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复分析阿尔福斯4-7章练习答案.pdf
Selected Solutions to Complex Analysis by Lars Ahlfors
Matt Rosenzweig
1
Contents
Chapter 4 - Complex Integration
Cauchy’s Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.2 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.3 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.3 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Local Properties of Analytical Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.2 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.2 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Calculus of Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.2 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.3 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.3 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.2 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.2 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.4 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.4 Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.4 Exercise 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.5 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.5 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Harmonic Functions
Chapter 5 - Series and Product Developments
Power Series Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Partial Fractions and Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.2 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.3 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.3 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.4 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.4 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.5 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.5 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.2 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.2 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.5 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.5 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.5 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Entire Functions
Normal Families
4
4
4
4
5
6
6
6
7
7
7
8
10
10
10
11
12
13
13
13
14
14
14
15
16
16
16
17
18
18
19
19
20
20
21
22
22
23
23
24
24
25
2
Conformal Mapping, Dirichlet’s Problem
The Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.1 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.1 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Elliptic Functions
Weierstrass Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.2 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.3 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.3 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.3 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.3 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.3 Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.3 Exercise 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.5 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
27
27
27
28
28
29
29
30
30
31
31
32
32
3
Chapter 4 - Complex Integration
Cauchy’s Integral Formula
4.2.2 Exercise 1
Applying the Cauchy integral formula to f (z) = ez,
|z|=1
ez
z
dz
dz ⇐⇒ 2πi =
1 = f (0) =
1
2πi
f (z)
|z|=1
z
Section 4.2.2 Exercise 2
Using partial fractions, we may express the integrand as
1
z2 + 1
=
i
2(z + i)
−
i
2(z − i)
Applying the Cauchy integral formula to the constant function f (z) = 1,
1
1
1
2πi
1
dz =
i
2
|z|=2
z2 + 1
2πi
|z|=2
z + i
2πi
|z|=2
1
dz − i
2
1
z − i
dz = 0
4.2.3 Exercise 1
1. Applying Cauchy’s differentiation formula to f (z) = ez,
1 = f (n−1)(0) =
(n − 1)!
2πi
|z|=1
ez
zn dz ⇐⇒ 2πi
(n − 1)!
=
|z|=1
ez
zn dz
2. We consider the following cases:
(a) If n ≥ 0, m ≥ 0, then it is obvious from the analyticity of zn(1 − z)m and Cauchy’s theorem that
the integral is 0.
(b) If n ≥ 0, m < 0, then by the Cauchy differentiation formula,
|z|=2
zn(1−z)mdz = (−1)m
0
(−1)m2πi
(|m|−1)!
(c) If n < 0, m ≥ 0, then by a completely analogous argument,
(z − 1)|m| dz =
|z|=2
zn
zn(1−z)mdz =
(1 − z)m
z|n|
dz =
|z|=2
0
(−1)|n|−12πi
(|n|−1)!
n ≥ |m|
n < |m| − 1
n!
(n−|m|+1)! = (−1)|m|2πi n|m|−1
(m−|n|+1)! = (−1)|n|−12πi m|n|−1
m!
m < |n| − 1
m ≥ n
(d) If n < 0, m < 0, then sincen(|z| = 2, 0) = n(|z| = 2, 1) = 1, we have by the residue formula that
|z|=2
|z|=2
(1 − z)mzn = 2πires(f ; 0) + 2πires(f ; 1) =
4
(1 − z)mzndz +
(1 − z)mzndz
|z|= 1
2
|z−1|= 1
2
Using Cauchy’s differentiation formula, we obtain
(1 − z)−|m|
z−|n|
(1 − z)mzndz =
2
=
2πi
dz +
|z|=2
z|n|
|z|= 1
· (|m| + |n| − 2)!
|m| + |n| − 2
(|m| − 1)!
3. If ρ = 0, then it is trivial that
1
|z|=ρ |z − a|−4 |dz| = 0, so assume otherwise. If a = 0, then
|m| + |n| − 2
(1 − z)|m| dz
· (−1)|m|−1(|n| + |m| − 2)!
(−1)|m|2πi
(|m| − 1)!
(|n| − 1)!
(|n| − 1)!
|n| − 1
|n| − 1
|z−1|= 1
= 2πi
= 0
−
+
2
|z|−4 |dz| =
ρ−42πiρdt =
2πi
ρ3
0
|z|=ρ
Now, assume that a = 0. Observe that
|z − a|−4 |dz| =
|z|=ρ
|z|=ρ
1
1
(z − a)2(z − a)2 |dz| =
(ρe2πit − a)2(ρe−2πit − a)2 ρ
2πie4πit
ie4πit dt
1
|z − a|4 =
(z − a)2(z − a)
2
1
1
0
−i
ρ
1
−iρ
a2
0
ρ2πie4πit
(ρe2πit − a)2(ρ − ae2πit)2 dt =
= −i
dz
a )2(z − a)2
We consider two cases. First, suppose |a| > ρ. Then z(z − a)−2 is holomorphic on and inside {|z| = ρ}
and ρ2
a lies inside {|z| = ρ}. By Cauchy’s differentiation formula,
ρ z)2(z − a)2
(z − ρ2
(ρ − a
(z − a)−2 − 2z(z − a)−3
|z − a|−4 |dz| = 2πi
1 − 2
|z|=ρ
|z|=ρ
dz =
2πρ
=
z
z
z= ρ2
a
a2( ρ2
a − a)2
ρ2
a − a)
a( ρ2
−iρ
a2
−2πρ(ρ2 + |a|2)
(ρ2 − |a|2)3
=
=
2πρ(ρ2 + |a|2)
(|a|2 − ρ2)3
|z|=ρ
|z|=ρ
Now, suppose |a| < ρ. Then ρ2
inside {|z| = ρ}. By Cauchy’s differentiation formula,
a lies outside |z| = ρ, so the function z(z − ρ2
|z − a|−4 |dz| = 2πi
−iρ
a2
−2πρ
=
(|a|2 − ρ2)2
a )−2 is holomorphic on and
1 − 2
a
(a − ρ2
a )
(z − ρ2
a
)−2 − 2z(z − ρ2
a
)−3
z=a
(a + ρ2
a )
a − ρ2
a
=
−2πρ(|a|2 + ρ2)
(|a|2 − ρ2)3
=
=
2πρ
a2(a − ρ2
a )2
2πρ(|a|2 + ρ2)
(ρ2 − |a|2)3
4.2.3 Exercise 2
Let f : C → C be a holomorphic function satisfying the following condition: there exists R > 0 and n ∈ N
such that |f (z)| < |z|n ∀|z| ≥ R. For every r ≥ R, we have by the Cauchy differentiation formula that for
all m > n,
f (m)(a)
≤ m!
2π
|z|n
|z|m+1 |dz| ≤ m!
rm−n
|z|=r
Noting that m − n ≥ 1 and letting r → ∞, we have that f (m)(a) = 0. Since f is entire, for every a ∈ C, we
may write
f (z) = f (a) + f(a)(z − a) + ··· +
(z − a)n + fn+1(z)(z − a)n+1 ∀z ∈ C
f (n)(a)
n!
where fn+1 is entire. Since fn+1(a) = f (n+1)(a) = 0 and a ∈ C was arbitary, we have that fn+1 ≡ 0 on C.
Hence, f is a polynomial of degree at most n.
5
+
cn−2
h(0)
1!
g(z) − cn−1
z→∞ g(z) − cn−1
lim
z→0
zn−1 +
Local Properties of Analytical Functions
4.3.2 Exercise 2
Let f : C → C be an entire function with a nonessential singularity at ∞. Consider the function g(z) = f 1
at z = 0. Let n ∈ N be minimal such that limz→0 zng(z) = 0. Then the function zn−1g(z) has an analytic
continuation h(z) defined on all of C. By Taylor’s theorem, we may express h(z) as
z
zn−1g(z) = h(z) = h(0)
cn−1
h(0)
2!
z2 + ··· +
h(n−1)(0)
(n − 1)!
z +
c0
zn−1 + hn(z)zn ∀z = 0
where hn : C → C is holomorphic. Hence,
= lim
z→0
zhn(z) = 0
cn−2
zn−2 + ··· + c0
cn−2
zn−2 + ··· + c0
And
since f is entire. Note that we also obtain that c0 = f (0). Hence, g(z) − cn−1
zn−1 +
f (z) = f (0)
= lim
z→0
lim
are abusing notation to denote the continuation to all of C) is a bounded entire function and is therefore
identically zero by Liouville’s theorem. Hence,
zn−1 + cn−2
zn−2 + ··· + c0
(we
∀z = 0, f (z) = cn−1zn−1 + cn−2zn−2 + ··· + c0
Since f (0) = c0, we obtain that f is a polynomial.
k=1 of C. By definition, the function ˜f (z) = f 1
has either a removable singularity or a pole at z = 0.
4.3.2 Exercise 4
Let f : C ∪ {∞} → C ∪ {∞} be a meromorphic function in the extended complex plane. First, I claim that
f has finitely many poles. Since the poles of f are isolated points, they form an at most countable subset
{pk}∞
In either case, there exists r > 0 such that ˜f is holomorphic on D(0; r). Hence, {pk}∞
k=1 ⊂ D(0; r). Since
this set is bounded, {pk}∞
k=1 has a limit point p. By continuity, f (p) = ∞ and therefore p is a pole. Since p
is an isolated point, there must exist N ∈ N such that ∀k ≥ N, pk = p.
Our reasoning in the preceding Exercise 2 shows that for any pole pk = ∞ of order mk, we can write in a
neighborhood of pk
(z − pk)mk−1 + ··· +
(z − pk)mk
c1
z − pk
cmk−1
+gk(z)
f (z) =
+ c0
cmk
+
z
and
cm∞
fk(z)
where gk is holomorphic in a neighborhood of pk. If p = ∞ is a pole, then analogously,
c1
z
+ c0
˜f∞(z)
˜f (z) =
+˜g∞(z)
zm∞ +
where ˜g∞ is holomorphic in a neighborhood of 0. For clarification, the coefficients cn depend on the pole,
cm∞−1
zm∞−1 + ··· +
but we omit the dependence for convenience. Set f∞(z) = ˜f∞ 1
h(z) = f (z) − f∞(z) − n
i=k fk(z) and in a neighborhood of z∞ as h(z) = g∞(z) −n
− f∞ 1
h can be written as h(z) = gk(z) −
which are sums of holomorphic functions. ˜h(z) = h 1
are polynomials and f 1
is evidently bounded in a neighborhood of 0 since
= ˜g∞(z), which is holomorphic in a neighborhood of 0. By
I claim that h is (or rather, extends to) an entire, bounded function. Indeed, in a neighborhood of each zk,
k=1 fk(z),
the fk
Liouville’s theorem, h is a constant. It is immediate from the definition of h that f is a rational function.
1
fk(z)
k=1
z
z
z
z
z
6
Calculus of Residues
4.5.2 Exercise 1
Set f (z) = 6z3 and g(z) = z7 − 2z5 − z + 1. Clearly, f, g are entire, |f (z)| > |g(z)| ∀|z| = 1, and
f (z) + g(z) = z7 − 2z5 + 6z3 − z + 1. By Rouch´e’s theorem, f and f + g have the same number of zeros,
which is 3 (counted with order), in the disk {|z| < 1}.
Section 4.5.2 Exercise 2
Set f (z) = z4 and g(z) = −6z + 3. Clearly, f, g are entire, |f (z)| > |g(z)| ∀|z| = 2. By Rouch´e’s theorem,
z4−6z +3 has 4 roots (counted with order) in the open disk {|z| < 2}. Now set f (z) = −6z and g(z) = z4 +3.
Clearly, |f (z)| > |g(z)| ∀|z| = 1. By Rouch´e’s theorem, z4 − 6z + 3 = 0 has 1 root in the in the open disk
{|z| < 1}. Observe that if z ∈ {1 ≤ |z| < 2} is root, then by the reverse triangle inequality,
3 = |z|z3 − 6 ≥ |z||z|3 − 6
So |z| ∈ (1, 2). Hence, the equation z4 − 6z + 3 = 0 has 3 roots (counted with order) with modulus strictly
between 1 and 2.
4.5.3 Exercise 1
1. Set f (z) =
1
z2+5z+6 =
2. Set f (z) =
1
(z2−1)2 =
(z+3)(z+2) . Then f has poles z1 = −2, z2 = −3 and by Cauchy integral formula,
1
res(f ; z1) =
res(f ; z2) =
1
2πi
1
2πi
|z+2|= 1
2
|z+3|= 1
2
(z + 3)−1
(z + 2)
(z + 2)−1
(z + 3)
dz =
1
z + 3
|z=−2 = 1
dz =
1
z + 2
|z=−3 − 1
(z−1)2(z+1)2 . Then f has poles z1 = −1, z2 = −1. Applying Cauchy’s differentia-
1
tion formula, we obtain
res(f ; z1) =
res(f ; z2) =
1
2πi
1
2πi
(z − 1)−2
(z + 1)2 dz = −2(z − 1)−3|z=−1 =
(z + 1)−2
(z − 1)2 dz = −2(z + 1)−3|z=1 = − 1
1
4
4
|z+1|=1
|z−1|=1
3. sin(z) has zeros at kπ, k ∈ Z, hence sin(z)−1 has poles at zk = kπ. We can write sin(z) = (z −
zk) [cos(zk) + gk(z)], where gk is holomorphic and gk(zk) = 0. By the Cauchy integral formula,
res(f ; zk) =
1
2πi
|z−zk|=1
[f(zk) + gk(z)]
−1
(z − zk)
dz =
1
f(zk) + g(zk)
= (−1)k
4. Set f (z) = cot(z). Since sin(z) has zeros at zk = kπ, k ∈ Z and cos(zk) = 0, cot(z) has poles at
zk, k ∈ Z. We can write sin(z) = (z − zk) [cos(zk) + gk(z)], where gk is holomorphic and gk(zk) = 0.
By Cauchy’s integral formula,
res(f ; zk) =
1
2πi
|z−zk|=1
cos(z) [cos(zk) + gk(z)]
(z − zk)
−1
dz =
cos(zk)
cos(zk) + gk(zk)
= 1
5. It follows from (3) that f (z) = sin(z)−2 has poles at zk = kπ, k ∈ Z. We remark further that
gk(z) = − cos(zk)(z − zk)2 + hk(z), where hk(z) is holomorphic. By the Cauchy differentiation formula,
res(f ; zk) =
1
2πi
|z−zk|=1
−2
[cos(zk) + gk(z)]
(z − zk)2
dz = −2
g
k(zk)
(cos(zk) + gk(zk))3 = 0
7
6. Evidently, the poles of f (z) =
zm(1−z)n are z1 = 0, z2 = 1. By Cauchy’s differentiation formula,
1
1
2πi
res(f ; z1) =
(1 − z)−n
dz =
(n + m − 2)!
(n − 1)!(m − 1)!
=
|z|= 1
2
zm
res(f ; z2) =
(−1)n
2πi
z−m
(z − 1)n dz =
|z−1|= 1
2
(−1)n(−1)n−1(m + n − 2)!
(m − 1)!
n + m − 2
n + m − 2
m − 1
= −
n − 1
4.5.3 Exercise 3
(a) Since a + sin2(θ) = a + 1−cos(2θ)
= 2 [(2a + 1) − cos(2θ)], we have
π
2
2
π
2
dθ
dθ
0
0
= 2
(2a + 1) − cos(2θ)
a + sin2(θ)
π
the integrand. Ahlfors p. 155 computes π
=
0
π
0
dτ
=
dt
(2a + 1) − cos(t)
=
0
dτ
−π
(2a + 1) + cos(τ )
where we make the change of variable τ = θ − π, and the last equality follows from the symmetry of
(2a + 1) + cos(τ )
π
2
dx
α+cos(x) = π√
α2−1
0
for α > 1. Hence,
dθ
a + sin2(θ)
0
(2a + 1)2 − 1
π
=
(b) Set
f (z) =
For R >> 0,
z4 + 5z2 + 6
z2
z2
=
(z2 + 3)(z2 + 2)
3i)(z − √
γ1 : [−R, R] → C, γ1(t) = t; γ2 : [0, π] → C, γ2(t) = Reit
(z − √
3i)(z +
√
=
z2
√
2i)
2i)(z +
and let γ be the positively oriented closed curve formed by γ1, γ2. By the residue formula and applying
the Cauchy integral formula to eiz
z+ai to compute res(f ; ai),
γ
f (z)dz = 2πires(f ;
3i) + 2πires(f ;
√
√
2i)
It is immediate from Cauchy’s integral formula that
√
2πires(f ;
3i) =
√
2i) =
2πires(f ;
3)−1(z2 + 2)−1
√
z2(z + i
√
(z − i
√
2)−1(z2 + 3)−1
√
(z − i
3)
2)
z2(z + i
|z−i
√
3|=
|z−i
√
2|=
√
(i
dz = 2πi ·
√
((i
3)2
√
3)2 + 2)(2i
√
3π
=
3)
dz = 2πi ·
√
((i
√
(i
2)2
2)2 + 3)(2i
√
= −
2π
√
2)
Using the reverse triangle inequality, we obtain the estimate
≤
∞
γ2
f (z)dz
x2
Hence,
∞
2
x2
=
x4 + 5x2 + 6
0
−∞
x4 + 5x2 + 6
πR3
|R2 − 3||R2 − 2| → 0, R → ∞
√
dx = (
3 −
√
2)π ⇒
∞
x2dx
x4 + 5x2 + 6
0
√
3 − √
(
2)π
2
=
8
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