270
Instructor’s Manual
10.
n
0
(
z
)
2
n
4
(
z
)
2
• • •
,
0
z
R
12.
3
i
(
z
i
)
14.
n
0
z
• • •
,
0
z
R
16.
We obtain
`
n
0
(
z
1)
n
1
,
0
z
1
2,
`
n
0
,
z
1
2
18.
n
0
(
1)
n
(
z
1)
n
,
0
z
1
1,
n
0
,
z
1
1
20.
n
0
(
z
1)
n
4
,
a
n
sinh 1
(
n
even),
a
n
cosh 1 (
n
odd),
z
1
0
22.
`
n
0
(
)
,
z
i
1,
n
0
(
)
,
z
i
1
24.
Team Project.
(a) Let
a
n
(
z
z
0
)
n
and
c
n
(
z
z
0
)
n
be two Laurent series of
the same function ƒ(
z
) in the same annulus. We multiply both series by (
z
z
0
)
k
1
and integrate along a circle with center at
z
0
in the interior of the annulus. Since the
series converge uniformly, we may integrate term by term. This yields 2
ia
k
2
ic
k
.
Thus,
a
k
c
k
for all
k
0,
1,
• • •
.
(b)
No, because tan (1/
z
) is singular at 1/
z
/2,
3
/2,
• • •
, hence at
z
2/
,
2/3
,
• • •
, which accumulate at 0.
(c)
These series are obtained by termwise integration of the integrand. The second
function is Si(
z
)/
z
3
, where Si(
z
) is the sine integral [see (40) in App. A.31].
Answer:
• • •
,
• • •
.
z
2
5!5
1
3!3
1
z
2
z
2
4!4
z
3!3
1
2!2
1
z
i
n
(
z
i
)
n
2
2
n
1
(
z
i
)
2
(
1
z
i
i
)
2
1
z
2
(
z
i
)
n
i
n
2
2
n
1
[
i
(
z
i
)]
2
1
z
2
a
n
n
!
(
1)
n
(
z
1)
n
1
(
1)
n
1
2
n
(
z
1)
n
2
1
(
z
1)
2
(
1
z
2
1
)
1
(
z
1)
2
2(
z
1)
1
1
(
z
1
1)
2
1
1
z
2
(
1)
n
1
2
n
1
1
120
z
3
1
6
z
1
(2
n
1)!
z
2
n
1
3
z
i
i
(
z
i
)
2
1
6!
1
4!
1
2!(
z
)
2
1
(
z
)
4
(
1)
n
1
(2
n
)!
im16.qxd
9/21/05
1:05 PM
Page 270