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Complex Analysis
Fall 2007
Homework 2: Solutions
1.3.6.
(a) We have log
 
i

= log 1 = 0 and arg(

i
)
∈ {
π/
2 + 2
nπ

n
∈
Z
}
. Hence, the values
of log(

i
) are
i
±

π
2
+ 2
nπ
¶
for
n
∈
Z
.
(b) We have log

1 +
i

= log
√
2 = (1
/
2)log 2 and arg(1 +
i
)
∈ {
π/
4 + 2
nπ

n
∈
Z
}
. Hence,
the values of log(1 +
i
) are
1
2
log 2 +
i
‡
π
4
+ 2
nπ
·
.
1.3.8.
(a) Since
i
is not a rational number, we know that (

1)
i
has inﬁnitely many values. To
compute them we ﬁrst compute the values of log(

1). Since log
 
1

= log 1 = 0 and
arg(

1)
∈ {
π
+2
nπ

n
∈
Z
}
the values of log(

1) are
i
(
π
+2
nπ
) for
n
∈
Z
. Hence, the
values of (

1)
i
are
e
i
log(

1)
=
e
i
2
(
π
+2
nπ
)
=
e

(
π
+2
nπ
)
for
n
∈
Z
.
(b) Again,
i
is not rational so we expect inﬁnitely many values for 2
i
. As above, we have
log

2

= log 2 and arg 2
∈ {
2
nπ

n
∈
Z
}
so that log 2 = log 2 + 2
nπi
and
2
i
=
e
i
log 2
=
e

2
nπ
+
i
log 2
for
n
∈
Z
.
1.3.12.
We begin by observing that for any
α
∈
C
we have (by deﬁnition)
tan
α
=
sin
α
cos
α
=

i
e
iα

e

iα
e
iα
+
e

iα
.
If we multiply the numerator and denominator by
e
iα
this becomes
tan
α
=

i
(
e
iα
)
2

1
(
e
iα
)
2
+ 1
.
Given
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This note was uploaded on 02/29/2012 for the course MATH 4364 taught by Professor Ryandaileda during the Fall '11 term at Trinity University.
 Fall '11
 RyanDaileda

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