theorem to 2πi dζ (regarding z as a parameter and fζ−z |ζ|=1 ζ−z R g(ζ) 1 an analytic function of ζ) and apply Cauchy’s integral formula to 2πi dζ. |ζ|=1 ζ− 1 Proof. We note

g(ζ) zζ−1

=

g(ζ) ζ− z1

z

4. Proof. For z ∈ {z : |z| < 3}, by Cauchy’s integral formula, f (z) = 2πi(3z 3 + 7z + 1). So f 0 (1 + i) = 2πi[6(1 + i) + 7] = −12π + 26πi. 5.

15

¶(1) ¯¯ ¯ ¯ ¯

# z=2

Proof. We note

Z

1 dz = (z + )2n z z |z|=1

Z

Z

2π

iθ

(e + e

−iθ 2n

2n

) idθ = 2 i

0

2π

cos2n θdθ.

0

On the other hand, 2n

2n

k=0

k=0

k 1 1 X k k −1 2n−k −1 X C2n (z + )2n · = C2n z (z ) ·z = . z z z −2n+2k−1

Therefore Z 2π cos2n θdθ =

1

Z

1 dz 1 = 2n (z + )2n 22n i |z|=1 z z 2 i

0

Z |z|=1

n n C2n dz 2πC2n 1 · 3 · 5 · · · · · (2n − 1) = . = 2π · 2n z 2 2 · 4 · 6 · · · · · 2n

6. Proof. This is straightforward from Cauchy’s integral formula. 7. n! 2πi

Proof. For any R > 0, we have f (n) (0) =

R

f (ξ) dξ. |z|=R ξ n+1

So

¯Z ¯ ¯ (n) ¯ ¯Z 2π ¯ iθ ¯ f (0) ¯ ¯ |f (Reiθ )| ¯¯ 1 ¯¯ 2π M e|Re | ¯¯ eR ¯ ¯= 1 ¯ dθ dθ . ≤ = M ¯ ¯ ¯ n! ¯ 2π ¯ ¯ 2π ¯ ¯ Rn Rn Rn 0 0 Let R = n, we get the desired inequality. R

e Remark 3. Note h(R) = ln R n = R − n ln R takes minimal value at n. This is the reason why we choose ¯ (n) ¯ ¯ ¯ 1 R = n, to get the best possible estimate. Another perspective is to assume f (z) = ez . Then ¯ f n!(0) ¯ = n! ∼ ¡ ¢ R n 1 1 e e √ by Stirling’s formula. So the bound M Rn is not “far” from the best one. n = √ 2πn n 2πn( n e)

8. (1) Proof. For any z ∈ D2 , we can find R > 0 sufficiently large, so that z is in the region enclosed by γ and {ζ : |ζ − z| = R}. By Cauchy’s integral formula Z Z 1 f (ζ) 1 f (ζ) f (z) = dζ + dζ. 2πi −γ ζ − z 2πi |ζ|=R ζ − z So 1 2πi

Z γ

f (ζ) dζ ζ −z

= =

We note

−f (z) + −f (z) +

1 2πi 1 2πi

Z |ζ|=R

f (ζ) dζ ζ −z

|ζ|=R

1 f (ζ) − f (∞) dζ + ζ −z 2πi

Z

Z |ζ|=R

f (∞) dζ. ζ −z

¯ ¯ ¯ 1 Z f (ζ) − f (∞) ¯¯ ¯ dζ ¯ ≤ sup |f (ζ) − A| → 0, as R → ∞, ¯ ¯ 2πi |ζ|=R ¯ ζ∈D(z,R) ζ −z

and Cauchy’s integral formula implies 1 2πi

Z |ζ|=R

f (∞) dζ = f (∞) · 1 = A. ζ −z 16

So by letting R → ∞ in the above equality, we have Z 1 f (ζ) dζ = −f (z) + A. 2πi γ ζ − z (ζ) If z ∈ D1 , h(ζ) = fζ−z is a holomorphic function in D2 , with z a parameter. So for R sufficiently large, Cauchy’s integral theorem implies Z Z f (ζ) 1 f (ζ) 1 dζ + dζ. 0= 2πi −γ ζ − z 2πi |ζ|=R ζ − z

An argument similar to (1) can show

1 2πi

R

f (ζ) dζ γ ζ−z

= A.

(2) Proof. We note

z zζ−ζ 2

=

1 ζ

−

1 ζ−z .

For R sufficiently large, Cauchy’s integral theorem implies 1 2πi

0= So

1 2πi

R γ

f (ζ) ζ dζ

=

1 2πi

R |ζ|=R

f (ζ) ζ dζ.

1 2πi Therefore 1 2πi

Z γ

Z γ

Z −γ

f (ζ) 1 dζ + ζ 2πi

Z |ζ|=R

f (ζ) dζ. ζ

Similarly,

f (ζ) dζ = ζ −z

zf (ζ) 1 dζ = 2 zζ − ζ 2πi

(

1 2πi 1 2πi

Z · γ

R

f (ζ) ζ−z dζ f (ζ) dζ |ζ|=R ζ−z

R|ζ|=R

− f (z)

z ∈ D2 z ∈ D1 .

( ¸ f (z) z ∈ D2 f (ζ) f (ζ) − dζ = ζ ζ −z 0 z ∈ D1 .

9. Proof. Since f (z) 6= 0 in D, 1/f (z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f (z), we conclude |f (z)| achieves its minimum on ∂D. Applying Maximum Modulus Principle to f (z), we conclude ¯ By Chapter |f (z)| achieves its maximum on ∂D. Since |f (z)| ≡ M on ∂D, |f (z)| must be a constant on D. 1, exercise problem 21 (iii), we conclude f (z) is a constant function. 10. Proof. For R large enough, |a|, |b| < R. So we can find ε > 0 such that (assume a 6= b) Z Z Z f (z) f (z) f (z) 2πi dz = dz + dz = [f (a) − f (b)]. a−b |z|=R (z − a)(z − b) |z−a|=ε (z − a)(z − b) |z−b|=ε (z − a)(z − b) Meanwhile, denote by M a bound of f , we have ¯Z ¯ ¯ ¯ f (z) ¯ ¯ dz ¯ ≤ ¯ ¯ |z|=R (z − a)(z − b) ¯ ≤ = Combined, we conclude 0 =

2πi a−b [f (a)

Z M |z|=R

Z

|dz| |(z − a)(z − b)|

|dz| (R − |a|)(R − |b|) |z|=R 2πM R → 0, as R → ∞. (R − |a|)(R − |b|)

M

− f (b)], ∀a, b ∈ C. This shows f is a constant function. 17

11. Proof. By Cauchy’s integral formula, n! |f (n) (0)| ≤ 2π

¯ ¯Z Z 1 ¯ f (ξ) ¯¯ n! 2π 1−r n! ¯ dξ ≤ dθ = n . ¯ ¯ ¯ |z|=r ξ n+1 ¯ 2π 0 rn r (1 − r)

12. Proof. We note 1 Φ(z) − Φ(z0 ) = lim lim z→z0 z − z0 2πi z→z0

Z γ

ϕ(ζ) 1 dζ = (ζ − z)(ζ − z0 ) 2πi

Z γ

ϕ(ζ) dζ, (ζ − z0 )2

where the exchange of limit and integration can be seen as an application of Lebesgue’s Dominated Conver1 gence Theorem (We can find ρ > 0 so that ∀z in a neighborhood of z0 , dist(γ, z) ≥ ρ. Then (ζ−z)(ζ−z ) ≤ ρ12 ). 0) This shows Φ is holomorphic in any region D which does not contain any point of γ. The formula for n-th derivative of Φ can be proven similarly and by induction. For details, we refer the reader to Fang [3], Chapter 4, §3, Lemma 2. Remark 4. The result still holds if ϕ is piecewise continuous on γ. See, for example, Fang [3], Chapter 7, proof of Theorem 7. 13. Proof. Since f (z) 6= 0 in D, 1/f (z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f (z), we conclude |f (z)| does not achieve its minimum in D unless it’s a constant function. Applying Maximum Modulus Principle to f (z), we conclude f (z) does not achieve its maximum in D unless it’s a constant function. Combined, we have m < |f (z)| < M for any point z ∈ D. 14. Proof. Define f (z) = pnz(z) and g(z) = f ( z1 ). Then g(z) is holomorphic and bounded on D(0, 1) \ {0}, since n pn (z) limz→0 g(z) = limz→∞ zn is a finite number and g(z) is clearly continuous in a neighborhood of ∂D(0, 1). By Theorem 2.11, Riemann’s Theorem, g(z) can be analytically continued to D(0, 1). By a corollary of Theorem 2.18, Maximum Modulus Principle, the maximum value of |g(z)| can only be reached on ∂D(0, 1). So |g(z)| ≤ max|z|=1 |g(z)| = max|z|=1 |f (z)| = max|z|=1 |pn (z)| ≤ M , ∀z ∈ D(0, 1), i.e. |f (z)| = |p|zn (z)| ≤M n| for 1 ≤ |z| < ∞. 15. Proof. Define g(z) = f (Rz) (z ∈ D(0, 1)) and apply Theorem 2.19, Schwarz Lemma, to g(z), we have M z z |g(z)| ≤ |z| (∀z ∈ D(0, 1)) and |g 0 (0)| ≤ 1. So for any z ∈ D(0, R), f (z) = f (R · R ) = M g( R ) ≤ M |z| R and M ≤ . |f 0 (0)| = |g 0 (0)| · M R R 16. Proof. Define ρ = dist(C \ V, K). Then ρ ∈ (0, ∞). By Theorem 2.7, for any z ∈ K, we have Z Z n! n! |f (ξ)| |dξ| |dξ| ≤ · sup |f (z)|. |f (n) (z)| ≤ 2π ∂V |ξ − z|n+1 2π ∂V ρn+1 z∈V R |dξ| n! Define cn = 2π and take supremum on the left side of the above inequality, we have ∂V ρn+1 sup |f (n) (z)| ≤ cn sup |f (z)|.

z∈K

z∈V

18

17. R Proof. We have to assume γ has finite length, that is, L(γ) := γ |dz| < ∞. Then by the definition of uniform ε convergence, for any ε > 0, there exists n0 such that ∀n ≥ n0 , supz∈D |f (z) − fn (z)| ≤ L(γ) . So for any n ≥ n0 , ¯Z ¯ Z Z Z ¯ ¯ ε ¯ f (z)dz − fn (z)dz ¯ ≤ |dz| = ε. |f (z) − fn (z)||dz| ≤ ¯ ¯ L(γ) γ γ γ γ R R Therefore limn→∞ γ fn (z)dz = γ f (z)dz. 18. Proof. We note the linear fractional transformation ω(z) = z−α z+α maps {z : Rez > 0} to D(0, 1). So h(z) = ω(f (z)) maps D(0, 1) to itself (see, for example, Fang [3], Chapter 3, §7) and h(0) = 0. By Schwarz Lemma, h(z)| = |z| and |h0 (0)| ≤ 1. This is equivalent to ¯ ¯ ¯ f (z) − α ¯ 0 ¯ ¯ ¯ f (z) + α ¯ ≤ |z|, and |f (0)| ≤ 2α.

19. Proof. The linear fractional transformation ω(z) = ω(f (z)) =

f (z) f (z)−2A

z z−2A

maps {z : Rez < A} to D(0, 1). So h(z) =

maps D(0, 1) to itself and h(0) = 0. By Schwarz Lemma, |h(z)| ≤ |z|. So ∀z ∈ D(0, 1) |f (z)| ≤ |z| · |f (z) − 2A| ≤ |z||f (z)| + 2A|z|, i.e. |f (z)| ≤

2A|f (z)| . 1 − |z|

20. (1) Proof. We note ez + e−z + 2 cos z 4

=

∞ ∞ X 1 z 1X 1 k z 4m [e + e−z + eiz + e−iz ] = [z + (−z)k + (iz)k + (−iz)k ] = . 4 4 k! (4m)! m=0 k=0

The radius of convergence is ∞. (2) h i(n) 1 −(n+3) Proof. By induction, it is easy to show (1−ζ) = (n+2)! . So the Taylor expansion of 3 2! (1 − ζ) P ∞ (n+1)(n+2) 1 is (1−ζ) ζ n , with radius of convergence equal to 1. This implies 3 = n=0 2

1 (1−ζ)3

∞ X ζ + 4ζ 2 + ζ 3 4 = ζ + 7ζ + (3n2 − 3n + 1)ζ n . (1 − ζ)3 n=3

P∞ Replace ζ with z 2 , we can get the Taylor expansion at z = 0: z 2 + 7z 8 + n=3 (3n2 − 3n + 1)z 2n , and the 1 radius of convergence is equal to 1 (this can be verified by calculating limn→∞ (3n2 − 3n + 1) n = 1). (3)

19

£ ¤(n) −(n+4) Proof. By induction, it is easy to show (1 − ζ)−4 = (n+3)! . So for ζ ∈ D(0, 1), (1 − ζ)−4 = 3! (1 − ζ) P∞ (n+1)(n+2)(n+3) n ζ . Replace ζ with z 5 , we get for z ∈ D(0, 1) n=0 6 ∞ X (n + 1)(n + 2)(n + 3) 5n 1 = z . (1 − z 5 )4 6 n=0

Therefore (1 − z −5 )−4 =

∞ X (n + 1)(n + 2)(n + 3) 5n+20 z 20 = z . (1 − z 5 )4 6 n=0

By Hadamard’s formula, the radius of convergence is 1. (4) £ ¤(n) Proof. By induction, we can show (1 + z)−2 = (−1)n (n + 1)!(1 + z)−(n+2) . So ∀z ∈ D(0, 1), ∞ X 1 = (−1)n (n + 1)z n . (1 + z)2 n=0

It is also easy to verify that ∞ ∞ ∞ 1 1 1 1 1X 1X n 1X n = − + − = − (−z) − z − (−1)n (n + 1)z n . (z + 1)2 (z − 1) 4(z + 1) 4(z − 1) 2(z + 1)2 4 n=0 4 n=0 2 n=0

Therefore ∞ ∞ z6 z6 X 1X n n n = − [(−1) + 1 + 2 · (−1) (n + 1)] z = − [(−1)n (2n + 3) + 1] z n+6 . (z + 1)(z 2 − 1) 4 n=0 4 n=0

By Hadamard’s formula, the radius of convergence is 1. 21. (1) Proof. We note 1 2π

Z

π

iθ

2

|f (re )| dθ

=

−π

= = =

1 2π

Z

π

Ã

−π

∞ 1 X 2π m,n=0 ∞ X

∞ X

n=0 Z π

! Ã n inθ

an r e

·

∞ X

! m −imθ

a ¯m r e

dθ

m=0

an a ¯m rn+m ei(n−m)θ dθ

−π

an a ¯m rn+m δnm

m,n=0 ∞ X

|an |2 r2n ,

n=0

(

1 if m = n and the exchange of integration and summation is justified by Theorem 1.14 0 if m 6= n (Abel Theorem) and Problem 17. where δmn =

(2)

20

Proof. In the result of part (1), multiply both sides by r and integrate with respect to r from 0 to 1, we have ∞ X

Z |an |2

P∞

|an |2 n=0 n+1

r2n+1 dr =

0

n=0

So

1

=

1 π

RR D

Z 1Z π Z Z ∞ X 1 1 |an |2 = |f (reiθ )|2 rdrdθ = |f (z)|2 dA. 2n + 2 2π 2π 0 −π D n=0

|f (z)|2 dA.

22. ¯ R), then ∀z ∈ D(0, ¯ 1), v(z) := u(zR) is Proof. Suppose u is harmonic on D(0, R) and continuous D(0, ¯ harmonic on D(0, 1) and continuous on D(0, 1). By formula (2.31), for any z ∈ D(0, R), ¯ z ¯2 Z 2π Z 2π 2 2 ¯ 1 − ¯R z 1 1 iτ iτ R − |z| u(z) = v( ) = dτ = v(e ) ¯ u(Re ) ¯ 2 dτ R 2π 0 2π 0 ¯1 − z¯ eiτ ¯2 |R − z¯eiτ | R Z 2π Z 2π 2 1 R2 − |z| R2 − |z|2 1 = u(Reiτ ) u(ζ) dτ. dτ = 2 2π 0 2π 0 |ζ − z|2 |R − ze−iτ | This is formula (2.32). To verify formula (2.33), note Re

Reiτ + z (Reiτ + z)(Re−iτ − z¯) R2 − |z|2 + R(ze−iτ − z¯eiτ ) R2 − |z|2 = Re = Re = . iτ iτ 2 iτ 2 Re − z |Re − z| |Re − z| |Reiτ − z|2

Therefore, u(z)

Z

2π

R2 − |z|2 1 dτ = |Reiτ − z|2 2π 0 # " Z ζ + z dζ 1 u(ζ) . = Re 2πi |ζ|=R ζ −z ζ

=

1 2π

Z

u(Reiτ )

2π

u(Reiτ )Re

0

· ¸ Z 2π iτ Reiτ + z 1 iτ Re + z dτ = Re u(Re ) dτ Reiτ − z 2π 0 Reiτ − z

This is formula (2.33). 23. Proof. On ∂D(0, 1), |(−6z) − (z 4 − 6z + 3)| = |z 4 + 3| ≤ 4 < | − 6z|. By Rouch´e Theorem, z 4 − 6z + 3 and −6z have the same number of zeros in D(0, 1), which is one. On ∂D(0, 2), |z 4 −(z 4 −6z +3)| = |6z −3| ≤ 15 < |z 4 |. So z 4 − 6z + 3 and z 4 have the same number of zeros in D(0, 2), which is four. Combined, we conclude z 4 − 6z + 3 = 0 has one root in D(0, 1) and three roots in the annulus {z : 1 < |z| < 2}. 24. Proof. On ∂D(0, 1), |(−5z 4 ) − (z 7 − 5z 4 − z + 2)| = |z 7 − z + 2| ≤ 4 < | − 5z 4 |. So z 7 − 5z 4 − z + 2 and −5z 4 have the same number of zeros in D(0, 1), which is four (counting multiplicity). 25. Proof. Let P (z) = z 4 + 2z 3 − 2z + 10. We note P (z) = (z 2 − 1)(z + 1)2 + 11. If z ∈ R and |z| ≥ 1, P (z) ≥ 11; if z ∈ R and |z| ≤ 1, P (z) ≥ −1 · (1 + 1)2 + 11 = 7. So P (z) = 0 has no root on the real axis. If z = iy with y ∈ R, we have P (iy) = y 4 + 10 − 2iy(y 2 + 1) 6= 0. So P (z) = 0 has no root on the imaginary axis. Consider the region D enclosed by the curve γ = γ1 ∪ γ2 ∪ γ3 , where (R is a positive number) π γ1 = {z : 0 ≤ Rez ≤ R, Imz = 0}, γ2 = {z : |z| = ³ R, arg z ∈ [0, ´2 ]}, and γ3 = {z : 0 ≤ Imz ≤ R, Rez = 0}. On γ1 , ∆γ1 arg P (z) = 0. On γ2 , P (z) = z 4 1 +

2z 3 −2z+10 z4

. So ∆γ2 arg P (z) = 4 · 4

π 2

+ o(1) = 2π + o(1)

(R ³ → ∞). On γ3´ , ∆γ3 arg P (z) = arg P (0) − arg P (iR) = arg 10 − arg(R + 10 − 2iR(R2 + 1)) = 0 − 2 P3 +1) = o(1) (R → ∞). Combined, we have ∆γ arg P (z) = k=1 ∆γk arg P (z) = 2π + o(1) arg 1 − 2i R(R R4 +10 (R → ∞). 21

By The Argument Principle, P (z) = 0 has only one root in the first quadrant. The root conjugate to this root must lie in the fourth quadrant. The other two roots are conjugate to each other and are located in the second and third quadrant. So one must lie in the second quadrant and the other must lie in the third quadrant. Remark 5. The above solution can be found in Fang [3], Chapter 6, §3 Example 5. 26. Proof. On ∂D(0, 1), |(−8z + 10) − (z 4 − 8z + 10)| = |z 4 | = 1 < |10 − 8|z|| ≤ | − 8z + 10|. So z 4 − 8z + 10 and −8z + 10 have the same number of zeros in D(0, 1), which is zero. On ∂D(0, 3), |z 4 − (z 4 − 8z + 10)| = |8z − 10| < 34 < |z 4 |. So z 4 − 8z + 10 and z 4 have the same number of zeros in D(0, 3), which is four (counting multiplicity). So z 4 − 8z + 10 = 0 has no root in D(0, 1) and four roots in the annulus {z : 1 < |z| < 3}. 27. Proof. If a > e, on ∂D(0, 1), |az n − (az n − ez )| = |ez | ≤ e|z| = e < |az n |. By Rouch´e Theorem, ez − az n and az n have the same number of zeros in D(0, 1), which is n. 28. Proof. On ∂D(0, 1), |z − (z − f (z))| = |f (z)| < 1 = |z|. So z − f (z) and z have the same number of zeros in D(0, 1), which is one. So there is a unique fixed point of f in D(0, 1). Remark 6. This is a special case of Brower’s Fixed Point Theorem. 29. (1) Proof. Re(zez ) = ex (x cos y − y sin y). (2) Proof. Let u(x, y) = ax3 + bx2 y + cxy 2 + dy 3 . Then ∆u = (6a + 2c)x + (2b + 6d)y. So u is harmonic if and only if 3a + c = 0 and b + 3d = 0. So the most general harmonic function of the form ax3 + bx2 y + cxy 2 + dy 3 is ax3 − 3dx2 y − 3axy 2 + dy 3 . 30. ¡ ∂ ¢ 1 ∂2 ∂ Proof. Let u(r, θ) = ln(1 − 2r cos θ + r2 ). Then using the fact ∆ = 1r ∂r r ∂r + r2 ∂θ2 , we can verify ∆u = 0. R 2π 1 So by the mean-value property of harmonic functions, we have 2π u(r, θ)dθ = u(0, 0) = 0. Since cos θ 0 R 2π Rπ 2 is an even function of θ, it’s not hard to show 0 ln(1 − 2r cos θ + r )dθ = 2 0 ln(1 − 2r cos θ + r2 )dθ. So Rπ ln(1 − 2r cos θ + r2 )dθ = 0. 0 31. Proof. Let M = supz∈C |u(z)|. Then ∀z ∈ C and R > |z|, the holomorphic function defined in formula (2.34) satisfies ¯ iθ ¯ Z 2π Z 2π ¯ Re + z ¯ 1 R + |z| ¯ dθ ≤ M |u(Reiθ )| ¯¯ iθ dθ → M, as R → ∞. |f (z)| ≤ 2π 0 Re − z ¯ 2π 0 R − |z| So f (z) is bounded on the entire complex plane, and by Liouville’s Theorem, f is a constant function. Therefore u = Re(f ) is a constant function. 32.

22

Proof. If the arc is ∂D(0, 1), the desired harmonic function u ≡ 1. So without loss of generality, we assume the arc γ = {z ∈ ∂D(0, 1) : θ1 ≤ arg z ≤ θ2 } with 0 < θ2 − θ1 < 2π. For n sufficiently large, we consider γn = {z ∈ ∂D(0, 1) : θ1 − n1 ≤ arg z ≤ θ2 + n1 }. By the Partition of Unity Theorem for R1 , we can find ϕn ∈ C(∂D(0, 1)) such that 0 ≤ ϕn ≤ 1, ϕn ≡ 1 on γ, and ϕn ≡ 0 on R 2π ∂D(0, 1) \ γn . Define un (z) = 0 P (ζ, z)ϕn (ζ)dθ where P (·, ·) is the Poisson kernel and ζ = Reiθ . Then un is the unique solution of the Dirichlet problem with boundary value ϕn . So ϕn is harmonic in D(0, 1). It is easy to see that if (ϕn )n uniformly converges to the indicator function 1γ (z), (un )n uniformly R 2π converges to u(z) := 0 P (ζ, z)1γ (ζ)dθ on D(0, ρ) (ρ ∈ (0, 1)). Since each un satisfies the local mean value property, u must also satisfies the local mean value property in D(0, 1) and is continuous. By Theorem 2.22, u is harmonic on D(0, 1). This suggests the harmonic function we are looking for is exactly u. What is left to prove is that for any z0 ∈ ∂D(0, 1), lim|z|=1,z→z0 u(z) = 1γ (z0 ). Indeed, in general, we have the following classical result: Suppose function ϕ is piecewise continuous on ∂D(0, 1), then the R 2π function u(z) := 0 P (ζ, z)ϕ(ζ)dθ is harmonic in D(0, 1) and for any continuity point z0 of ϕ, we have limz→z0 ,|z| 0, there exists N , such that for any n ≥ N , p ∈ N, we have ¯ n+p ¯ ¯ ¯ X ¯ ¯ fk (ζ)¯ < ε. sup ¯ ¯ ζ∈∂U ¯ k=n+1

By Maximum Modulus Principle, ¯ n+p ¯ ¯ n+p ¯ ¯ X ¯ ¯ X ¯ ¯ ¯ ¯ ¯ sup ¯ fk (z)¯ ≤ sup ¯ fk (ζ)¯ < ε. ¯ ¯ ¯ ¯ ¯ ζ∈∂U z∈U k=n+1

k=n+1

P∞

¯. So Cauchy criterion implies n=1 fn (z) converges uniformly on U Remark 7. This result is the so-called Weierstrass’ Second Theorem. 34. (z0 ) f (z) ¯ R) \ {z0 }. Then g(z) can be analytically continued to Proof. Define g(z) = f (z)−f = z−z , ∀z ∈ D(0, z−z0 0 ¯ R). By Cauchy’s integral D(0, R) (Theorem 2.11, Riemann Theorem) and is therefore continuous on D(0, formula, we have Z Z f (0) 1 g(ζ) 1 f (ζ) = g(0) = dζ = dζ. −z0 2πi |ζ|=R ζ 2πi |ζ|=R ζ(ζ − z0 )

Therefore

¯ ¯ Z 2π ¯ f (0) ¯ M |f (Reiθ ) ¯ ¯≤ 1 dθ ≤ , ¯ z0 ¯ 2π iθ − z | |Re R − |z0 | 0 0

which is equivalent to R|f (0)| ≤ (M + |f (0)|)|z0 |. Remark 8. The point of this problem is to give an estimate of a holomorphic function’s modulus at z = 0 in terms of its zeros. 35. Qn Qn Proof. This problem seems suspicious. For example, when z0 = 0, of course k=1 |z0 − zk | = k=1 |zk | > 1. Qn When z0 is sufficiently large, it is also clear k=1 |z0 − zk | > 1. So the point z0 can be both inside and outside D(0, 1). I’m not sure what proof is needed for such a trivial claim.

23

36. Proof. By Maximum Modulus Principle, M (r) = max|z|≤r |f (z)|. So M (r) is an increasing function on [0, R). 37. Proof. Suppose there is an n-th degree polynomial p(z) = a0 + a1 z + · · · + an z n such that n ≥ 1, an 6= 0, and p(z) 6= 0 for any z ∈ C. Then q(z) = 1/p(z) is holomorphic over the whole complex plane. For any R > 0, we have by Maximum Modulus Principle max |q(z)| ≤ max |q(z)|.

|z|≤R

|z|=R

Since on {z : |z| = R}, µ ¶ |a1 | |an−1 | |a0 | − · · · − n−1 − n → ∞ as R → ∞, |p(z)| ≥ |an |Rn −|an−1 |Rn−1 −· · ·−|a1 |R−|a0 | = Rn |an | − R R R we must have limR→∞ max|z|=R |q(z)| = 0. This implies supz∈C |q(z)| = 0, i.e. p(z) ≡ ∞. This is a contradiction and our assumption must be incorrect. 38. Proof. Apply the Maximum Modulus Principle to the function g(z) = 1/f (z), which is holomorphic on U. 39. Proof. Note the Laplace operator ∆ = 4 ∂∂¯ ∂∂ , so log |z| is a harmonic function on C \ {0} and log |f (z)| is a harmonic function outside the zeros of f (z). Define K = {z ∈ U : f (z) = 0} and Vε = ∪z∈K D(z, ε). Then ¯ \ Vε . By Maximum Modulus for any α ∈ R, α log |z| + log |f (z)| is harmonic in U \ V¯ε and continuous on U Principle for harmonic functions, maxz∈U¯ \Vε (α log |z| + log |f (z)|) = maxz∈∂(U¯ \Vε ) (α log |z| + log |f (z)|). When z approaches to zeros of f (z), log |f (z)| → −∞, so by letting ε → 0, we can further deduce that maxz∈U (α log |z| + log |f (z)|) = maxz∈∂U (α log |z| + log |f (z)|). Therefore α log |z| + log |f (z)| ≤ max{α log r1 + log M (r1 ), α log r2 + log M (r2 )}, ∀z ∈ U, which is the same as α log r + log M (r) ≤ max{α log r1 + log M (r1 ), α log r2 + log M (r2 )}, r ∈ [r1 , r2 ]. Now let α be such that the two values inside the parentheses on the right are equal, that is α=

log M (r2 ) − log M (r1 ) . log r1 − log r2

Then from the previous inequality, we get log M (r) ≤ α log r1 + log M (r1 ) − α log r, which upon substituting value for α gives log M (r) ≤ (1 − s) log M (r1 ) + s log M (r2 ), where s =

log r1 −log r log r2 −log r1 .

40.

24

Proof. Fix r ∈ (0, 1). For any ρ ∈ (0, r) and any z ∈ D(0, ρ), we have z n ∈ D(0, ρ) (∀n ∈ N). By Cachy’s integral formula, ∀n, p ∈ N, we have ¯ · ¸ ¯¯ n+p n+p n+p X X X 1 ¯¯Z f (ζ) f (ζ) ¯ |f (z k )| = |f (z k ) − f (0)| = − dζ ¯ ¯ k ¯ ¯ 2π ∂D(0,r) ζ − z ζ k=n+1 k=n+1 k=n+1 ¯ ¯ n+p n+p ¯ X 1 ¯¯Z X 1 Z 2π |f (reiθ )|dθ f (ζ) k¯ ≤ dζ · z ¯ ≤ · ρk ¯ ¯ 2π ¯ ∂D(0,r) ζ(ζ − z k ) 2π 0 r − ρk k=n+1

≤

1 2π

Z

2π 0

k=n+1

iθ

|f (re )|dθ r−ρ

n+p X

ρk → 0

k=n+1

P∞

¯ ρ). as n → ∞. So n=1 f (z n ) converges absolutely and uniformly on D(0, Remark 9. In general, there is no Mean Value Theorem for holomorphic functions. For example, f (z) = ³ ´ 2z−(z1 +z2 ) exp i z2 −z1 π is analytic but f (z2 ) − f (z1 ) 6= f 0 (z)(z2 − z1 ), ∀z ∈ C (for more details, see Qazi [5]). But as the proof of Theorem 2.7 shows, the formula Z f (ζ)dζ z − z0 f (z) − f (z0 ) = 2πi ∂U (ζ − z)(ζ − z0 ) more or less fills the void. In particular, it shows holomorphic functions are locally Lipschitz. 41. Proof. This problem is the same as Problem 15. 42. Proof. This problem is the same as Problem 19. 43. (1) ¯ ¯ ¯ ¯ Proof. By Problem 18, ¯ ff (z)−1 (z)+1 ¯ ≤ |z|. So |f (z)| − 1 ≤ |f (z) − 1| ≤ |z||f (z) + 1| ≤ |z||f (z)| + |z|. Since |z| < 1, 1+|z| . the above inequality implies |f (z)| ≤ 1−|z| ¯ ¯ ¯ f (z)−1 ¯ From the same inequality ¯ f (z)+1 ¯ ≤ |z|, we have

(Ref (z) − 1)2 + (Imf (z))2 ≤ |z|2 [(Ref (z) + 1)2 + (Imf (z))2 ] ≤ |z|2 (Ref (z) + 1)2 + (Imf (z))2 . 1−|z| 1+|z| 1−|z| 1+|z| ≤

So |Ref (z) − 1| ≤ |z|(Ref (z) + 1). If Ref (z) > 1, we can deduce from this inequality

≤ 1 < Ref (z). If

Ref (z) ≤ 1, we can deduce from this inequality 1 − Ref (z) ≤ |z|Ref (z) + |z|, i.e.

Ref (z).

(2) Proof. We note f (z) = Lemma.

1+eiθ z 1−eiθ z

if and only if

f (z)−1 f (z)+1

= eiθ . So the claim is straightforward from Schwarz

44.

25

Proof. Because D(0, 1) is pre-compact, by Theorem 1.11 (Bolzano-Weierstrass Theorem), it suffices to find a ∞ sequence (zn )∞ n=1 ⊂ D(0, 1), such that limn→∞ zn = z0 ∈ ∂D(0, 1) and (f (zn ))n=1 is bounded. Assume this does not hold. Then ∀z0 ∈ ∂D(0, 1) and ∀n ∈ N, there exists δ(z0 ) > 0 such that ∀z ∈ D(z0 , δ(z0 )) ∩ D(0, 1), |f (z)| ≥ n. The family of these open sets (D(z, δ(z))z∈∂D(0,1) is an open covering of the compact set ∂D(0, 1). By Theorem 1.10 (Heine-Borel Theorem), there is a finite sub-covering (D(zk , δ(zk ))pk=1 of ∂D(0, 1). Consequently, for each n ∈ N, we can find εn > 0 such that ∀z ∈ {z : 1 − εn ≤ |z| < 1}, |f (z)| ≥ n. Without loss of generality, we can assume (εn )∞ n=1 monotonically decreases to 0. Since f (z) uniformly approaches to Q ∞ as z → ∂D(0, 1), by Theorem 2.13, f has finitely many zeros in m D(0, 1). So f can be written as f (z) = i=1 (z − zi )ki · h(z), where h is a holomorphic function on ∂D(0, 1) with no zeros in D(0, 1). So h satisfies the Minimum Modulus Principle: ∀ρ > 0, |h(z)| ≥ min|ζ|=ρ |h(ζ)| for any z ∈ D(0, ρ). Therefore, for any z ∈ D(0, 1 − εn ), |h(z)| ≥

min

|ζ|=1−εn

|h(ζ)| =

min

|ζ|=1−εn

|

|f (ζ)| n n Qm ≥ ≥ Pm k . ki | ki i=1 i (ζ − z ) max |ζ − z | 2 i i |ζ|=1−ε n i=1 i=1

Qm

Fix z ∈ D(0, 1 − εn ) and let n → ∞, we get h(z) = ∞. Contradiction. 45. Qn Proof. By assumption, we can write f (z) as f (z) = j=1 (z − zj )kj · h(z), where h(z) is a holomorphic Qn function in D(0, 1) with no zeros in D(0, 1). Define G(z) = h(z) · j=1 (1 − z¯j z)kj . Then G is holomorphic on D(0, 1). Apply Maximum Modulus Principle to G(z) on {z : |z| ≤ 1 − ε} (ε > 0), we get |G(z)| ≤ max |h(z)| |z|=1−ε

Since each

z−zj 1−¯ zj z

n Y

|1 − z¯j z|kj = max

|f (z)| 1 ¯ ¯kj ≤ max Q ¯ ¯ . |z|=1−ε ¯ z−zj ¯ ¯ z−zj ¯kj n ¯ ¯ ¯ j=1 1−¯ j=1 1−¯ zj z zj z ¯

|z|=1−ε Qn

j=1

maps ∂D(0, 1) to ∂D(0, 1), by letting ε → 0, we get G(z) ≤ 1, i.e. |f (z)| ≤

¯ ¯ ¯ z−zj ¯kj ¯ j=1 1−¯ zj z ¯ .

Qn

46. Proof. Let a = f (0) and define ϕa (ζ) = −ζ+a 1−¯ aζ . Then h(z) = ϕa (f (z)) maps D(0, 1) to D(0, 1) with h(0) = ϕa (a) = 0. So Schwarz Lemma implies |h0 (0)| ≤ 1. We note h0 (z) = ϕa (f (z)) · f 0 (z). Since ϕ0a (ζ) = we have h0 (0) = ϕ0a (a)f 0 (0) =

f 0 (0) |a|2 −1 .

−(1 − a ¯ζ) − (−ζ + a)(−¯ a) −1 + |a|2 = , 2 (1 − a ¯ζ) (1 − a ¯ζ)2 So we have

|f 0 (0)| ≤ 1, i.e. |f 0 (0)| ≤ 1 − |a|2 . 1 − |a|2

48. + to D = D(0, 1). So f ∈ Aut(C+ ) if and only if Proof. The M¨obius transformation w(z) = z−i z+i maps C w ◦ f ◦ w−1 ∈ Aut(D). By Theorem 2.20, ∃a ∈ D and τ ∈ R, such that w ◦ f ◦ w−1 (ζ) = ϕa ◦ ρτ (ζ). Plain calculation shows

f (z) = w−1 ◦ ϕa ◦ ρτ ◦ w(z) =

(1 + a)(z + i) − (1 + a ¯)eiτ (z − i) . (1 − a)(z + i) − (¯ a − 1)eiτ (z − i)

50. Proof. This problem is the same as Problem 10. 26

3

Theory of Series of Weierstrass

Throughout this chapter, all the integration paths will take the following convention on orientation: all the arcs take counterclockwise as their orientation and all the segments take the natural orientation of R1 as their orientation. 1. Proof. This is just Mittag-Leffler Theorem (Theorem 3.8). 2. Proof. To prove Theorem 3.1, choose varepsilon > 0 sufficiently small so that the circles {z : |z − zk | = ε} (1 ≤ k ≤ n) don’t intersect with each other or with ∂U . Then by Cauchy’s integration theorem, Z n Z X f (z)dz − f (z)dz = 0, ∂U

R

k=1

Pn

|z−zk |=ε

which is ∂U f (z)dz = 2πi k=1 Res(f, zk ). ¯ ⊂ D(0, R). Then To prove Theorem 3.12, we choose R > 0 sufficiently large so that U Z Z Z n X f (z)dz = 2πi Res(f, zk ) and f (z)dz − f (z)dz = 0. ∂U

Since

R ∂D(0,R)

∂D(0,R)

k=1

f (z)dz = −2πiRes(f, ∞), we conclude

Pn k=1

∂U

Res(f, zk ) + Res(f, ∞) = 0.

3. (i) P∞ 1 has Laurent series n=−∞ cn (z + i)n . Then by Theorem 3.2 and Cauchy integral Proof. Suppose z3 (z+i) theorem, for ε ∈ (0, 1), ¯ Z 1 1 1 1 dn+1 −3 ¯¯ cn = · dζ = (ζ )¯ 2πi |ζ+i|=ε (ζ + i)n+1 ζ 3 (ζ + i) (n + 1)! dζ n+1 ζ=−i (n + 3)! (n + 2)(n + 3) n 1 (−1)n+1 (−i)−(n+4) = (−1)n+1 i . (n + 1)! 2! 2

=

(ii) Proof. We note z2 (z + 1)(z + 2)

(z + 1)(z + 2) − 3z − 2 3(z + 1) − 1 1 4 =1− =1+ − (z + 1)(z + 2) (z + 1)(z + 2) z+1 z+2 ∞ ∞ k X X 1 1 1 (−1) (−1)k k = 1+ · −2· −1− z . z = 1 k+1 z 1+ z 1+ 2 z 2k−1

=

k=0

k=1

(iii) Proof. We note µ log

z−a z−b

¶

Ã

!

µ ¶ b a´ = log − log 1 − = log 1 − z z 1− µ ¶ ∞ ∞ ∞ n X X 1 ³ a ´n X 1 b 1 n = − = (b − an )z −n . + n z n z n n=1 n=1 n=1 1−

a z b z

³

27

(iv) Proof. We note 1

z2e z

z2

=

∞ X z −k k=0

k!

∞

= z2 + z +

1 X z −k + . 2 (k + 2)! k=1

(v) P∞ P∞ (−1)k z2k+1 z 2k+1 z z Proof. We note sin z = k=0 (−1)k (2k+1)! . So sin z+1 = k=0 (2k+1)! . Suppose sin z+1 has Laurent (z+1)2k+1 P∞ n series n=−∞ cn (z + 1) . Then by Theorem 3.2, for ε > 0, Z ∞ X 1 1 (−1)k ζ 2k+1 cn = dζ n+1 2πi |ζ+1|=ε (ζ + 1) (2k + 1)! (ζ + 1)2k+1 k=0 Z ∞ X (−1)k 1 ζ 2k+1 = · dζ (2k + 1)! 2πi |ζ+1|=ε (ζ + 1)2k+n+2 k=0

X

=

k≥ −n−1 2

1 d2k+n+1 ¡ 2k+1 ¢¯¯ (−1)k ζ . ζ=−1 (2k + 1)! (2k + n + 1)! dζ 2k+n+1

Since ¡ 2k+1 ¢¯ d ¯ ζ 2k+n+1 ζ=−1 dζ 2k+n+1

we have cn

=

   0P

(−n)!

n ≥ 1, n = 0, n ≤ −1,

n ≥ 1,

(−1)k k≥0 (2k+1)!    (−1)−n P −n−1 (−1)k k≥ 2 (−n)! (2k+n+1)!

  0    sin 1 = 1   (−n)! cos 1    1 sin 1 Therefore

  0 = (2k + 1)!   (2k+1)! (−1)−n (−n)!

n = 0, n ≤ −1

n ≥ 1, n = 0, n ≤ −1, n is odd, n ≤ −1, n is even.

¸ ∞ · X z sin 1 cos 1 −2k −(2k+1) sin = (z + 1) + (z + 1) . z+1 (2k)! (2k + 1)! k=0

4. (i) Proof. z = 0 is a removable singularity. (ii) Proof. z = 1 is a pole of order 1. The Laurent series of the function in a neighborhood of z = −1 can be obtained by µ ¶2n ∞ X πz 1 πz 1 (−1)n cos = . z2 − 1 z+1 (z − 1)(z + 1) n=0 (2n)! z + 1 By discussion (3) of page 91, z = −1 is an essential singularity. 28

(iii) 1

Proof. Since e z =

P∞ n=0

1

z −n , we have z(e z − 1) =

P∞ n=1

z −n+1 . Therefore z = 0 is an essential singularity.

(iv) Proof. z = 1 is an essential singularity. (v) Proof. z = 1 is an essential singularity. z = 0 is a pole of order 1. (vi) Proof. z = kπ +

π 2

(k ∈ Z) are poles of order 1.

5. (1) 1 exists. By discussion (3) on page Proof. The limit limz→a f (z) exists (finite or ∞) if and only if limz→a f (z) 1 93, a is an essential singularity of f (z) if and only if a is an essential singularity of f (z) .

(2) Proof. Clearly, a is an isolated singularity of P (f (z)). We note by Fundamental Theorem of Algebra, P (ζ) maps C to C. Consequently, P (ζ) maps a dense subset of C to a dense subset of C. By Weierstrass Theorem (Theorem 3.3), f (z) maps a neighborhood of a to a dense subset of C. Therefore, P (f (z)) maps a neighborhood of a to a dense subset of C. So it is impossible for a to become a removable singularity or a pole of P (f (ζ)). Hence a must be also an essential singularity of P (f (z)). 6. (i) P∞ Proof. We note any positive integer can be uniquely represented in the ak · 2k , where each k=0 Q∞form 2n ak ∈ {0, 1} and only finitely many ak ’s are non-zero. In the expansion of n=0 (1 + z ), each representation P∞ k z k=0 ak ·2 appears once and only once. Therefore, after a rearrangement of the terms, we must have ∞ Y

n

(1 + z 2 ) = 1 + z + z 2 + z 3 + z 4 + · · · =

n=0

1 . 1−z

(ii) Proof. sinh πz = 12 [eπz − e−πz ] has zeros an = ni (n ∈ Z), where a0 = 0 is a zero of order 1. Since for any ³ ´2 P P∞ 2 ∞ R > 0, we have n=−∞,n6=0 |aRn | = n=−∞,n6=0 R n2 < ∞, by Weierstrass Factorization Theorem, we can find an entire function h(z) such that sinh πz = ze

h(z)

n=∞ Y n=−∞,n6=0

Since limz→0

sinh πz z

n³

¶ ∞ µ Y z2 z´ zo h(z) ni e = ze 1+ 2 . 1− ni n n=1

= π, we conclude eh(0) = π. Meanwhile we have π coth πz

=

" ¶#0 ∞ µ 2 Y (sinh πz)0 1 z = zeh(z) 1+ 2 sinh πz sinh πz n n=1

=

∞ ∞ 2z X X 1 2z 1 0 n2 + h0 (z) + = + h (z) + . 2 2 z z z n + z2 1 + n2 n=1 n=1

29

Let γn be the rectangular path [n + (n + 12 )i, −n + (n + 21 )i, −n − (n + 12 )i, n − (n + 21 )i, n + (n + 12 )i]. Then for any given a, when n is large enough, we have Z coth πz dz = I + II + III + IV, 2 2 γn z − a where

1

Z

e2π[x+(n+ 2 )i] +1

−n

I=

Z

1 e2π[x+(n+ 2 )i] −1

dx = £ ¤2 x + (n + 21 )i − a2

n

Z II =

−(n+ 12 )

n+ 21

Z

−n

Z

1 e2π[x−(n+ 2 )i] −1

[x − (n + 12 )i]2 − a2

and

Z IV =

n+ 12

−(n+ 12 )

We note

ex −1 ex +1

−e2πx +1 −e2πx −1 ¤2 (n + 12 )i

− a2

dx,

e2π(−n+yi) +1 e2π(−n+yi) −1 idy, (−n + yi)2 − a2

1

III =

£ x+

n

e2π[x−(n+ 2 )i] +1

n

−n

n

dx = −n

[x −

−e2πx +1 −e2πx −1 (n + 12 )i]2

− a2

dx

e2π(n+yi) +1 e2π(n+yi) −1 idy. (n + iy)2 − a2

∈ (−1, 1) for any x ∈ R. So we have Z n 2 n 1 dx = q |I| ≤ arctan q → 0, 2 + (n + 1 )2 − a2 1 x 2 2 −n 2 (n + ) − a (n + 1 )2 − a2 2

2

as n → ∞. Similarly we can conclude limn→∞ III = 0. We also note for x ∈ R large enough, for n large enough, Z |II| ≤

n+ 12

−(n+ 12 )

ex +1 ex −1

< 2. So

1+e−2πn

n + 12 4 e 1−e−2πn √ √ dy ≤ arctan → 0, n2 + y 2 − a 2 n2 − a2 n2 − a2

as n → ∞. Similarly we can conclude limn→∞ IV = 0. Combined, we have shown Z coth πz lim dz = 0. n→∞ γ z 2 − a2 n Define f (z) =

coth πz z 2 −a2 .

By Residue Theorem, for a 6∈ iZ, we have

Z f (z)dz = 2πiRes(f ; a) + 2πiRes(f ; −a) + 2πi γn

Res(f ; an ).

k=−n

It’s easy to see Res(f ; a) = Res(f ; −a) = lim (z − an )f (z) = lim

z→an

n X

z→an

coth πa 2a .

Since

2 2π(z − an ) 1 e2πz + 1 z − an = lim =− , z 2 − a2 e2πz − 1 2π(a2n − a2 ) z→an e2πz − e2πan π(n2 + a2 )

we must have Res(f ; an ) = − π(n21+a2 ) . Therefore à ! Z n ∞ X X coth πa 1 coth πa 1 1 lim f (z)dz = lim − = − 0= n→∞ 2πi n→∞ γn a π(a2 + k 2 ) a π(a2 + k 2 ) k=−n

This implies for a 6∈ iZ,

k=−∞

∞

π coth πa =

1 X 2a + . a a2 + n2 k=1

30

Plugging this back into the formula π coth πz = By Theorem 2.13, h0 (z) ≡ 0 for any z ∈ C. So sinh πz = ze

h(0)

1 z

+ h0 (z) +

P∞

2z n=1 n2 +z 2 ,

we conclude h0 (z) = 0 for z 6∈ iZ.

¶ ¶ ∞ µ ∞ µ Y Y z2 z2 1 + 2 = πz 1+ 2 . n n n=1 n=1

Remark 10. The above solution is a variant of the proof for factorization formula of sin πz. See Conway[2] VII, §6 for more details. (iii) Proof. From problem (v), we have

π 2

Q∞ ¡ n=1 1 −

1 4n2

¢

= 1 and

à ¡1 ¢2 ! ¶ µ ¶Y ∞ 1 1 2 −z −z =π −z 1− sin π 2 2 n2 n=1 µ ¶Y ¶µ ¶¾ ∞ ½µ 1 1 −z −z 1 1+ 2 π −z 1− 2 2 n n n=1 µ ¶µ ¶¾ ¶Y ¶µ ¶µ ∞ ½µ 1 1 z 1 z π 1+ 1− −z 1− 1+ 2 2n 2n n + 21 n − 12 n=1 ¶ N µ ¶ Y ¶ µ ¶ Y N µ N µ π z Y 1 z z lim · 1− 1 1− 2 · 1 + 1− N →∞ 2 4n n + 12 n − 12 2 n=1 n=1 n=1 ( N µ ¶) µ ¶ Y ¸ N · 1 z z2 π Y lim 1− 2 · 1− · 1− N →∞ 2 n=1 4n N + 12 (n − 12 )2 n=1 " # µ ¶2 ∞ Y z 1− . n + 12 n=0 µ

cos πz

= = = = = =

(iv) Proof. The function eπz − 1 has zeros an = 2ni (n ∈ Z), where a0 = 0 is a zero of order 1. Since for any ³ ´2 P P∞ ∞ R2 R > 0, we have n=−∞,n6=0 |aRn | = n=−∞,n6=0 4n 2 < ∞, by Weierstrass Factorization Theorem, we can find an entire function h(z) such that eπz − 1 = zeh(z)

n³

∞ Y

1−

n=−∞,n6=0

¶ ∞ µ Y z ´ z o z2 e 2ni = zeh(z) 1+ 2 . 2ni 4n n=1

To determine h(z), we note ∞ ∞ 2z X X πeπz (eπz − 1)0 1 1 2z 0 0 4n2 = = + h (z) + = + h (z) + . 2 πz πz 2 z e −1 e −1 z z 4n + z 2 1 + 4n2 n=1 n=1

Let γn be the rectangular path [2n + (2n + 1)i, −2n + (2n + 1)i, −2n − (2n + 1)i, 2n − (2n + 1)i, 2n + (2n + 1)i]. πz Define f (z) = (z2 −aπe 2 )(eπz −1) . Then for any given a, when n is large enough, we have Z f (z)dz = I + II + III + IV, γn

31

where

Z Z

Z

−2n

I=

πx

−2n

πee dx, [(x + (2n + 1)i)2 − a2 ](eπx + 1)

f (x + (2n + 1)i)dx = 2n

II =

2n

Z

−(2n+1)

−(2n+1)

f (−2n + yi)idy = 2n+1

Z

2n+1

Z

2n

III = −2n

and

2n

f (x − (2n + 1)i)dx =

Z

−2n

Z

2n+1

πx

πee dx, [(x − (2n + 1)i)2 − a2 ](eπx + 1)

2n+1

f (2n + yi)idy =

IV = −(2n+1)

πeπ(−2n+yi) idy, [(−2n + yi)2 − a2 ](eπ(−2n+yi) − 1)

−(2n+1)

πeπ(2n+yi) idy. [(2n + yi)2 − a2 ](eπ(2n+yi) − 1)

It is easy to see Z

2n

2π πdx 2n ≤p arctan p →0 2 2 x2 + (2n + 1)2 − a2 (2n + 1) − a (2n + 1)2 − a2

|I| ≤ −2n

as n → ∞. Similarly, we can show limn→∞ III = 0. Also, we note Z 2n+1 2π 2n + 1 πdy √ |II| ≤ ≤ arctan √ → 0, 2 + y 2 − a2 )(e2nπ − 1) 2nπ − 1) 4n2 − a2 2 − a2 (4n (e 4n −(2n+1) as n → ∞. For n sufficiently large, Z

2n+1

|IV | ≤ −(2n+1)

(4n2

e2nπ e2nπ −1

≤ 2, so

4π 2n + 1 πe2πn dy ≤√ arctan √ → 0, + − a2 )(e2πn − 1) 4n2 − a2 4n2 − a2 y2

as n → ∞. Combined, we have shown

Z lim

f (z)dz = 0.

n→∞

γn

By Residue Theorem, for a 6∈ 2Zi, we have Z n X f (z)dz = 2πiRes(f ; a) + 2πiRes(f ; −a) + 2πi Res(f ; an ). γn

k=−n

It’s easy to see Res(f ; a) =

πeπa 2a(eπa −1)

and Res(f ; −a) =

lim (z − an )f (z) = lim

z→an

z→an

π 2a(eπa −1) .

Since

z − an 1 πeπz · =− 2 , − a2 ) eπz − 1 4n + a2

(z 2

Res(f ; an ) = − 4n21+a2 . Therefore 1 0= lim 2πi n→∞ So

π eπa +1 2a eπa −1

=

1 a2

+

P∞

2 n=1 4n2 +a2

Ã

Z f (z)dz = lim

n→∞

γn

n X π eπa + 1 1 − πa 2 2a e − 1 4n + a2

! .

k=−n

for a 6∈ 2Zi. Therefore for z 6∈ 2Zi, we have ∞

h0 (z) = So h(z) = So

πz 2

πeπz 1 X 2z πeπz π eπz + 1 π − − = − = . eπz − 1 z n=1 4n2 + z 2 eπz − 1 2 eπz − 1 2

+ h(0). It’s easy to see limz→0 πz

e

eπz −1 z

= π. So eh(0) = π. Combined, we conclude eh(z) = πe

− 1 = πze

πz 2

∞ µ Y n=1

32

z2 1+ 2 4n

¶ .

πz 2

.

Equivalently, we have

∞ µ Y e − 1 = ze 1+ z 2

z

n=1

¶

z2 4π 2 n2

.

(v) Proof. Using the formula sin πz = −i sinh(iz) and the factorization formula for sinh πz, we have ¶ ¶ ∞ µ ∞ µ Y Y z2 z2 sin πz = −iπ(iz) 1 − 2 = πz 1− 2 . n n n=1 n=1

7. For a clear presentation of convergence of infinite products, we refer to Conway [2], Chapter VII, §5. In particular, we quote the following theorem (Conway [2], Chapter VII, §5, Theorem 5.9) Let G be an open subset of the complex plane. Denote by H(G) the collection of analytic functions on G, equipped with the topology determined by uniform convergence on compact subsets of G. Then H(G) is a complete metric space. Furthermore, we have the following theorem (proof omitted). ∞ Theorem no fn is identically P 11. Let G be a region in C and let (fn )n=1 be a sequence in H(G) such that Q∞ zero. If [fn (z)−1] converges absolutely and uniformly on compact subsets of G, then n=1 fn (z) converges in H(G) to an analytic function f (z). If a is a zero of f then a is a zero of only a finite number of the functions fn , and the multiplicity of the zero of f at a is the sum of the multiplicities of the zeros of the functions fn at a. P∞ Proof. (Proof of Blaschke Product) Fix r ∈ (0, 1). Since k=1 (1 − |ak |) < ∞, limk→∞ |ak | = 1. So there exists k0 ∈ N, such that for any k ≥ k0 , 1+r 2 < |ak | < 1. So for any k ≥ k0 , ¯ ¯ ¯ ak − z |ak | ¯ |ak |(1 − |ak |) + |z||ak |(1 − |ak |) 2 (1 + r)(1 − |ak |) ¯ = · − 1¯¯ ≤ ≤ (1 − |ak |). 1+r ¯1 − a ¯ k z ak |1 − a ¯k z||ak | 1−r 2 (1 − r) ¯ P∞ ¯¯ ak −z |ak | P∞ ¯ · − 1 Since k=1 (1 − |ak |) < ∞, we conclude k=1 ¯ 1−¯ ¯ is absolutely and uniformly convergent on ak z ak Q∞ ak −z |ak | {z : |z| ≤ r}. Therefore k=1 1−¯ak z · ak is absolutely and uniformly convergent on {z : |z| ≤ r}. By Q∞ ak −z |ak | Weierstrass Theorem (Theorem 3.1), f (z) = k=1 1−¯ ak z · ak represents a non-zero holomorphic function on ak −z {z : |z| < 1}. Since the mapping z 7→ 1−¯ (0 < |a | < 1) maps D(0, 1) to D(0, 1), we conclude |f (z)| ≤ 1. k ak z By the theorem quoted at the beginning of the solution, it’s clear that (ak )∞ k=1 are the only zeros of f (z).

8. Proof. Let Rε = R − ε, ε ∈ (0, R). Define Qt Fε (z) = f (z) ·

Rε (z−bj ) j=1 Rε2 −¯ bj z Qs Rε (z−ai ) i=1 Rε2 −¯ ai z

, ∀z ∈ D(0, Rε ).

R2 (z−a )

R2 (z−b )

It’s easy to verify that each of Rε2 −¯bj zj and Rε2 −¯ai zi maps D(0, Rε ) onto itself and takes the boundary to ε ε the boundary. Therefore Fε (z) is analytic on D(0, Rε ), has no zeros in D(0, Rε ), and |Fε (z)| = |f (z)| for |z| = Rε . Since log |Fε (z)| is a harmonic function, by formula (2.33) on page 71, 1 log |Fε (z)| = 2π

Z

2π 0

Rε eiφ + z 1 log |Fε (Rε e )|Re dφ = Rε eiφ − z 2π

Z

iφ

33

0

2π

log |f (Rε eiφ )|Re

Rε eiφ + z dφ. Rε eiφ − z

Plugging the formula of Fε into the above equality, we get ¯ 2 ¯ ¯ 2 ¯ Z 2π s t X ¯ Rε − ¯bj z ¯ X ¯ Rε − a ¯i z ¯¯ 1 Rε eiφ + z iφ ¯ ¯ ¯ log |f (z)| = log |f (Rε e )|Re dφ + log ¯ − log ¯ . 2π 0 Rε eiφ − z Rε (z − bj ) ¯ i=1 Rε (z − ai ) ¯ j=1 Letting ε → 0 yields the desired formula. 9. Proof. By Theorem 3.4 (correction: “holomorphic function f (z)” in the theorem’s statement should be 1 2 3 7 “meromorphic function f (z)”), f (z) has the form of z+1 + z−2 + (z−2) 2 + c, where c is a constant. f (0) = 4 implies c = 1. Since for any ζ ∈ D(0, 1), !0 µ ¶0 à X ∞ ∞ X 1 1 n = ζ = (n + 1)ζ n , = (1 − ζ)2 1−ζ n=0 n=0 the Laurent expansion of f (z) in 1 < |z| < 2 is f (z)

= = = =

1 2 3 + + +1 z + 1 z − 2 (z − 2)2 1 1 1 3 1 − + · +1 z 1 + z1 1 − z2 4 (1 − z2 )2 ¶n X ∞ µ ∞ ³ ´n ∞ ³ z ´n 1X 1 3X z − − + (n + 1) +1 z n=0 z 2 4 n=0 2 n=0 −

∞ ∞ X (−1)n 3 X 3n − 1 n + + z . zn 4 n=1 2n+2 n=1

10. Proof. We follow the construction outlined in the proof of Mittag-Leffler Theorem (Theorem 3.9). For each n ∈ N, when |z| < n2 , ψn (z) has Taylor expansion ³ z ´k n 1 1 1X (k + 1) = · = . ¡ ¢ (z − n)2 n 1− z 2 n n k=0 n ∞

ψn (z) =

Let λn be a positive integer to be determined later. We estimate the tail error obtained by retaining only the first λn terms of the Taylor expansion of ψn (z): ¯ ¯ Ã ∞ ! λn ∞ ∞ ¯ ³ z ´k ¯ 1X 1 X (k + 1) 1 X (k + 2) 1 X k 1 ¯ ¯ (k + 1) = = + λn −1 ¯ ≤ ¯ψn (z) − ¯ n n ¯ n 2k n 2k+1 n 2k+1 2 k=0 k=λn +1 k=λn k=λn µ ¶x Z Z ∞ 1 1 1 X k+1 x 1 1 ∞ x ≤ + dx = + 1d 2 x λn −1 n · 2λn −1 n 2 n · 2 n log λn 2 k=λn k · ¸ 1 1 λn 1 = + + . n · 2λn −1 n · 2λn log 2 (log 2)2 i h i h λn 1 1 1 n 1 If we let λn = n, then εn := n·2λ1n −1 + n·21λn log + = + + satisfies 2 n−1 n 2 2 (log 2) n·2 n·2 log 2 (log 2) P∞ n=1 εn < ∞. By the proof of Mittag-Leffler Theorem, we can write f (z) as " # ∞ n ³ z ´k X n 1X f (z) = U (z) + − (k + 1) , (z − n)2 n n n=1 k=0

where U (z) is an entire function. 34

11. (i) 1 Proof. Define f (z) = (z2 −a2 )(e πz −1) , where a ∈ C \ 2iZ. Let γn be the rectangular path [2n + (2n + 1)i, −2n + (2n + 1)i, −2n − (2n + 1)i, 2n − (2n + 1)i, 2n + (2n + 1)i]. Then for any given a, when n is large enough, we have Z f (z)dz = I + II + III + IV, γn

where

Z

Z

−2n

2n

Z

2n

dx , [(x + (2n + 1)i)2 − a2 ](eπx + 1)

f (x + (2n + 1)i)dx =

I=

Z

−(2n+1)

II =

−2n

−(2n+1)

f (2n + yi)idy = 2n+1

Z

2n

III =

2n

f (x − (2n + 1)i)dx = −2n

and

[(−2n +

2n+1

Z

Z

−2n

Z

2n+1

IV =

idy , − a2 ](eπ(−2n+yi − 1)

−dx , [(x − (2n + 1)i)2 − a2 ](eπx + 1)

2n+1

f (2n + iy)idy = −(2n+1)

yi)2

−(2n+1)

[(2n +

iy)2

idy . − a2 ](eπ(2n+iy) − 1)

It’s easy to see Z

2n

|I| ≤ −2n

dx 2 2n ≤p arctan p →0 2 2 x2 + (2n + 1)2 − a2 (2n + 1) − a (2n + 1)2 − a2

as n → ∞. Similarly, we can show limn→∞ III = 0. Also, we note Z 2n+1 2 2n + 1 dy √ = arctan √ →0 |II| ≤ 2 + y 2 − a2 )(1 − e−2nπ ) −2nπ ) 4n2 − a2 2 − a2 (4n (1 − e 4n −(2n+1) as n → ∞, and Z

2n+1

dy 2n + 1 2 √ arctan 2 = →0 2 2πn 2πn 2 2 + − a )(e − 1) 4n − a2 (e − 1) 4n − a −(2n+1) R as n → ∞. Combined, we have shown limn→∞ γn f (z)dz = 0. By Residue Theorem, for a 6∈ 2iZ, we have |IV | ≤

(4n2

y2

Z f (z)dz = 2πiRes(f ; a) + 2πiRes(f ; −a) + 2πi γn

It’s easy to see Res(f ; a) =

n X

Res(f ; 2ni).

k=−n 1 2a(eπa −1)

and Res(f ; −a) =

lim (z − 2ni)f (z) = lim

z→2ni

z→2ni

π(z 2

1 2a(1−e−πa ) .

Since

π(z − 2ni) 1 =− , 2 πz 2 − a )(e − 1) π(4n + a2 )

we have Res(f ; 2ni) = − π(4n21+a2 ) . Therefore 1 lim 0= 2πi n→∞

Ã

Z f (z)dz = lim γn

n→∞

n 1 eπa + 1 1 X 1 · πa − 2a e − 1 π 4k 2 + a2 k=−n

! =

∞ 1 eπa + 1 1 X 1 · πa − . 2a e − 1 π 4k 2 + a2

Replacing πa with z, we get after simplification ∞

1 2z 1 1 X = − + , ∀z ∈ C \ 2Zπi. z e −1 z 2 4k 2 π 2 + z 2 k=1

This is the partial fraction of 1/(ez − 1). 35

k=−∞

(ii) ¡ ¢ ¡ ¢ 1 1 1 Proof. Define f (z) = (z−a) sin 2 (πz) , where a ∈ C\Z. Let γn be the rectangular path [ n + 2 +ni, − n + 2 + ¡ ¢ ¡ ¢ ¡ ¢ ni, − n + 12 − ni, n + 12 − ni, n + 12 + ni]. Then when n is large enough, we have Z f (z)dz = I + II + III + IV, γn

where Z

−(n+ 12 )

I=

n+ 21

Z

dx = (x + ni − a) sin2 [π(x + ni)]

Z

(n+ 21 )

−(n+ 21 )

4dx , (x + ni − a)(e−2nπ+2ixπ + e2nπ−2ixπ − 2)

Z n idy −4idy = , 2 1 1 1 [−(n + ) − a + yi](e2πy + e−2πy + 2) [−(n + ) + yi − a] sin [−π(n + ) + πyi] n −n 2 2 2 Z (n+ 12 ) Z (n+ 12 ) dx −4dx III = = , 2 2nπ+2iπx + e−2nπ−2iπx − 2) −(n+ 12 ) (x − ni − a) sin [π(x − ni)] −(n+ 12 ) (x − a − ni)(e −n

II =

and Z

n

IV = −n

idy = [(n + 12 ) + yi − a] sin2 [π(n + 12 ) + πyi]

Z

n −n

4idy . [(n + 12 ) − a + yi](e−2yπ + e2yπ + 2)

It’s easy to see Z |I| ≤

(n+ 21 )

4dx

p

(x − a)2 + n2 (e2nπ − e−2nπ − 2)

−(n+ 21 )

4(2n + 1) →0 n(e2nπ − e−2nπ − 2)

≤

as n → ∞, Z |II|

n

q

≤

(n +

−n

Z

n

≤ 0

Z

4dy 1 2

n

q (n +

=

+ a)2 + y 2 (e2πy + e−2πy + 2)

0

8dy 1 2

+ a)2 + y 2 (e2πy + e−2πy + 2)

8dy 8 1 − e−2πn = →0 2π (n + 12 + a)e2πy (n + 12 + a)

as n → ∞, Z

(n+ 21 )

|III| ≤ as n → ∞, and lastly Z n q |IV | ≤ −n

Z

4dx

p

−(n+ 12 )

(x −

a)2

+

n2 (e2nπ

−

e−2nπ

1 2

− 2) Z

4dy (n +

≤

n

8dy

q

=

− a)2 + y 2 (e2yπ + e−2yπ + 2)

4(2n + 1) →0 n(e2nπ − e−2nπ − 2)

(n +

0

1 2

− a)2 + y 2 (e2yπ + e−2yπ + 2)

n

8 1 − e−2πn 8 = →0 1 2yπ 2π (n + 21 − a) 0 (n + 2 − a)e R as n → ∞. Combined, we have shown limn→∞ γn f (z)dz = 0. By Residue Theorem, we have ≤

Z f (z)dz = 2πiRes(f ; a) + 2πi γn

n X k=−n

36

Res(f ; k).

2

(z−k) 1 It’s easy to see Res(f ; a) = sin21(πa) . Define hk (z) = sin 2 (πz) . Since limz→k hk (z) = π 2 , hk is holomorphic in a neighborhood of k and for ε > 0 small enough, and we have · ¸¯ Z Z 1 dz 1 hk (z) dz d hk (z) ¯¯ = . Res(f ; k) = = · 2πi |z−k|=ε (z − a) sin2 (πz) 2πi |z−k|=ε (z − a) (z − k)2 dz z − a ¯z=k

We note d dz

·

¸¯ hk (z) ¯¯ z − a ¯z=k

=

lim

z→k

h0k (z)(z − a) − h(z) h0 (z) 1 = lim k − , 2 z→k z − a (z − a) (k − a)2 π 2

and lim

z→k

h0k (z) z−a

= = = = =

2(z − k) sin(πz) − 2π(z − k)2 cos(πz) z→k sin3 (πz) 2 sin(πz) + 2π(z − k) cos(πz) − 4π(z − k) cos(πz) + 2π 2 (z − k)2 sin(πz) lim z→k 3 sin2 (πz) · π cos(πz) 2 sin(πz) − 2π(z − k) cos(πz) lim z→k 3π sin2 (πz) 2π cos(πz) − 2π cos(πz) + 2π 2 (z − k) sin(πz) lim z→k 6π 2 sin(πz) cos(πz) 0. lim

So for a ∈ C \ Z,

Z f (z)dz = γn

Let n → ∞, we get

n X 2πi 2πi − . sin2 (πa) k=−n (k − a)2 π 2

∞ X π2 1 . = sin2 (πz) n=−∞ (z − n)2

Remark 12. Another choice is to let f (z) = cot(πz) (z+a)2 , see Conway [2], page 122, Exercise 6. The trick used in this problem and problem 6 (ii) will be generalized in problem 12. (iii) Proof. By the solution of problem 6 (ii), we have ∞

coth z =

1 X 2z + . 2 z n=1 z + π 2 n2

Since coth z = i cot(iz), we have ∞

cot z Therefore πβ π cot α α

= = = =

= i coth(iz) =

∞

1 X 2z 1 X − = − 2 2 2 z n=1 π n − z z n=1

µ

1 1 − πn − z πn + z

¶ .

" ¶ ¶# ∞ µ N µ 1 X 1 X 1 1 1 1 − − = lim − − N →∞ β β n=1 αn − β αn + β αn − β αn + β n=1 · µ ¶ µ ¶¸ 1 1 1 1 1 1 1 − + + ··· + + + + ··· + lim N →∞ β α−β 2α − β Nα − β α+β 2α + β Nα + β ·µ ¶ µ ¶ µ ¶ ¸ 1 1 1 1 1 1 1 lim − + − + ··· + − + N →∞ β α−β α+β 2α − β (N − 1)α + β Nα − β Nα + β ¸ X ∞ · ∞ X 1 1 α − 2β − = . nα + β nα + (α − β) (nα + β)[nα + (α − β)] n=0 n=0 37

By letting α = 3 and β = 1, we get ∞ X π π π 1 √ = cot = . 3 3 (3n − 2)(3n − 1) 3 3 n=1

Remark 13. For a direct proof without using problem 6 (ii), we can let f (z) = cot(πz) z 2 −a2 and apply Residue Theorem to it. For details, see Conway [2], page 122, Exercise 8. The line of reasoning used in this approach will be generalized in problem 12 below. 12. (1) Proof. Since z = ∞ is a zero of f (z) with multiplicity p, f (z) can be written as h(z) z p where h is holomorphic in a neighborhood of ∞. Consequently, h is bounded in a neighborhood of ∞. In the below, we shall work with a sequence of neighborhoods of ∞ that shrinks to ∞. So without loss of generality, we can assume h is bounded and denote its bounded by M . Let γn be the rectangular path [(n + 12 ) + ni, −(n + 12 ) + ni, −(n + 12 ) − ni, (n + 12 ) − ni, (n + 21 ) + ni]. Then Z f (z) cot(πz)dz = I + II + III + IV, γn

where Z I=

−(n+ 21 )

(n+ 12 )

h(x + ni) (x + ni)p Z

II

eiπ(x+ni) +e−iπ(x+ni) 2 eiπ(x+ni) −e−iπ(x+ni) 2i

−n

= n

Z

−n

= n

Z dx =

h(−(n + 21 ) + yi) [−(n + 12 ) + yi]p

−(n+ 21 ) (n+ 12 )

h(x + ni) eiπx−nπ + e−iπx+nπ idx, (x + ni)p eiπx−nπ − e−iπx+nπ

1

1

1

1

eiπ[−(n+ 2 )+yi] +e−iπ[−(n+ 2 )+yi] 2 eiπ[−(n+ 2 )+yi] −e−iπ[−(n+ 2 )+yi] 2i n −yπ n yπ

idy

h(−(n + 12 ) + yi) (−1) e (−i) + (−1) e i (−1)dy, 1 n −yπ p (−i) − (−1)n eyπ i [−(n + 2 ) + yi] (−1) e Z

III =

(n+ 12 )

−(n+ 12 )

h(x − ni) eiπx+nπ + e−iπx−nπ idx, (x − ni)p eiπx+nπ − e−iπx−nπ

and Z

n

IV = −n

We note

Z

(n+ 21 )

|I| ≤

−(n+ 12 )

as n → ∞,

Z

n

|II| ≤ −n

h((n + 12 ) + yi) (−1)n e−yπ i + (−1)n eyπ (−i) (−1)dy. [(n + 12 ) + yi]p (−1)n e−yπ i − (−1)n eyπ (−i) n+ M e2nπ + 1 2M e−nπ + enπ dx = arctan p nπ −nπ e2nπ − 1 n n (x2 + n2 ) 2 e − e

¯ yπ ¯ ¯ e − e−yπ ¯ n 2M ¯ ¤ p2 ¯ eyπ + e−yπ ¯¯ dy ≤ £ 1 arctan 1 2 n + n + 2 (n + 2 ) + y 2

1 2

M

1 2

→0

→0

as n → ∞, Z

(n+ 12 )

|III| ≤

−(n+ 21 )

n+ enπ + e−nπ enπ + e−nπ 2M M dx = nπ arctan p nπ −nπ −nπ 2 2 e −e n n (x + n ) 2 e − e

1 2

→0

as n → ∞, and Z

n

|IV | ≤ −n

¯ yπ ¯ ¯ e − e−yπ ¯ n 2M ¯ £ ¤ p2 ¯ eyπ + e−yπ ¯¯ dy ≤ 1 arctan 1 2 n + n + 2 (n + 2 ) + y 2 M

38

1 2

→0

R as n → ∞. Combined, we can conclude limn→∞ γn f (z) cot(πz)dz = 0. Meanwhile, for n sufficiently large, by Residue Theorem, we have Z n m X X Res(f (z) cot(πz), αk ) + 2πi Res(f (z) cot(πz), k). f (z) cot(πz)dz = 2πi γn

k=−n

k=1

We note 1 Res(f (z) cot(πz), k) = 2πi So

Z

Z

cos(πz) 1 f (z) dz = sin(πz) 2πi |z−k|=ε

Z f (z) cot(πz)dz = 2πi γn

m X

|z−k|=ε

(−1)k f (z) cos(πz)(z − k) f (k) · dz = . z−k sin[π(z − k)] π

Res(f (z) cot(πz), αk ) + 2πi

k=1

Letting n → ∞ gives us

n X

lim

n→∞

f (k) = −π

k=−n

n X f (k) . π

k=−n n X

Res(f (z) cot(πz), αk ).

k=1

(2) Proof. As argued in the solution of (1), we can assume f (z) = h(z) z p where h is holomorphic in a neighborhood of ∞ and has bound M in that neighborhood. Let γn denote the same rectangular path as in our solution of (1). Then Z f (z) dz = I + II + III + IV, sin(πz) γn where

Z I= Z

−(n+ 12 )

n+ 12

2idx h(x + ni) , p iπx−nπ (x + ni) e − e−iπx+nπ

h(−(n + 12 ) + yi) 2i idy, 1 p e−yπ (−1)n (−i) − eyπ (−1)n i [−(n + ) + yi] n 2 Z n+ 12 h(x − ni) 2idx III = , p eiπx+nπ − e−iπx−nπ 1 (x − ni) −(n+ 2 )

II =

and

−n

Z

n

IV = −n

h((n + 21 ) + yi) 2i £ ¤p −πy idy. n 1 (−1) i − eπy (−1)n (−i) (n + 2 ) + yi e

We note 2M |I| ≤ nπ e − e−nπ

Z

n+ 12

−(n+ 12 )

4M dx p ≤ nπ − e−nπ 2 2 2 e (x + n )

Z

n+ 12

0

n+ dx 2M = nπ arctan x2 + n2 (e − e−nπ )n n

1 2

→0

as n → ∞, and Z

n

|II| ≤ −n

2M dy ≤ £ ¤ yπ −yπ (n + 21 )2 + y 2 e + e 1

p 2

Z

n

−n

2M M dy n = 1 2 1 arctan 2 (n + 2 ) + y n+ 2 n+

1 2

→0

as n → ∞. By argument, we can prove limn→∞ III = limn→∞ IV = 0. Combined, we can conclude R similar f (z) that limn→∞ γn sin(πz) dz = 0. Meanwhile, for n sufficiently large, by Residue Theorem, we have Z γn

m

X f (z) dz = 2πi Res sin(πz) k=1

µ

f (z) , αk sin(πz) 39

¶ + 2πi

n X k=−n

µ Res

¶ f (z) ,k . sin(πz)

It’s easy to see µ Res Therefore,

f (z) ,k sin(πz)

Z

¶ =

Z

1 2πi

|z−k|=ε

µ

m

(−1)k f (k) (−1)k f (z)(z − k) · dz = . z − k sin[π(z − k)] π ¶

n X (−1)k f (k) . π γn k=−n k=1 ³ ´ Pn Pm f (z) Let n → ∞, we get limn→∞ k=−n (−1)k f (k) = −π k=1 Res sin(πz) , αk .

X f (z) Res dz = 2πi sin(πz)

f (z) , αk sin(πz)

+ 2πi

(3) (i) Proof. Let f (z) =

1 (z+a)2 .

Then by result of (1),

∞ X

1 = −πRes (n + a)2 n=−∞

µ

¶ cot(πz) d π2 , −a = −π [cot(πz)]|z=−a = . 2 2 (z + a) dz sin (πa)

This result is verified by problem 11 (ii). (ii) Proof. Let f (z) =

1 z 2 +a2 .

Then by result of (2),

∞ X

(−1)n n2 + a2 n=−∞

=

· µ −π Res

=

−

¶ µ ¶¸ 1 1 , ai + Res , −ai (z 2 + a2 ) sin(πz) (z 2 + a2 ) sin(πz)

π πcsch(πa) = . ai · sin(πai) a

This result can be verified by Mathematica. 13. (i) Proof. f (z) =

1 z 2 −z 4

has poles 0, 1, and −1. ∞ is a removable singularity. µ ¶ Z 1 dz d 1 −(−2z) Res(f ; 0) = = |z=0 = |z=0 = 0, 2 2 2 2πi |z|=ρ z (1 − z ) dz 1 − z (1 − z 2 )2 Z 1 dz 1 1 Res(f ; 1) = =− |z=1 = − , 2 2 2πi |z−1|=ρ (1 − z)(1 + z)z (1 + z)z 2

and Res(f ; −1) =

1 2πi

Z |z+1|=ρ

dz 1 = . (1 + z)(1 − z)z 2 2

(ii) Proof. We note f (z) :=

z2 + z + 2 2 −1 + i −4 − i −1 − i −4 + i = + + + + . 2 2 2 2 z(z + 1) z 4(z − i) 4(z − i) 4(z + i) 4(z + i)

Therefore, the function f (z) has poles 0, i and −i. ∞ is a removable singularity of f (z). Furthermore, we have i −4 + i i −4 − i = −1 − , Res(f ; −i) = = −1 + . Res(f ; 0) = 2, Res(f ; i) = 4 4 4 4

40

(iii) Proof. Suppose a1 , · · · , an are the roots of the equation z n + an = 0. Then each ai is a pole of order 1 for n−1 the function f (z) = zzn +an . ∞ is a removable singularity of f (z). Then an−1 (z − ai ) an−1 (z − ai ) 1 an−1 i = lim Qin = lim i n = . n z→a z→a z +a n i i j=1,j6=i (ai − aj ) j=1 (z − aj )

Res(f ; ai ) = Qn

(iv) Proof. f (z) =

1 sin z

has πZ as poles of order 1. ∞ is not an isolated singularity of f (z). And

Res(f, nπ) =

1 2πi

Z |z−nπ|=ε

1 dz = sin z 2πi

Z |z−nπ|=ε

1 (−1)n (z − nπ) dz = (−1)n . z − nπ sin(z − nπ)

(v) ³ ´ ³ i ´ ³ ´ P∞ (−1)n i 1 1 3 Proof. First, we note as cos z−2 = 12 e z−2 + e− z−2 = . So f (z) := z cos = n=0 (z−2)2n z−2 P∞ (−1)n z3 n=0 (z−2)2n . This shows z = 2 is an essential singularity of f (z), and Res(f ; 2) =

Z ∞ X (−1)n z3 1 dz = (−1) · 3z 2 |z=2 + (−1)2 · · 6|z=2 = −11. 2n 2πi |z−2|=ε (z − 2) 3! n=0

Also, ∞ is an isolated singularity of f (z), and we have µ ¶ µ ¶ Z Z 1 1 1 1 3 3 Res(f ; ∞) = − z cos dz = − z cos dz 2πi |z|=R z−2 2πi |z−2|=R z−2 µ ¶3 µ ¶ Z Z 1 1 dζ 1 (ζ + 2)3 = − + 2 cos ζ − 2 = cos ζdζ 2πi |ζ|= R1 ζ ζ 2πi |ζ=ε| ζ5 =

¤ 8 1 d4 £ (ζ + 2)3 cos ζ |ζ=0 = − . 4 4! dζ 3

(vi) Proof. Let f (z) =

ez z(z+1)

= ez

³

1 z

−

1 z+1

´ . Then Res(f ; 0) = 1, Res(f ; −1) = −e−1 , and Res(f ; ∞) =

1 − e−1 . 14. Proof. We can write g(z) as (z − a)2 h(z) where h is holomorphic and h(a) 6= 0. Then h(a) = h0 (a) =

g 0 (a)(z − a) − 2g(a) . (z − a)3

41

g(a) (z−a)2

and

Therefore µ Res

f (z) ,a g(z)

¶

µ = Res

1 f (z) · ,a (z − a)2 h(z)

¶ =

·

d dz

¸¯ f (z) ¯¯ f 0 (a)h(a) − f (a)h0 (a) = ¯ h(z) z=a h2 (a)

0

g(a) g (a)(z−a)−2g(a) f 0 (a) (z−a) 2 − f (a) (z−a)3

=

g 2 (a) (z−a)4

g(a)(z − a)[f 0 (a)(z − a) + 2f (a)] − f (a)g 0 (a)(z − a)2 . g 2 (a)

=

15. We make an observation of some simple rules that facilitate the evaluation of integrals via Residue Theorem. Rule 1 (rule for integrand function). The poles of the integrand function should be easy to find, such f (z) that the integrand function can be easily represented as (z−a) n , where f is holomorphic. The holomorphic function f (z) can be multi-valued function like logarithm function and power function. When it’s difficult to make f holomorphic in the desired region, check if it is the real or imaginary part of a holomorphic function. Rule 2 (rule for integration path). The choice of integration path is highly dependent on the properties of the integrand function, where symmetry and multi-valued functions (log and power functions) are often helpful. Oftentimes, the upper and lower limits of integration either form a full circle (i.e. [0, 2π]) so that substitution for trigonometric functions can be easily done or have ∞ as one of the end points. (i) Proof. Let γ1 = {z : −R ≤ Rez ≤ R, Imz = 0}, γ2 = {z : |z| = R, 0 ≤ arg z ≤ π}, and f (z) = Z γ1

and

Then

· ¸ d z2 π = , 2 z→i dz (z + i) 2

Z f (z)dz +

z2 (z 2 +1)2 .

f (z)dz = 2πi · Res(f, i) = 2πi · lim γ2

¯ Z π ¯ R3 R2 e2θi iθ ¯≤ Re · idθ dθ → 0 ¯ 2 2θi 2 2 2 + 1) 0 (R − 1) γ2 0 (R e R ∞ 2 dx R 1 ∞ x2 dx as R → ∞. So by letting R → ∞, we have 0 (xx2 +1) = π4 . 2 = 2 −∞ (x2 +1)2 ¯Z ¯ ¯ ¯

¯ ¯Z ¯ ¯ f (z)dz ¯¯ = ¯¯

π

(ii) Proof. We note sin2 x =

1−cos 2x , 2

Z

π 2

0

so

dx 1 = 2 2 a + sin x

Z

π

0

dx 1 = 2 2 a + sin x

Z

2π

dθ . (2a + 1) − cos θ

0

Let b = 2a + 1, then Z 0

π 2

dx 1 = 2 2 a + sin x

Z |z|=1

dz iz

b−

1 2 (z

+

z −1 )

The equation − 12 z 2 + bz − 12 = 0 has two roots: z1 = b − ¯ 1). Therefore, have z1 ∈ D(0, 1) and z2 6∈ D(0, Z 0

π 2

dx =i a + sin2 x

Z |z|=1

=

√

1 2i

Z 1 2 |z|=1 − 2 z

dz . + bz − 12

b2 − 1 and z2 = b +

√

b2 − 1. Since b > 1, we

1 π π dz = i · 2πi · =√ = p . 2 (z − z1 )(z − z2 ) z1 − z2 b −1 2 a(a + 1)

42

Remark 14. If R(x, y) is a rational function of two variables x and y, for z = eiθ , we have µ µ ¶ µ ¶¶ 1 1 1 1 dz R(sin θ, cos θ) = R z+ , z− , dθ = . 2 z 2i z iz Therefore

Z

µ

Z

2π

R(sin θ, cos θ)dθ = 0

R |z|=1

¶ 1 1 1 1 dz (z + ), (z − ) . 2 z 2i z iz

Remark 15. When the integrand function is a rational function of trigonometric functions, in view of the previous remark, it is desirable to have [0, 2π] as the integration interval. For this reason, we first use symmetry to expand the integration interval from [0, π2 ] to [0, π], and then use double angle formula to expand [0, π] to [0, 2π]. This solution is motivated by Rule 2. (iii) Proof. Let γ1 = {z : −R ≤ Rez ≤ R, Imz = 0}, γ2 = {z : |z| = R, 0 ≤ arg z ≤ π}, and f (z) = Z

Z

Z

f (z)dz +

f (z)dz = 2πiRes(f, i) =

γ1

γ2

and

|z−i|=ε

zeiz z 2 +1 .

Then

ie−1 π zeiz dz = 2πi · = i, (z − i)(z + i) 2i e

¯ Z ¯ ¯¯Z π ¯ π iReiθ ¯ ¯ R2 e−R sin θ Re ¯ iθ Re · idθ¯ ≤ dθ → 0 f (z)dz ¯¯ = ¯ 2 2θi ¯ ¯ 0 R e +1 R2 − 1 γ2 0

¯Z ¯ ¯ ¯

as R → ∞ by Lebesgue’s Dominated Convergence Theorem. Therefore by letting R → ∞, we have Z ∞ Z ∞ Z ∞ π xeix x cos x x sin x dx = dx + i dx = i, 2 2 2 e −∞ x + 1 −∞ x + 1 −∞ x + 1 R ∞ x sin x π which implies 0 x2 +1 dx = 2e . (iv) Proof. Let r, R be two positive numbers such that r < 1 < R. Let γ1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ |z| ≤ R, arg z = π}, γR = {z : |z| = R, 0 ≤ arg z ≤ π}, and γr = {z : |z| = r, 0 ≤ arg z ≤ π}. log z Define f (z) = (1+z 2 )2 where log z is defined on C \ [0, ∞) and takes the principle branch of Logz with arg z ∈ (0, 2π) (see page 23). By Residue Theorem, · ¸ Z d log z π π2 f (z)dz = 2πiRes(f, i) = 2πi lim = − + i. z→i dz (z + i)2 2 4 γ1 +γR +γ2 −γr We note ¯Z ¯ ¯ ¯

γR

¯ ¯Z ¯ ¯ f (z)dz ¯¯ = ¯¯

π

0

as R → ∞, ¯Z ¯ ¯Z ¯ ¯ ¯ ¯ f (z)dz ¯¯ = ¯¯ ¯ γr

as r → 0, and

0

π

¯ Z π p ¯ R (log R)2 + θ2 πR(log R + π) log(Reiθ ) iθ Re · idθ¯¯ ≤ dθ ≤ →0 2 2iθ 2 2 − 1)2 (R e + 1) (R (R2 − 1)2 0 ¯ Z π p ¯ r (log r)2 + θ2 πr(− log r + π) log(reiθ ) iθ ¯≤ re · idθ dθ ≤ →0 ¯ 2 )2 (r2 e2iθ + 1)2 (1 − r (1 − r2 )2 0

Z

Z

−r

f (z)dz = γ2

−R

log x dx = (1 + x2 )2

Z

R r

43

log(−x) dx = (1 + x2 )2

Z

R r

log x + πi dx. (1 + x2 )2

Therefore by letting r → 0 and R → ∞, we have Z ∞ π π2 2 log x + πi dx = − + i, (1 + x2 )2 2 4 0 R ∞ log x π i.e. 0 (1+x 2 )2 dx = − 4 . (v) 1−α

z 1−α Proof. Let γ1 , γ2 , γr and γR be defined as in (iv). Define f (z) = 1+z = e(1−α) log z is defined 2 , where z on C \ [0, ∞) with log z taking the principle branch of Logz. By Residue Theorem,

Z f (z)dz = 2πiRes(f, i) = 2πi · γ1 +γR +γ2 −γr

We note

¯Z ¯ ¯ ¯

γR

as R → ∞,

¯Z ¯ ¯ ¯

¯ ¯Z ¯ ¯ f (z)dz ¯¯ = ¯¯

γr

π

0

¯ ¯Z ¯ ¯ f (z)dz ¯¯ = ¯¯

π e(1−α) log z |z=i = πe(1−α) 2 i . z+i

¯ 2−α ¯ R1−α eiθ(1−α) iθ ¯≤ R · Re · idθ ¯ R2 − 1 π → 0 1 + R2 e2iθ

π 0

¯ 2−α ¯ r1−α eiθ(1−α) iθ ¯≤ r · re π→0 · idθ ¯ 2 2iθ 1+r e 1 − r2

as r → 0, and Z

Z

−r

f (z)dz = γ2

−R

x1−α dx = 1 + x2

Z

R

r

e(1−α) log(−x) dx = 1 + x2

Z

R

x1−α ei(1−α)π dx. 1 + x2

r

Therefore by letting r → 0 and R → ∞, we have Z ∞ 1−α Z ∞ 1−α π x (1 − α)π i(1−α) π x (1 + ei(1−α)π ) 2 = πei(1−α) 2 , dx = dx · 2 cos e 2 2 1 + x 1 + x 2 0 0 i.e.

R∞ 0

x1−α 1+x2 dx

=

π 2

csc

¡ απ ¢ 2 .

(vi) Proof. This problem is a special of problem (x). See the solution there. (vii) Proof. We use the method outlined in the remark of Problem (ii). Z 0

π

dθ 1 = a + cos θ 2

Z

π

−π

dθ 1 = a + cos θ 2

Z |z|=1

a+

dz iz z+z −1 2

The equation 12 z 2 + az + 12 = 0 has two roots: z1 = −a + |z1 | = a+√1a2 −1 < 1 and |z2 | > 1. So Z 0

π

dθ 1 = a + cos θ i

Z |z|=1

√

1 = 2i

Z 1 2 |z|=1 2 z

dz . + az + 21

a2 − 1 and z2 = −a −

√

a2 − 1. Clearly,

dz 1 π = 2π · =√ . (z − z1 )(z − z2 ) z1 − z2 a2 − 1

Remark 16. The expansion of integration interval from [0, π] to [−π, π] is motivated by Rule 2. (viii)

44

Proof. Using the result on Dirichlet integral: Z

∞ 0

µ

sin x x

¶2

Z

∞

dx = 0

R∞ 0

sin x x dx

=

π 2,

we have

µ ¶ Z ∞ Z ∞ 1 2 sin x cos xdx sin(2x) π sin xd − = = d(2x) = . x x 2x 2 0 0 2

(ix) Proof. It is easy to see that when λ = p = 0, the integral is equal to 1. So without loss of generality, we only consider the cases where λ and p are not simultaneously equal to 0. Choose r, R so that 0 < r < R, and r is sufficiently small and R is sufficiently large. Define γ1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ |z| ≤ R, arg z = 2π}, γr = {z : |z| = r, 0 < arg z < 2π}, and γR = {z : p log z p |z| = R, 0 < arg z < 2π}. Let ρ = cos λ, then ρ ∈ (−1, 1]. Finally, define f (z) = z2 +2zzcos λ+1 = z2e+2ρz+1 . For R sufficiently large and r sufficiently small, the two roots √ of z 2 + 2rz + 1 = 0 are contained in {z : r √ < |z| < R}. These two roots are, respectively, z1 = −ρ + i 1 − r2 = − cos λ + i| sin λ| and z2 = −ρ − i 1 − r2 = − cos λ − i| sin λ|. So |z1 | = |z2 | = 1, and arg z1 + arg z2 = 2π. Denote arg z1 by θ. If λ 6= 0, we have by Residue Theorem · p log z1 ¸ Z ep log z2 e f (z)dz = 2πi[Res(f, z1 ) + Res(f, z2 )] = 2πi + z1 − z2 z2 − z1 γ1 +γR −γ2 −γr π 2πi = [epθi − ep(2π−θ)i ] = − sin(λp)epπi . | sin λ| sin λ If λ = 0, we have by Residue Theorem Z f (z)dz = 2πiRes(f, −1) = −2pπi · epπi . γ1 +γR −γ2 −γr

Meanwhile, we have the estimates ¯Z ¯ ¯ ¯ Rp ¯ ¯≤ f (z)dz ¯ ¯ R2 − 2R − 1 · R · 2π → 0 γR as R → ∞,

¯Z ¯ ¯ ¯

as r → 0, and

¯ ¯ rp · r · 2π → 0 f (z)dz ¯¯ ≤ 1 − 2ρr − r2 γr

Z

Z f (z)dz = − γ2

Since 1 − e

2pπi

0

R

(xe2πi )p dx =− x2 + 2ρx + 1

Z 0

R

xp dx e2pπi . x2 + 2ρx + 1

2

= 2 sin (pπ) − 2 sin(pπ) cos(pπ)i = −2i sin(pπ)epπi , by letting R → ∞ and r → 0, we have ( π sin(λp) Z ∞ if λ 6= 0 xp dx = sin λ sin(pπ) 2 x + 2x cos λ + 1 pπ csc (pπ) if λ = 0 and p 6= 0. 0

If we take the convention that single one: sinπ λsin(λp) sin(pπ) .

α sin α

= 1 for α = 0, then the above three formulas can be unified into a

Remark 17. We choose the above integration path in order to take advantage of the multi-valued function xp (Rule 1). The no symmetry in x2 + 2x cos λ + 1 is handled by using the full circle instead of half circle. (x)

45

Proof. Choose two positive numbers r and R such that 0 < r < 1 < R. Let γ1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ |z| ≤ R, arg z = 2π}, γR = {z : |z| = R, 0 < arg z < 2π} and γr = {z : |z| = r, 0 < arg z < 2π}. log z 1 z 1/p where z 1/p = e p is defined on C \ [0, ∞). Note by substituting y p for x, we get Define f (z) = p(z+1)z Z

∞ 0

dx = 1 + xp

Z 0

∞

1

y p dy . p(y + 1)y

By Residue Theorem, Z

1

f (z)dz = 2πRes(f, −1) = 2πi · γ1 +γR −γ2 −γr

where α =

π p.

2πi logpeπi (−1) p =− e = −2iαeαi , −p p

We have the estimates ¯Z ¯ ¯ ¯

as R → ∞,

¯ ¯ ¯¯Z 2π iθ 1 1 ¯ p log(Re ) ¯ ¯ e 2πR p ¯ iθ ¯ f (z)dz ¯ = ¯ Re · idθ¯ ≤ →0 ¯ 0 p(Reiθ + 1)Reiθ ¯ p(R − 1) γR

¯Z ¯ ¯ ¯

as r → 0, and

¯ ¯ ¯¯Z 2π iθ 1 1 ¯ p log(re ) ¯ ¯ e 2πr p ¯ iθ f (z)dz ¯¯ = ¯ ≤ →0 re · idθ ¯ ¯ 0 p(reiθ + 1)reiθ ¯ p(1 − r) γr Z

Z

R

f (z)dz = − γ2

r

1

(xe2πi ) p dx =− p(x + 1)x

Z

R r

1

x p dx · e2αi . p(x + 1)x

Therefore by letting r → 0 and R → ∞, we have Z 0

∞

dx = 1 + xp

Z 0

∞

1

x p dx −2iαeαi −2iαeαi α π = = = = csc π p(x + 1)x 1 − e2αi sin α p −2 sin αe( 2 +α)i

µ ¶ π . p

Remark 18. The roots of 1 + xp = 0 are not very expressible. So following Rule 1, we make the change of variable xp = y. Then we choose the above integration path to take advantage of the multi-valued function 1 y p . Since there is no symmetry in the denominator (y + 1)y, we used a full circle instead of a half circle. The change of variable xp = y is equivalent to integration on an arc of angle 2π/p instead of 2π. (xi) Proof. (Oops! looks like my solution has a bug. Catch it if you can.) By the change of variable x = have Z 1 1−p Z ∞ x (1 − x)p y p dy dx = . 1 + x2 (y + 1)3 + (y + 1) 0 0

1 y+1 ,

we

The equation (y + 1)3 + (y + 1) = 0 has three roots: z0 = −1, z1 = −1 + i, and z2 = −1 − i. Define p f (z) = (z+1)3z+(z+1) . Then 2 X

Res(f, zi ) =

i=0

¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 ¯ ¯ ¯ + + ¯ ¯ (z + 1)(z − z1 ) z=z2 (z + 1)(z − z2 ) z=z1 (z − z1 )(z − z2 ) ¯z=z0

1 1 1 + + −i · (−2i) i · 2i −i · i = 0.

=

46

Let R and r be two positive numbers with R > r. Define γ1 = {z : r < |z| ≤ R, arg z = 0}, γ2 = {z : r < |z| ≤ R, arg z = 2π}, γR = {z : |z| = R, 0 < arg z < 2π}, and γr = {z : |z| = r, 0 < arg z < 2π}. Then by Residue Theorem Z 2 X f (z)dz = 2πi Res(f, zi ) = 0. γ1 +γR −γ2 −γr

i=0

We note ¯Z ¯ ¯ ¯

ΓR

¯ ¯Z ¯ ¯ f (z)dz ¯¯ = ¯¯

2π

0

as R → ∞, and

¯Z ¯ ¯ ¯

2π

0

¯ ¯ 2πRp+1 (Reiθ )p iθ ¯≤ Re · idθ ¯ R3 − 3R2 − 4R − 2 → 0 (Reiθ + 1)3 + (Reiθ + 1)

¯ ¯ (reiθ )p 2πrp+1 iθ ¯≤ re · idθ ¯ 2 − 4r − 3r2 − r3 → 0 (reiθ + 1)3 + (reiθ + 1)

as r → 0. So by letting r → 0 and R → ∞, we have Z ∞ Z ∞ yp yp dy + 0 − dy · e2pπi − 0 = 0. 3 3 (y + 1) + (y + 1) (y + 1) + (y + 1) 0 0 Therefore..., “Houston, we got a problem here.” (xii) Proof. Choose r, R such that 0 < r < R and r is sufficiently small and R is sufficiently large. Let γ1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ |z| ≤ R, arg z = π}, γR = {z : |z| = R, 0 < arg z < π} and log z γr = {z : |z| = r, 0 < arg z < π}. Define f (z) = z2 +2z+2 . By Residue Theorem, µ ¶ Z log 2 3 f (z)dz = 2πi · Res(f, −1 + i) = π + πi . 2 4 γ1 +γR +γ2 −γr We note

¯Z ¯ ¯ ¯

¯ ¯Z ¯ ¯ f (z)dz ¯¯ = ¯¯

γR

as R → ∞,

2π

0

¯Z ¯ ¯ ¯

γr

¯ ¯Z ¯ ¯ f (z)dz ¯¯ = ¯¯

as r → 0, and

¯ ¯ log R + 2π log(Reiθ ) iθ Re · idθ¯¯ ≤ 2 2πR → 0 2 2iθ iθ R e + 2Re + 2 R − 2R − 2

2π 0

¯ ¯ log(reiθ ) iθ ¯ ≤ log r + 2π 2πr → 0 re · idθ ¯ 2 − 2r − r2 2 2iθ iθ r e + 2re + 2

Z

Z

R

f (z)dz = γ2

r

log(−x) dx = 2 x − 2x + 2

Z

R r

log x + πi dx. x2 − 2x + 2

Therefore by letting r → 0 and R → ∞, we have ¸ Z ∞· Z ∞ log x log x πdx π 3 + 2 dx + i = log 2 + π 2 i, 2 − 2x + 2 2 − 2x + 2 x x + 2x + 2 x 2 4 0 0 i R ∞ h log x log x dx = π2 log 2. i.e. 0 x2 −2x+2 + x2 +2x+2 i h R∞ R ∞ log y R∞ log x log x 4x dx = 0 log x (x2 +2) Meanwhile, we note 0 x2 −2x+2 − x2 +2x+2 dy. To calculate 2 −4x2 dx = 0 y 2 +4 R ∞ log y log z the value of 0 y2 +4 dy, we apply again Residue Theorem. Let’s define g(z) = z2 +4 and then we have Z πh π i g(z)dz = 2πi · Res(g, 2i) = log 2 + i . 2 2 γ1 +γR +γ2 −γr We note

¯Z ¯ ¯ ¯

γR

¯ ¯Z ¯ ¯ g(z)dz ¯¯ = ¯¯

2π 0

¯ ¯ log R + 2π log(Reiθ ) iθ ¯≤ Re · idθ 2πR → 0 ¯ 2 2iθ R e +4 R2 − 4 47

as R → ∞,

¯Z ¯ ¯ ¯

γr

as r → 0, and

¯ ¯Z ¯ ¯ g(z)dz ¯¯ = ¯¯

2π 0

Z

¯ ¯ log r + 2π log(reiθ ) iθ re · idθ¯¯ ≤ 2πr → 0 r2 e2iθ + 4 4 − r2

Z g(z)dz =

γ2

r

R

log(−x) dx = x2 + 4

Z

R

r

log x + πi dx. x2 + 4

So by letting r → 0 and R → ∞, we have Z ∞ Z ∞ log x dx π π2 2 dx + πi = log 2 + i. 2 2 x +4 x +4 2 4 0 0 R∞ x π Hence 0 xlog 2 +4 dx = 4 log 2. Solving the equation (R ∞ R0∞ 0

log x x2 −2x+2 dx log x x2 −2x+2 dx

we get

Z 0

∞

R∞ + 0 R∞ − 0

log x x2 +2x+2 dx log x x2 +2x+2 dx

= =

π 2 π 4

log 2 log 2,

π log x dx = log 2. x2 + 2x + 2 8

Remark 19. Note how we handled the combined difficulty caused by no symmetry in the denominator of the integrand function and log function: because x2 + 2x + 2 is not symmetric, we want to use full circle; but this will cause the integrals produced by different integration paths to cancel with each other. Therefore we are forced to choose half circle and use two equations to solve for the desired integral. (xiii) Proof. We choose r and R such that 0 < r < R, and r is sufficiently small and R is sufficiently large. Let γ1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ √|z| ≤ R, arg z = π}, γR = {z : |z| = R, 0 < arg z < π}, and z γr = {z : |z| = r, 0 < arg z < π}. Define f (z) = zz2log +1 . Then Z f (z)dz = 2πiRes(f, i) = γ1 +γR +γ2 −γr

We note

π2 π i π2 ie 4 = √ (−1 + i). 2 2 2

¯ ¯ ¯¯Z 2π ¯ 2πR√R(log R + 2π) iθ iθ 12 ¯ ¯ log(Re ) (Re ) ¯ iθ ¯ f (z)dz ¯ = ¯ Re · idθ¯ ≤ →0 2 e2iθ + 1 ¯ ¯ R R2 − 1 γR 0

¯Z ¯ ¯ ¯

as R → ∞,

¯ ¯ ¯¯Z 2π ¯ 2πr√r(log r + 2π) iθ iθ 12 ¯ ¯ log(re ) (re ) ¯ iθ f (z)dz ¯¯ = ¯ re · idθ →0 ¯≤ 2 e2iθ + 1 ¯ ¯ r 1 − r2 γr 0

¯Z ¯ ¯ ¯

as r → 0, and Z

Z f (z)dz =

γ2

r

R

√ Z R √ πi Z R√ Z R √ π π −x log(−x) xe 2 (log x + πi) x log x x 2i 2 i πi dx = dx = e dx + e dx. 2 2 2 2 x +1 x +1 x +1 r r r x +1

By letting r → 0 and R → ∞, we have Z ∞√ Z ∞ √ π2 x log x x (1 + i) dx − π dx = √ (−1 + i). 2 2 x +1 x +1 2 2 0 0 √ R∞ 2 x π √ Therefore, 0 xx2log +1 dx = 2 2 . 48

(xiv) P∞ n P∞ n Proof. We note the power series expansion ln(1 − z) = − n=1 zn and ln(1 + z) = n=1 (−1)n+1 zn hold in |z| < 1 and the convergence is uniform on |z| ≤ 1 − ε, for any ε ∈ (0, 1). Therefore, for any δ > 0, we have # # µ x ¶ Z ∞ Z ∞ "X Z ∞ "X ∞ ∞ ∞ ∞ −nx −nx −(2k−1)x X X e e e e−(2k−1)δ e +1 n+1 log dx = (−1) + dx = 2 dx = 2 . ex − 1 n n 2k − 1 (2k − 1)2 δ δ δ n=1 n=1 k=1

Therefore

Z

µ

∞

log 0

ex + 1 ex − 1

¶ dx = lim 2 δ→0

∞ X e−(2k−1)δ k=1

(2k −

1)2

=2

∞ X k=1

k=1

1 , (2k − 1)2

where the last equality can be seen as an example of Lebesgue’s Dominated Convergence Theorem applied P∞ 1 π2 to counting measure. By Problem 11 (ii) with z = 21 , we know k=1 (2k−1) 2 = 8 . So the integral is equal π2 4 .

to

(xv) Proof. Since we have argued rather rigorously in (xiv), we will argue non-rigorously in this problem. Z

∞

x dx = ex + 1

0

Z 0

∞

e−x x dx = 1 + e−x

Z

∞ 0

e−x x

∞ X

(−1)n e−nx dx =

n=0

∞ X

Z

∞

(−1)n

n=0

e−(n+1)x xdx =

0

π Using the function π2 sin πz and imitating what we did in Problem 12, we can easily get R∞ x 2 Therefore 0 ex +1 dx = π12 .

P∞ n=1

∞ X (−1)n . (n + 1)2 n=0 (−1)n−1 n2

=

π2 12 .

(xvi) Proof. We provide two solutions, which are essentially the same one. Solution 1. In problem (xvii), we shall prove (let a = 1) Z π x sin x dx = π log 2. 2 − 2 cos x 0 Note

Z

π 0

So

R

π 2

0

x sin x dx = 2 − 2 cos x

Z

π 0

x · 2 sin x2 cos x2 dx =2 2 · 2 sin2 x2

Z 0

π 2

θ cos θdθ = −2 sin θ

Z

π 2

log(sin θ)dθ.

0

log(sin θ)dθ = − π2 log.

Solution 2. We provide a heuristic proof which discloses the essence of the calculation. Note how Rule 1-3 lead us to this solution. Z π2 Z π2 Z π 1 2 log(sin θ)dθ = log(cos α)dα = log(cos α)dα 2 − π2 0 0 µ ¶ Z π 1 2 1 + cos 2α = log dα 4 − π2 2 Z 1 π π = log(1 + cos β)dβ − log 2. 8 −π 4 (Note how Rule 2 leads us to extended the integration interval from [0, π2 ] to [−π, π].) Motivated by the similar difficulty explained in the remark of Problem (xvii), we try Rule 1 and note Z π Z log(1 + z) dz = 2π log 1 = 0. log(1 + eiβ )dβ = iz −π |z|=1 49

For the above application of Cauchy’s integral formula to be rigorous, we need to take a branch of logarithm function that is defined on C \ (−∞, 0] and take a small bypass around 0 for the integration contour, and then take limit. Using the above result, we have Z Z Z π £ ¤ 1 π 1 π 2 2 iβ 0 = Re log(1 + e )dβ = log (1 + cos β) + sin β dβ = [log 2 + log(1 + cos β)] dβ. 2 −π 2 −π −π Rπ Therefore −π log(1 + cos β)dβ = −2π log 2 and Z π2 Z π 1 π π log(sin θ)dθ = log(1 + cos β)dβ − log 2 = − log 2. 8 −π 4 2 0 (xvii) Proof. We first assume a > 1. Then by integration-by-parts formula we have Z π Z x sin x 1 π x sin xdx dx = 2 1 − 2a cos x + a 2 1 − 2a cos x + a2 0 −π Z π 1 xd(1 − 2a cos x + a2 ) = 2 −π (1 − 2a cos x + a2 ) · 2a · ¸ Z π 1 2 2 = 2π log(1 + a) − log(1 − 2a cos x + a )dx . 4a −π Since a > 1, a − ζ ∈ {z : Rez > 0} for any ζ ∈ ∂D(0, 1). Therefore we can take a branch of the logarithm ¯ 1). For example, function such that log(a − z) is a holomorphic function on D(0, 1) and is continuous on D(0, iπ we can take the branch log z such that it is defined on C \ (−∞, 0] with log(e ) = π and log(e−iπ ) = −π. By Cauchy’s integral formula, Z π Z log(a − z) log(a − eiθ )dθ = dz = 2π log a. iz −π |z|=1 Rπ Rπ Therefore, −π log(1 − 2a cos x + a2 )dx = 2Re −π log(a − eiθ )dθ = 4π log a. Plug this equality into the ¡ ¢ calculation of the original integral, we get πa log 1+a . a If 0 < a < 1, we have µ ¶ Z π Z π 1 + a1 x sin x 1 x sin x 1 π π dx = 2 = log(a + 1). 2 1 dx = a2 · 1 log 1 2 1 − 2a cos x + a a a 0 0 1 − a cos x + a2 a a Assuming the integral as a function of a is continuous at 1, we can conclude for a = 1, the integral is equal to π log 2. The result agrees with that of Mathematica. To prove the continuity rigorously, we split Rπ R π x sin xdx π x sin xdx the integral into 02 1−2a cos x+a2 and π 1−2a cos x+a2 . For the second integral, we have (x ∈ [ 2 , π]) 2

x sin x x sin x x sin x = ≤ ≤ x sin x. π 2 ix 1 − 2a cos x + a |a − e | |a − ei 2 | So by Lebesgue’s dominated convergence theorem, the second integral is a continuous function of a for a ∈ (0, ∞). For the first integral, we note (1 − 2a cos x + a2 ) takes its minimum at cos x. So x x sin x ≤ . 1 − 2a cos x + a2 sin x Again by Lebesgue’s dominated convergence theorem, the first integral is a continuous function of a for a ∈ (0, ∞). Combined, we conclude the integral Z π x sin x dx 2 0 1 − 2a cos x + a as a function of a ∈ (0, ∞) is a continuous function. 50

−1

Remark 20. We did not take the transform cos x = z+z2 because log(1 − 2a cos x + a2 ) would then become −1 log[w(a) − w(z)] − log(2a), where w(z) = z+z2 maps D(0, 1) to U := C \ [−1, 1] and we cannot find a holomorphic branch of logarithm function on U . The application of integration-by-parts formula and the extension of integration interval from [0, π] to [−π, π] are motivated by Rule 2. But here a new trick emerged: if the integrand function or part of it could come from the real or imaginary part of a holomorphic function, apply Cauchy’s integral formula (or integration theorem) to that holomorphic function and try to relate the result to our original integral (Rule 1). (xviii) Proof. The function log(z + i) is well-defined on C+ := {z : Imz ≥ 0}. We take the branch of log(z + i) that evaluates to log 2 + π2 i at i. Define γR = {z : |z| = R, 0 ≤ arg z ≤ π}. Then for R > 1, Residue Theorem implies Z R Z log(z + i) log(z + i) i+i π2 = π log 2 + i. dz + dz = 2πi · 1 + z2 1 + z2 i+i 2 −R γR Note

as R → ∞. So

¯Z ¯ ¯ ¯

γR

R∞

−∞

¯ ¯Z ¯ ¯ πR[log(R + 1) + 2π] log(z + i) ¯¯ ¯¯ π log(Reiθ + 1) iθ ¯≤ dz = Re · idθ →0 ¯ ¯ ¯ 2 2 2iθ 1+z 1+R e R2 − 1 0 log(x+i) 1+x2 dx

Z

= π log 2 + ∞

π log 2 = Re −∞

π2 2 i.

Hence

log(x + i) dx = 1 + x2

Z

∞

−∞

√ Z ∞ log x2 + 1 log(1 + x2 ) dx = dx. 1 + x2 1 + x2 0

1 Remark 21. The main difficulty of this problem is that the poles of 1+x 2 coincide with the branch points 2 of log(x + 1), so that we cannot apply Residue Theorem directly. An attempt to avoid this difficulty 2 +1) might be writing log(x as the sum of log(x+i) and log(x−i) and integrate them separately along different 1+x2 1+x2 1+x2 paths which do not contain their respective branch points. But log(x + i) and log(x − i) cannot be defined simultaneously in a region√of a Riemann surface which contains both integration paths. But the observation that Re[log(x ± i)] = log x2 + 1 gives us a simple solution (Rule 1).

16. Proof. The power series if z 6= 1 and z = eiθ , N X n=1

n

z =

P∞ n=0

N X n=0

z n is divergent on every point of ∂D(0, 1): if z = 1,

cos(nθ) + i

N X

sin(nθ) =

sin (N +1)θ cos N2θ 2

n=0

sin θ2

+i

P∞ n=0

zn = 1 + 1 + 1 + · · · ;

sin (N +1)θ sin N2θ 2 sin θ2

.

P∞ Similarly, the series n=0 z −n−2 is divergent on every point of ∂D(0, 1). So the functions represented by these two power series cannot be analytic continuation of each other. 17. P∞ Proof. It’s clear we need the assumption that α 6= 0. The series n=0 (αz)n is convergent in U1 = {z : |z| < n P ∞ 1 1 n [(1−α)z] n=0 (−1) |α| }. The series 1−z (1−z)n is convergent in U2 = {z : |1 − α||z| < |1 − z|}. On U3 = U1 ∩ U2 , 1 both of the series represent the analytic function 1−αz . So the functions represented by these two series are analytic continuation of each other. 18.

51

Proof. On the line segment {z : 0 < Rez < 1, Imz = 0}, Taylor series of ln(1 + z) is z − 12 z 2 + 13 z 3 − · · · . 1 On the same line segment, 0 < 1−z 2 < 2 . So Taylor series of ln(1 − x) (|x| ≤ 1, x 6= 1) gives µ ¶ 1−z (1 − z)2 (1 − z)3 1−z ln 2 − − − − · · · = ln 2 + ln 1 − = ln(1 + z). 2 2 · 22 3 · 23 2 Since the two holomorphic functions represented by these two series agree on a line segment, they must agree on the intersection of their respective domains (Theorem 2.13). Therefore they are analytic continuations of each other. 19. Proof. Clearly the series P∞ is nconvergent in D(0, 1) and is divergent for z = 1. So its radius of convergent R = 1. Let f (z) = n=0 z 2 . Then f (z) is analytic in D(0, 1) and z = 1 is a singularity of f (z). We note P∞ P∞ P∞ n n−1 n−1 f (z) = z + n=1 z 2 = z + n=1 z 2·2 = z + n=1 (z 2 )2 = z + f (z 2 ). Therefore, we have f (z) = z + f (z 2 ) = z + z 2 + f (z 4 ) = z + z 2 + z 4 + f (z 8 ) = · · · . n

So the roots of equations z 2 = 1, z 4 = 1, z 8 = 1,· · · , z 2 = 1, · · · , etc. are all singularities of f . These roots form a dense subset of ∂D(0, 1), so f (z) can not be analytically continued to the outside of its circle of convergence D(0, 1).

4

Riemann Mapping Theorem

3. P∞ Proof. Let g(z) = n=0 c¯n z n . Then g(z) is holomorphic and has the same radius of convergence as f (z). ∀z∗ ∈ R ∩ convergence domain of g(z), we have g(z∗ ) =

∞ X

c¯n z∗n =

n=0

∞ X

n c¯n (z¯∗ ) = f¯(¯ z∗ ) = f¯(z∗ ) = f (z∗ ),

n=0

where the next to last equality is due to the fact that z∗ ∈ R and the last equality is due to the fact that f (R) ⊂ R. By Theorem 2.13, f ≡ g in the intersection of their respective domains of convergence. So cn = c¯n , i.e. cn (n = 0, 1, · · · ) are real numbers. 11. Proof. Let g(z) = f¯(¯ z ) (first take conjugate of z, then take conjugate of f (¯ z )). Then µ ¶ ∂g(z) ∂ f¯(¯ z) ∂f (¯ z) = = conjugate = 0. ∂ z¯ ∂ z¯ ∂z ¯) = So g is a holomorphic function. Clearly g is univalent. Since U is symmetric w.r.t. real axis, g(U ) = f¯(U f¯(U ) = conjugate(D(0, 1)) = D(0, 1). Furthermore, g(z0 ) = f¯(¯ z0 ) = f¯(z0 ) = 0, and f¯(¯ z ) − f¯(¯ z0 ) z→z0 z − z0 µ ¶ f (¯ z ) − f (¯ z0 ) = lim conjugate z→z0 z¯ − z¯0 = conjugate(f 0 (¯ z0 )) = conjugate(f 0 (z0 ))

g 0 (z0 ) =

=

lim

f 0 (z0 ) > 0.

By the uniqueness of Riemann Mapping Theorem, g ≡ f . 52

5

Differential Geometry and Picard’s Theorem

6

A First Taste of Function Theory of Several Complex Variables

There are no exercise problems for this chapter.

7

Elliptic Functions

There are no exercise problems for this chapter.

8

The Riemann ζ-Function and The Prime Number Theory

There are no exercise problems for this chapter.

References [1] Johann Boos. Classical and modern methods in summability, Oxford University Press, 2001. [2] John B. Conway. Functions of one complex variable, 2nd Edition, Springer, 1978. [3] Fang Qi-Qin. A course on complex analysis (in Chinese), Peking University Press, 1996. [4] James Munkres. Analysis on manifolds, Westview Press, 1997. [5] M. A. Qazi. The mean value theorem and analytic functions of a complex variable. J. Math. Anal. Appl. 324(2006) 30-38.

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