@Grognor One way to solve i^i is w/ Euler's formula e^(i x) = cos x + i sin x. With x = pi/2 it gives the surprising result e^(i pi/2) = i.
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@Grognor One way to solve i^i is w/ Euler's formula e^(i x) = cos x + i sin x. With x = pi/2 it gives the surprising result e^(i pi/2) = i.