Polar Coordinates And Complex Numbers Worksheets With Answers - Math 611b Page 43
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50De Moivre’s Theorem
θ
θ
=
=
n
n
n
th
z
[
rcis
]
r cisn
It states that:
; it is useful in finding the p
root of a complex number.
θ
1
1
1
⎛
⎞
θ
=
=
p
p
p
⎜
⎟
z
[
rcis
]
r cis
⎝
⎠
p
Without De Moivre’s Theorem problems like; find the cube roots of 8; will not yield all possible roots. A
th
p
root problem should yield p roots. To find all possible roots we can use the fact that:
θ
θ
π
θ
π
θ
π
=
+
=
+
=
=
+
sin
sin(
2
)
sin(
4
)
. . .
sin(
2 k
)
θ
θ
π
θ
π
θ
π
=
+
=
+
=
=
+
cos
cos(
2
)
cos(
4
)
. . .
cos(
2 k
)
1
1
θ
=
p
p
z
[
rcis
]
1
θ
π
=
+
p
[
rcis
(
2
k
)]
Therefore
θ
π
θ
π
θ
π
1
+
1
⎛
+
⎞
⎛
+
⎞
2
k
2
k
2
k
=
+
p
p
⎜
⎟
⎜
⎟
r cis
(
)
or r
(cos
i
sin
)
⎝
⎠
⎝
⎠
p
p
p
Evaluating this formula for k = 0,1,2,...,p ! 1 will yield the p roots.
1
3
8
Ex. Solve
for all roots and represent them on the Argand plane.
b
θ
−
=
1
tan
=
= 8
z
r
Solution: 8 means 8 + 0i where a = 8 and b = 0. Thus
and
a
θ
=
0
2
Using r = 8,
= 0, p = 3, k = 0,1,2
1
1
θ
π
θ
π
+
+
⎛
⎞
⎛
⎞
2
k
2
k
=
+
p
p
⎜
⎟
⎜
⎟
z
r
(cos
i
sin
)
⎝
⎠
⎝
⎠
n
n
1
1
=
+
=
=
3
3
8
8 (cos 0
i
sin 0)
2
k
0
π
π
1
1
2
2
=
+
= − +
=
3
3
8
8 (cos
i
sin
)
1
3
i k
1
3
3
π
π
1
1
4
4
=
+
= − −
=
3
3
8
8 (cos
i
sin
)
1
3
i k
2
3
3
43
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