P
ROBLEM SET
8
Partial solution
1. Note that
f
(
z
) =
1
z

1
1
z
+1
. Since
1
z
+1
is analytic at
z
= 1
it has a Taylor series expansion
near
z
= 1
.
The Taylor series can be determined in a standard way by computing
derivatives, but also more efficiently by using that when

a

<
1
then
1

a
+
a
2

a
3
+
a
2
 · · ·
= 1
/
(1 +
a
)
. In particular
1
1 +
z
=
1
2 + (
z

1)
=
1
2
1
1 +
z

1
2
=
1
2
(1

z

1
2
+
(
z

1)
2
2
2
 · · ·
)
when

z

1
2

<
1
that is when

z

1

<
2
. So the Taylor series converges when

z

1

<
2
.
This can also be verified by using the root or ratio tests for power series. Therefore the
Laurent series for
f
(
z
)
is
f
(
z
) =
1
z

1
∞
k
=0
(

1)
k
(
z

1)
k
2
k
+1
=
∞
l
=

1
(

1)
l
+1
(
z

1)
l
2
l
+2
and converges when

z

1

<
2
. The residue at
1
is the coefficient with the
(
z

1)

1
term, that is
1
2
.
Sec. 2.5 p14. Since
f
has a zero of order
m
at
z
0
there exists an analytic function
g
(
z
)
near
z
0
such
that
g
(
z
0
) = 0
and
f
(
z
) = (
z

z
0
)
m
g
(
z
)
near
z
0
. Therefore
Res
(
f
f
, z
0
) =
Res
m
(
z

z
0
)
m

1
g
(
z
) + (
z

z
0
)
m
g
(
z
)
(
z

z
0
)
m
g
(
z
)
, z
0
=
Res
(
m
z

z
0
, z
0
)+
Res
(
g
g
, z
0
)
By Cauchy’s formula Res
(
m
z

z
0
, z
0
) =
m
while since
g /g
is analytic at
z
0
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 Spring '09
 Power Series, Taylor Series, Taylor series expansion, Z0

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