Solutions for Math 311 Assignment #8
(1) Let C denote the positively oriented boundary of the square
whose sides lie along the lines x = ±2 and y = ±2. Evaluate
each Zof these integrals
zdz
(a)
;
ZC 2z + 1
cosh z
(b)
dz;
4
ZC z
tan(z/2)
(c)
dz.
2
C (z − π/2)
Solution. (a)
Z
C
Z
zdz
zdz
1
=
2z + 1
2 C z − (−1/2)
2πi πi
=
z
=− .
2 z=−1/2
2
(b) For 0 < r < 1,
Z
2πi
cosh z
(3) dz
=
(cosh
z)
=0
z4
3!
C
z=0
(c) For 0 < r < 2 − π/2,
Z
tan(z/2)
2πi
0
dz
=
(tan(z/2))
= 2πi.
2
1!
C (z − π/2)
z=π/2
(2) Find the value of the integral g(z) around the circle |z − i| = 2
oriented counterclockwise when
1
(a) g(z) = 2
;
z +4
1
(b) g(z) = 2
.
(z + 4)2
Solution. (a)
Z
|z−i|=2
1
dz =
z2 + 4
Z
|z−i|=2
1
dz
(z + 2i)(z − 2i)
2πi π
=
=
z + 2i z=2i
2
1
2
(b)
Z
|z−i|=2
1
dz =
2
(z + 4)2
Z
1
(z +
− 2i)2
0 2πi
π
=
=
2
(z + 2i)
16
z=2i
|z−i|=2
2i)2 (z
dz
(3) Let C be a simple closed curve oriented counterclockwise and
let
Z 3
s + 2s
g(z) =
ds.
3
C (s − z)
Show that g(z) = 6πiz when z is inside C and that g(z) = 0 if
z is outside C.
Proof. When z is inside C,
Z 3
2πi 3
s + 2s
00 dz
=
(s
+
2s)
= 6πiz
3
2!
C (s − z)
s=z
When z is outside C, f (s) = (s3 + 2s)/(s − z)3 is analytic inside
C and hence
Z 3
s + 2s
dz = 0
3
C (s − z)
(4) Show that if f is analytic inside and on a simple closed curve
C and z0 is not on C, then
Z
Z 0
f (z)dz
f (z)dz
=
.
2
C (z − z0 )
C z − z0
Proof. When z0 is inside C,
Z
f 0 (z)
dz = 2πif 0 (z0 )
z
−
z
0
C
and
Z
C
2πi 0
f (z)
dz =
f (z0 ) = 2πif 0 (z0 )
2
(z − z0 )
1!
by Cauchy Integral Formula. When z0 is outside C, both
f 0 (z)
f (z)
and
z − z0
(z − z0 )2
3
are analytic inside C and hence
Z
f 0 (z)
dz = 0
C z − z0
and
Z
f (z)
dz = 0
2
C (z − z0 )
by Cauchy Integral Theorem. Therefore, we always have
Z
Z
f (z)
f 0 (z)
dz =
dz
2
C (z − z0 )
C z − z0
(5) Let C be the unit circle z = eiθ (−π ≤ θ ≤ π). First show that
for any real constant a,
Z az
e
dz = 2πi.
C z
Then write this integral in terms of θ to derive the integration
formula
Z π
ea cos θ cos(a sin θ)dθ = π.
0
Proof. By Cauchy Integral Formula,
Z az
e
az dz = 2πi(e )
= 2πi
C z
z=0
Z
C
On the other hand,
Z π
exp(aeiθ ) iθ
e
dz =
d(e )
z
eiθ
−π
Z π
=i
exp(aeiθ )dθ
Z−π
π
=i
exp(a cos θ + ia sin θ)dθ
−π
Z π
=i
ea cos θ (cos(a sin θ) + i sin(a sin θ))dθ
Z−ππ
Z π
a cos θ
=−
e
sin(a sin θ)dθ + i
ea cos θ cos(a sin θ)dθ
az
−π
−π
Therefore,
Z az Z π
e
a cos θ
e
cos(a sin θ)dθ = Im
dz = 2π
−π
C z
4
And since ea cos θ cos(a sin θ) is even,
Z π
Z
1 π a cos θ
a cos θ
e
cos(a sin θ)dθ =
e
cos(a sin θ)dθ = π
2 −π
0
(6) Suppose that f (z) is entire and Re(f (z)) ≤ M for all z and a
real constant M . Show that f (z) must be a constant.
Proof. Let g(z) = ef (z) . Since f (z) is entire, g(z) is entire. Since
Re(f (z)) ≤ M ,
|g(z)| = exp(Re(f (z))) ≤ eM
Therefore, |g(z)| is bounded. By Louville’s theorem, g(z) = c
is a constant. Therefore,
g(z) = c ⇒ g 0 (z) = 0 ⇒ f 0 (z)ef (z) = 0 ⇒ f 0 (z) = 0
and hence f (z) is a constant.
(7) Let f (z) be an analytic function in a domain D. Suppose that
f (z) 6= 0 for all z in D. Show that if |f (z)| has a minimum
value in D, then f (z) must be constant in D.
Proof. Let g(z) = 1/f (z). Since f (z) 6= 0 in D, g(z) is analytic
in D. If |f (z)| achieves the minimum at z0 in D, then |f (z)| ≥
|f (z0 )| for all z ∈ D. Hence |g(z)| ≤ |g(z0 )| for all z ∈ D.
That is, |g(z)| achieves the maximum at z0 . By the principle of
maximum modulus, g(z) is a constant in D and hence f (z) is
a constant in D.
(8) Let f (z) = u(x, y)+iv(x, y) be an analytic function in a domain
D, where u(x, y) = Re(f (z)) and v(x, y) = Im(f (z)). Show
that if v(x, y) has a minimum value in D, then f (z) must be a
constant in D.
Proof. Let g(z) = exp(−f (z)). Since f (z) is analytic in D, g(z)
is analytic in D. Suppose that u(x, y) achieves the minimum at
z0 = x0 + y0 i in D. Then u(x0 , y0 ) ≤ u(x, y) for all z = x + yi
in D. Then
|g(z0 )| = exp(−u(x0 , y0 )) ≥ exp(−u(x, y)) = |g(z)|
5
for all z ∈ D. That is, |g(z)| achieves the maximum at z0 . By
the principle of maximum modulus, g(z) = c is a constant in
D. Therefore,
g(z) = c ⇒ g 0 (z) = 0 ⇒ −f 0 (z)e−f (z) = 0 ⇒ f 0 (z) = 0
and hence f (z) is a constant in D.
(9) Let f (z) = ez and R be the rectangular region 0 ≤ x ≤ 1 and
0 ≤ y ≤ π. Find the points in R where v(x, y) = Im(f (z))
reaches its maximum and minimum values.
Solution. For a nonconstant analytic function f (z) in R, we
see that v(x, y) = Im(f (z)) takes the maximum and minimum
only on the boundary of R by applying the principle of maximum modulus to exp(if (z)) and exp(−if (z)). So to find the
maximum and minimum of
v(x, y) = Im(ez ) = ex sin y
in R = {0 ≤ x ≤ 1, 0 ≤ y ≤ π}, it suffices to find its maximum
and minimum on the boundary of R. We restrict v(x, y) to the
boundary of R:
• restricting v(x, y) to x = 0 and 0 ≤ y ≤ π, we have
v(0, y) = sin y and 0 ≤ v(0, y) ≤ 1;
• restricting v(x, y) to x = 1 and 0 ≤ y ≤ π, we have
v(1, y) = e sin y and 0 ≤ v(1, y) ≤ e;
• restricting v(x, y) to 0 ≤ x ≤ 1 and y = 0, we have
v(x, 0) = 0;
• restricting v(x, y) to 0 ≤ x ≤ 1 and y = π, we have
v(x, π) = 0.
In conclusion, vmax = e when z = 1 + πi/2 and vmin = 0 when
Im(z) = 0 or Im(z) = π.
(10) Use the triangle inequality |zn | − |z| ≤ |zn − z| to show that
if limn→∞ zn = z, then limn→∞ |zn | = |z|.
Proof. Since limn→∞ zn = z, limn→∞ |zn − z| = 0. And since
0 ≤ |zn | − |z| ≤ |zn − z|
we conclude
lim |zn | − |z| = 0
n→∞
by comparison theorem. Therefore, limn→∞ |zn | = |z|.
6
(11) Show that if
∞
X
zn = S
n=1
then
∞
X
zn = S.
n=1
Proof. Since
P∞
n=1 zn
= S,
n
X
lim zk − S = 0
n→∞ k=1
Therefore,
n
n
X
X
zk − S = 0
lim zk − S = lim n→∞ n→∞ k=1
and hence
k=1
∞
X
zn = S
n=1