Find all complex roots of z3=−8i?
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Answer:In trigonometric form:
−
8
i
=
8
(
cos
(
−
π
2
)
+
i
sin
(
−
π
2
)
)
By de Moivre we have:
(
cos
θ
+
i
sin
θ
)
n
=
cos
n
θ
+
i
sin
n
θ
Hence one of the cube roots of
−
8
i
is:
2
(
cos
(
−
π
6
)
+
i
sin
(
−
π
6
)
)
=
√
3
−
i
The others can be found by adding multiples of
2
π
3
...
2
(
cos
(
−
π
6
+
2
π
3
)
+
i
sin
(
−
π
6
+
2
π
3
)
)
=
2
(
cos
(
π
2
)
+
i
sin
(
π
2
)
)
=
2
i
2
(
cos
(
−
π
6
+
4
π
3
)
+
i
sin
(
−
π
6
+
4
π
3
)
)
=
2
(
cos
(
7
π
6
)
+
i
sin
(
7
π
6
)
)
=
−
√
3
−
i
Here are the three roots in the complex plane...
graph{(x^2+(y-2)^2-0.005)((x-sqrt(3))^2+(y+1)^2-0.005)((x+sqrt(3))^2+(y+1)^2-0.005) = 0 [-5, 5, -2.5, 2.5]}
Step-by-step explanation:
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