Find all possible values of $(-i)^{i}$. Find all possible values of $(-i)^{i}$.


Here is my attempt:

$(-i)^{i}=\mathit{e}^{ilog(-i)}=\mathit{e}^{i(log|-i|+iarg(-i))}=\mathit{e}^{-\frac{3\pi}{2}}=\mathit{e}^{\frac{\pi}{2}+2\pi n}$, where $n$ is an integer.

However, the answer is $\mathit{e}^{\frac{\pi}{2}-2\pi n}$. Why is $2\pi n$ being subtracted?
Note that the solutions 

$$e^{\frac\pi2 - 2\pi n}\&gt;\&gt;\&gt;\&gt;\&gt; \text{and} \&gt;\&gt;\&gt;\&gt;\&gt;e^{\frac\pi2 + 2\pi n}$$

are equivalent for $n \in \mathbb{Z} $.
Sorry, I'm having trouble with Math.Jax but -i is exp(((4n+3)(pi)/2)i) so the answer is exp((-(4n+3)(pi)/2), where n is any integer.