SOLUTIONS OF HW7
MINGFENG ZHAO
April 03, 2011
1. [Problem 11, in Page 128] Evaluate
R
γ
zdz, where γ is a rectifiable closed curve.
Proof. Let γ : [0, 1] → C be the rectifiable closed curve, so γ(0) = γ(1). In particular, γ is continuous
R
on [0, 1]. Let w(z) = z, then w is continuous on C, which implies that
uniformly continuous on γ. Assume
R
γ
w(z)dz exists, and w(z) is
w(z)dz = A ∈ C.
For any > 0, we take n large enough such that
1
n
< δ. Consider the partition
P=
γ
0 = t0 < t1 < · · · < tn = 1 : tk =
k
n
.
Hence, we have
n
X
w(γ(tk ))(γ(tk ) − γ(tk−1 )) − A
<
w(γ(tk−1 ))(γ(tk ) − γ(tk−1 )) − A
<
.
k=1
n
X
k=1
And
|w(γ(tk )) − w(γ(tk−1 ))| < ,
for all k = 1, · · · , n.
So we get
|2A| =
n
X
w(γ(tk ))(γ(tk ) − γ(tk−1 )) − A +
k=1
−
n
X
w(γ(tk−1 ))(γ(tk ) − γ(tk−1 )) − A
k=1
n
X
w(γ(tk ))(γ(tk ) − γ(tk−1 )) −
k=1
≤
n
X
n
X
w(γ(tk−1 ))(γ(tk ) − γ(tk−1 ))
k=1
w(γ(tk ))(γ(tk−1 ) − γ(tk−1 )) − A +
k=1
n
X
k=1
1
w(γ(tk ))(γ(tk ) − γ(tk−1 )) − A
2
MINGFENG ZHAO
+
n
X
w(γ(tk ))(γ(tk−1 ) − γ(tk−1 )) +
k=1
<
++
n
X
w(γ(tk−1 ))(γ(tk ) − γ(tk−1 ))
k=1
n
X
(w(γ(tk )) + w(γ(tk−1 )))(γ(tk−1 ) − γ(tk−1 ))
k=1
=
2 +
n
X
(γ(tk ) + γ(tk−1 ))(γ(tk−1 ) − γ(tk−1 ))
k=1
=
2 +
n
X
((γ(tk ))2 − (γ(tk−1 ))2 )
k=1
=
2 + |(γ(tn ))2 − (γ(t0 ))2 |
=
2 + 0
=
2
Take → 0, we get |A| = 0, that is,
R
γ
zdz = 0.
2. [Problem 12, in Page 128] Evaluate the integral
R
|z − 1||dz| along the circle |z| = 1 described in
the positive direction.
Proof. Let z(t) = eit , 0 ≤ t ≤ 2π, then we have deit = ieit dt, and
Z
Z
2π
eit − 1 |ieit dt|
|z − 1||dz| =
|z|=1
0
Z
2π
eit − 1 dt
=
0
Z
2π
=
2 sin
0
Z
=
0
=
=
2π
t
dt
2
t
2 sin dt
2
−4 cos
t
2
2π
0
8.
SOLUTIONS OF HW7
3
3. [Problem 6, in Page 127] Prove the theorem on the term-by-term integration of uniformly convergent
series (Section 8.7).
Proof. Let sn (z) =
Pn
k=1
wk (z), and s(z) =
P∞
k=1
wk (z). And for all k ≥ 1, wk (z) is continuous on
the rectifiable curve γ, and sn (z) converges uniformly to s(z) on γ. Then we know sn (z) is continuous
on γ, and s(z) is continuous on γ, in particular,
Z
sn (z)dz =
γ
Z X
n
R
γ
s(z)dz exists. And also we know
n Z
X
wk (z)dz =
γ k=1
k=1
wk (z)dz.
γ
For any > 0, there exists N > 0 such that for all n ≥ N , we have
|sn (z) − s(z)| < ,
for all z ∈ γ.
So we get
Z
Z
Z
s(z)dz −
(s(z) − sn (z))dz
=
sn (z)dz
γ
γ
γ
Z
≤
|s(z) − sn (z)||dz|
γ
Z
≤ |dz|
γ
≤ L(γ).
So as n → ∞, we have
Z
n Z
X
Z
s(z)dz = lim
γ
n→∞
sn (z) = lim
n→∞
γ
k=1
wk (z)dz =
γ
∞ Z
X
k=1
wk (z)dz.
γ
Therefore, we get
∞ Z
X
k=1
Z
wk (z)dz
γ
=
γ
∞
X
wk (z)dz
k=1
4. [Problem 13, in Page 128] Evaluate the integral
Z
ez
dz
1 + z2
4
MINGFENG ZHAO
along the curve |z − 1| = 1 described in the positive direction.
Proof. Since w(z) =
ez
1+z 2
is analytic in the disc {z : |z − 1| ≤ 1}, and the disc {z : |z − 1| ≤ 1} is
convex, then by Cauchy’s theorem, we know w(z) is integrable, that is, there is a primitive function
v(z) such that v 0 (z) = w(z) on the disc {z : |z − 1| ≤ 1}.
Figure 1. |z − 1| = 1
Because the circle |z − 1| = 1 is smooth, then we get
Z
ez
dz = 0.
1 + z2
5. [Cartan, Problem 5, in Page 76] Sow that if f (z) is continuous in the closed disc |z| ≤ r, and
holomorphic in the open disc |z| < r, then
f (z) =
1
2πi
Z
|t|=r
f (t)
dt
t−z
where the integral is taken in the positive sense.
for all |z| < r.
SOLUTIONS OF HW7
5
Remark 1. Let D be a bounded smooth domain in C, and f (z) = u(x, y) + iv(x, y) is analytic in D,
and continuous on D, then we have
Z
Z
f (z)dz
=
∂D
(u(x, y) + iv(x, y))d(x + iy)
∂D
Z
Z
u(x, y)dx − v(x, y)dy + i
=
∂D
v(x, y)dx + u(x, y)dy
∂D
Z
=
Z
(uy (x, y) + vx (x, y))dxdy + i
D
Z
=
By Green’s formula
Z
0dxdy +
D
=
(vy (x, y) + ux (x, y))dxdy
D
0dxdy
By Cauchy-Riemann equation, ux = vy , uy = −vx
D
0.
Proof. For any z0 ∈ {z : |z| < r}. Since {z : |z| < r} is open, then there exists a r0 > 0 such that
{z : |z − z0 | < r} ⊂ {z : |z| < r}. For any 0 < < r0 , we know
1
2πi
Z
|t−z0 |=
1
dt = 1.
t − z0
Figure 2. |z − z0 | = Since f (z) is analytic in |z| < r, then
f (t)
t−z0
is analytic in {t : |t| < r, |t − z0 | > }, and
continuous on {t : |t| ≤ r, |t − z0 | ≥ }, then by Green’s formula, we can get
1
2πi
Z
|t|=r
f (t)
1
dt −
t − z0
2πi
Z
|t−z0 |=
f (t)
dt = 0.
t−z
f (t)
t−z0
is
6
MINGFENG ZHAO
On the other hand, we know
1
2πi
Z
|t−z0 |=
f (t)
dt − f (z0 )
t − z0
1
2πi
=
=
≤
Z
|t−z0 |=
f (t)
1
dt − f (z0 ) ·
t − z0
2πi
Z
|t−z0 |=
1
dt
t − z0
Z
1
f (t) − f (z0 )
dt
2πi |t−z0 |=
t − z0
Z
1
f (t) − f (z0 )
dt
2π |t−z0 |=
t − z0
Since f (t) is analytic at z0 , so when is very small,there exists some M > 0 such that
f (t) − f (z0 )
≤M
t − z0
So we get
1
2πi
Z
f (t)
dt − f (z0 )
t − z0
|z−z0 |=
≤
1
M
2π
Z
dt
|t−z0 |=
= M .
Therefore, we obtain
1
2πi
Z
|t|=r
f (t)
dt − f (z0 ) ≤ M .
t − z0
Take → 0, we get
1
2πi
Z
|t|=r
f (t)
dt = f (z0 ).
t − z0
6. [Cartan, Problem 8, in Page 77] Let f (z) = u(x, y) + iv(x, y) be a holomorphic function in a
connected open set D. If
au(x, y) + bv(x, y) = c in D,
where a, b and c are real constants which are not all zero, then f (z) is constant in D.
SOLUTIONS OF HW7
7
Proof. Since a, b and c are real constants which are not all zero, then we must have a2 + b2 > 0. Since
au(x, y) + bv(x, y) = c, and f is analytic, then we can get
aux (x, y) + bvx (x, y)
=
0
auy (x, y) + bvy (x, y)
=
0.
By the Cauchy-Riemann equation, we know ux = vy , uy = −vx . So we can get
aux (x, y) − buy (x, y)
=
0
auy (x, y) + bux (x, y)
=
0.
Since a2 + b2 > 0, then ux (x, y) = uy (x, y) = 0, which implies that vx (x, y) = vy (x, y) = 0. Since
the domain D is connected, then u(x, y) = A, v(x, y) = B for all (x, y) ∈ D, where A, B are real
constants. Therefore, f (z) is constant in D.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu