P
ROBLEM SET
7
Partial solution
Sec 2.4 p. 2 Let
f
(
z
) = (
e
z

1)
2
. Note that
f
(
z
) = 0
when
e
z
= 1
. Writing
z
=
x
+
iy
gives
e
x
e
iy
= 1
. Since the modulus of
e
x
e
iy
has to be 1 we get that
e
x
= 1
which implies
x
= 0
. Thus
e
iy
= 1
and hence
y
= 2
πk
where
k
is an integer. So there ane infinitely
many zeroes
z
k
= 2
πki
, where
k
in an integer.
Computing gives
f
(
z
k
) = 0
,
f
(
z
) = 2(
e
z

1)
e
z
,
f
(
z
k
) = 0
,
f
(
z
) = 4
e
2
z

2
e
z
, and
f
(
z
k
) = 2
. Since
f
(
z
k
) =
f
(
z
k
) = 0
and
f
(
z
k
) = 0
,
f
has a zero of order 2 at
z
k
for
all
k
’s.
Another way to solve the problem is to show that
e
z

1
has a zero of order 1 at
z
k
and
than use the fact that if a function
g
has zero of order
m
at
z
0
then
g
2
has zero of order
2
m
at
z
0
. This was a part of problem 17 on the problem set.
Sec 2.4 p. 10 Let
f
(
z
) =
e
z
and
z
0
=
πi
. By Taylor’s expansion
e
z
=
∞
k
=0
f
(
k
)
(
z
0
)
k
!
(
z

z
0
)
k
Note that
f
(
k
)
(
z
) =
e
z
and hence
f
(
k
)
(
πi
) =
e
πi
=

1
. Thus
e
z
=

∞
k
=0
1
k
!
(
z

πi
)
k
.
Since
e
z
is entire by Theorem 1 Sec 2.4 (a consequence of Cauchy’s Formula that gives
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 Spring '09
 ez, zk

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