Some Graded Homework 7 Problems
Chapter 14, Section 3, Problem 22.
Let
- :=
u
sin(2z) dz
C (6z ⇡)3 ,
where C is the (positively-oriented) circle {z 2 C : |z| = 3}. Using Problem
u sin(2z) dz
2!
#21 with n = 2, a = ⇡6 , and f (z) = sin(2z), we have f 00 ( ⇡6 ) = 2⇡i
C (z ⇡ )3 .
00
Since f (z) =
4 sin (2z) and sin
u sin(2z) dz
C (z ⇡ )3 =
6
So,
- =
1
63
R
⇡
3
p
6
3
2 , we
p
4 23 )
=
2⇡i
2! (
sin(2z) dz
C (z ⇡6 )3
must have
p
= 2 3⇡i.
p
3⇡i
108 .
=
3.
Let
- :=
R
z 2 ez
C 2z+i
dz,
where C is the negatively-oriented unit circle in C. Using Cauchy’s
u z 2 ez integral
i
i
1
2 z
formula with f (z) = z e and a = 2 , we have f ( 2 ) = 2⇡i C z+ i dz. Since
f(
i
2)
=
1
4e
i
2
2
, we must have
u
z 2 ez
C z+ 2i
e
4
dz = 2⇡i(
i
2
⇡
i/2
.
2 ie
) =
Remembering the orientation of C, we get
u z 2 ez
- = 21 C z+
i dz =
2
⇡
i/2
.
4 ie
Final answer:
- =
⇡ (⇡ 1)i/2
4e
=
⇡
4
1
z(z 1)(z 2)
=
sin 12 + ⇡4 cos 12 i.
Chapter 14, Section 4, Problem 3.
Let
f (z) =
1
1
z
1
z 2
1
z 1
.
If 0 < |z| < 1, then
⇣ 1
⌘
1
1
2
f (z) = z 1 z + 1 z
✓ 2
◆
1
1
1
= z1
z +
21
1 z
⇣ h 2
i ⇥
⇤⌘
2
3
1
1
z
z 2
z 3
= z
+ 2 + ··· + 1 + z + z + z + ···
2 1+ 2 + 2
⇣ h
i ⇥
⇤⌘
2
3
1
1
z
z2
z3
= z
2 1 + 2 + 4 + 8 + ··· + 1 + z + z + z + ···
⇣h
i ⇥
⇤⌘
2
3
1
1
z
z2
z3
= z
2
4
8
16 + · · · + 1 + z + z + z + · · ·
=
1
2z
+ 34 +
7z
8
+
15z 2
16
+ ··· .
So the residue of f at the origin is 1/2.
If 1 < |z| < 2, then
⇣ 1
⌘
1
1
2
z
f (z) = z 1 z 1 1
2
= ···
If |z| > 2, then
f (z) =
z
1
z3
1
z
⇣
1
z2
1
z
1
= ··· +
2
z
7
z5
1
2z
1
z
1
+
1
z
3
z4
1
4
⌘
+
z2
16
z
8
1
z2
=
1
z3 .
⇣
1
1
2
z
z3
32
1
1
··· .
1
z
⌘
Chapter 14, Section 4, Problem 10d.
When z is near ⇡,
cos
⇡
z ⇡
=
1
X
⇡
(
2n+1
( ⇡)
1)n z(2n+1)!
=
n=0
1
X
(
n=0
⇡
z ⇡
So, ⇡ is an essential singularity of cos
2
.
1)n ⇡ 2n+1
(2n+1)! (z
⇡)
(2n+1)
.