(2
n

2)!
2
·
1
·
2
·
2
· · ·
2
·
(
n

1)
= (

1)
n

1
1
2
n
n
!
(2
n

2)!
2
n

1
·
1
·
2
· · ·
(
n

1)
=
= (

1)
n

1
1
2
n
n
!
(2
n

2)!
2
n

1
(
n

1)!
= (

1)
n

1
1
2
n
n
!
(2
n

2)!
2
n

1
(
n

1)!
2
n
(2
n

1)
2
n
(2
n

1)
=
(

1)
n

1
1
2
n
n
!
(2
n
)!
2
n
n
!(2
n

1)
(

1)
n

1
(2
n
)!(2
n

1)
2
2
n
(
n
!)
2
= (

1)
n

1
(

1)
n

1
2
2
n
(2
n

1)
2
n
n
.
Thus
(1 +
z
)
1
2
=
∞
X
n
=0
(

1)
n

1
2
2
n
(2
n

1)
2
n
n
z
n
,

z

<
1
.
41.
(a) and (b)
There are several possible ways to derive the Taylor series expansion of
f
(
z
) = Log
z
about the point
z
0
=

1 +
i
. Here is one way. Let
z
0
=

1 +
i
, so

z
0

=
√
2.
The function Log
z
is analytic except on the negative real axis and 0. So it is guaranteed by
Section 4.3
Taylor Series
99
Theorem 1 to have a series expansion in the largest disk around
z
0
that does not intersect
the negative real axis. Such a disc, as you can easily verify, has radius Im
z
0
= 1. However,
as you will see shortly, the series that we obtain has a larger radius of convergence, namely

z
0

=
√
2 (of course, this is not a contradiction to Theorem 1).
Consider the function Log
z
in
B
1
(
z
0
), where it is analytic.
(The disk of radius 1,
centered at
z
0
is contained in the upper halfplane.) For
z
∈
B
1
(
z
0
), we have
d
dz
Log
z
=
1
z
.
Instead of computing the Taylor series of Log
z
directly, we will first compute the Taylor
series of
1
z
, and then integrate termbyterm within the radius of convergence of the series
(Theorem 3, Sec. 4.3). Getting ready to apply the geometric series result, we write
1
z
=
1
z
0

(
z
0

z
)
=
1
z
0
(
1

z
0

z
z
0
)
(
z
0
6
= 0)
=
1
z
0
·
1
1

z
0

z
z
0
=
1
z
0
∞
X
n
=0
z
0

z
z
0
n
=
1
z
0
∞
X
n
=0
(

1)
n
(
z

z
0
)
n
z
n
0
,
where the series expansion holds for
z
0

z
z
0
<
1
⇔ 
z
0

z

<

z
0

.
Thus the series representation holds in a disk of radius

z
0

=
√
2, around
z
0
. Within this
disk, we can integrate the series termbyterm and get
Z
z
z
0
1
ζ
dζ
=
1
z
0
∞
X
n
=0
(

1)
n
z
n
0
Z
z
z
0
(
ζ

z
0
)
n
dζ
=
1
z
0
∞
X
n
=0
(

1)
n
(
n
+ 1)
z
n
0
(
z

z
0
)
n
+1
.
Reindexing the series by changing
n
+ 1 to
n
, we obtain
Z
z
z
0
1
ζ
dζ
=
∞
X
n
=1
(

1)
n
+1
nz
n
0
(
z

z
0
)
n

z

z
0

<

z
0

=
√
2
.
Now we have to decide what to write on the left side. The function Log
z
is an antiderivative
of
1
z
in the disk of radius 1, centered at
z
0
. (Remember that Log
z
is not analytic on the
negative real axis, so we cannot take a larger disk.) So, for

z

z
0

<
1, we have
Z
z
z
0
1
ζ
dζ
= Log
ζ
z
z
0
= Log
z

Log
z
0
.
Thus, for

z

z
0

<
1, we have
Log
z
= Log
z
0
+
∞
X
n
=1
(

1)
n
+1
nz
n
0
(
z

z
0
)
n
,
even though the series on the right converges in the larger disk

z

z
0

<
√
2.
100
Chapter 4
Series of Analytic Functions and Singularities
Solutions to Exercises 4.4
1.
We have that
1
1 +
z
=
1
z
1
1 +
1
z
=
1
z
∞
X
n
=0

1
z
n
=
∞
X
n
=0
(

1)
n
z
n
+1
,
and this converges provided

1
z
<
1, that is, when 1
<

z

.
5.
Since
1
z
2
<
1, we can use a geometric series in
1
z
2
as follows. We have
1
1 +
z
2
=
1
z
2
1
1


1
z
2
=
1
z
2
∞
X
n
=0

1
z
2
n
=
1
z
2
∞
X
n
=0
(

1)
n
z
2
n
=
∞
X
n
=0
(

1)
n
z
2(
n
+1)
=
∞
X
n
=1
(

1)
n

1
z
2
n
,
where in the last series we shifted the index of summation by 1.
Note that (

1)
n

1
=
(

1)
n
+1
, and so the two series that we derived are the same.