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MA108C HW 3 SOLUTIONS
PAUL NELSON (AND SOME MODIFICATIONS BY FOKKO VAN DE BULT)
(1) By the generalized binomial theorem,
√
z
= [1 + (
z

1)]
1
/
2
=
∞
X
k
=0
±
1
/
2
k
²
(
z

1)
k
for

z

<
1
,
where
(
α
k
)
=
α
(
α

1)
···
(
α

k
+1)
k
!
for
α
∈
C
,k
∈
Z
.
The ratio of successive terms is
(
1
/
2
k
+1
)
(
z

1)
k
+1
/
(
1
/
2
k
)
(
z

1)
k
=
1
/
2

k
k
+1
(
z

1) which tends to 1

z
as
k
→ ∞
, so the radius of convergence of
our series is 1.
Let us now prove this via the general theory of holomorphic functions. Deﬁne
f
(
z
) =
e
1
2
log
z
for
z
∈
C
\
R
≤
0
, where log(
re
iθ
) = log
r
+
iθ
for
θ
∈
(

π,π
), as in the ﬁrst homework. Then
f
(
z
) =
√
z
for
z
∈
R
>
0
, so the power series for
f
at
z
= 1 is as above. Since
f
is holomorphic in
{
x
+
iy
:
x >
0
} ⊃ {
z
:

z

1

<
1
}
and the power series for a holomorphic function at a point
converges absolutely in any disc centered at that point in which the function is holomorphic, we see
that the radius of convergence of
f
at
z
= 1 is at least 1. The radius of convergence cannot exceed
1; otherwise,
f
would extend to a holomorphic function
F
on
{
z
:

z

1

<
1 +
±
}
for some
± >
0, so
that
F
is holomorphic in a neighborhood of the origin. By continuity,
F
(0) = 0. For small enough
h
∈
R
>
0
we have
F
(
h
) =
f
(
h
) =
√
h
, so
F
(
h
)

F
(0)
h
=
h

1
/
2
. This expression does not have a limit
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