Answer to Question #114669 in Complex Analysis for adebayo

Question

Solve (-1)^2i

Answer 1

We know that

$(-1) = \cos (2n+1)\pi \pm i\sin (2n+1)\pi = e^{\pm (2n+1)\pi i}, \text{for $n=1,2,3,\cdots$}$

Therefore,

$(-1)^{2i} = e^{(\pm(2n+1)\pi i) \cdot 2i}= e^{\mp 2(2n+1)\pi}, \text{for $n=1,2,3,\cdots$}\\ (-1)^{2i} = e^{\mp m\pi}, \text{for $m=2,4,6,\cdots$}\\$