Question 1028076
Find all complex numbers z such that z^2 = 2i.
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z^2 = 2cis90
z = sqrt(2)cis(45), sqrt(2)cis(225)
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z = sqrt(2)*sqrt(2)/2 + isqrt(2)*sqrt(2)/2
= 1 + i
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z = sqrt(2)cis(225)
sqrt(2)*(-sqrt(2)/2) + i*sqrt(2)*(-sqrt(2)/2)
= -1 - i