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MATH 135 Algebra, Solutions to Assignment 12
1:
Express each of the following complex numbers in cartesian form.
(a) 4
e
i
5
π/
3
Solution: We have 4
e
i
5
π/
3
= 4
(
cos
5
π
3
+
i
sin
5
π
3
)
= 4
±
1
2

√
3
2
i
²
= 2

2
√
3
i
(b) (1 +
i
√
3)
10
Solution: We have
(1 +
i
√
3)
10
=
±
2
e
i π/
3
²
10
= 2
10
e
i
10
π/
3
= 2
10
(
cos
10
π
3
+
i
sin
10
π
3
)
= 1024
±

1
2

√
3
2
i
²
=

512

512
√
3
i .
(c) 5
e
i θ
, where
θ
= tan

1
2
Solution: We have 5
e
i θ
= 5 (cos
θ
+
i
sin
θ
) = 5
±
1
√
5
+
2
√
5
i
²
=
√
5 + 2
√
5
i
.
2:
Express each of the following complex numbers in the polar form
re
i θ
.
(a)

2 + 2
i
Solution: We have

2 + 2
i
= 2
√
2
e
i
3
π/
4
.
(b)
(1

i
)
2
(1 +
i
√
3)
Solution: We have
(1

i
)
2
(1 +
i
√
3)
=
(
√
2
e

i π/
4
)
2
2
e
iπ/
3
=
2
e

i π/
2
2
e
i π/
2
=
e

i
(
π
2
+
π
3
)
=
e

i
5
π/
6
=
e
i π/
6
.
(c)

3

i
Solution: We have

3

i
=
√
10
e
i θ
, where
θ
=
π
+ tan

1 1
3
.
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View Full Document3:
Solve each of the following for
z
∈
C
. Express your answers in the polar form
re
i θ
.
(a)
z
3
+ 8
i
= 0
Solution: Write
z
=
r e
i θ
with
r >
0 and
θ
∈
[0
,
2
π
). Then
z
3
+ 8
i
= 0
⇐⇒
(
r e
i θ
)
3
=

8
i
⇐⇒
r
3
e
i
3
θ
= 8
e
i
3
π/
2
⇐⇒
r
3
= 8 and 3
θ
=
3
π
2
+ 2
π k
for some
k
∈
Z
⇐⇒
r
= 2 and
θ
=
π
2
+
2
π
3
k
where
k
= 0
,
1 or 2
.
Thus the solutions are
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 Fall '08
 ANDREWCHILDS
 Algebra, Complex Numbers

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