SOLUTION: find all complex numbers that are solution of give equation the answer leave in a + bi z^8

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Question 273613: find all complex numbers that are solution of give equation the answer leave in a + bi
z^8 - 16 = 0
Answer by CharlesG2(834) (Show Source):
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find all complex numbers that are solution of give equation the answer leave in a + bi
z^8 - 16 = 0
z^8 = 16
z^4 = 4 or -4
z^2 = +- 2 or +- 2i
(2*2=4, -2*-2=4)
(sqrt(4 * -1)=2i)
(2i * 2i = 4i^2 = 4 * -1 = -4)
(-2i * -2i = 4i^2 = 4 * -1 = -4)
z = +- sqrt(2) or +- sqrt(-2) or +- sqrt(2i) or +- sqrt(-2i)
(sqrt(2) * sqrt(2) = 2)
(-sqrt(2) * -sqrt(2) = 2)
z = +- sqrt(2) or +- sqrt(2)*i or +- sqrt(2i) or +- sqrt(-2i)
(sqrt(2)*i*sqrt(2)*i = 2 * -1 = -2)
(-sqrt(2)*i*-sqrt(2)*i = 2 * -1 = -2)
(sqrt(2i) * sqrt(2i) = 2i)
(-sqrt(2i) * -sqrt(2i) = 2i)
(sqrt(-2i) * sqrt(-2i) = -2i)
(-sqrt(-2i) * -sqrt(-2i) = -2i)
z = +- sqrt(2) or +- sqrt(2)*i or +- sqrt(2i) or +- sqrt(-2i)

SOLUTION: find all complex numbers that are solution of give equation the answer leave in a + bi z^8 - 16 = 0