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

Question

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?

Answer 1

Note that the solutions

$$e^{\frac\pi2 - 2\pi n}\>\>\>\>\> \text{and} \>\>\>\>\>e^{\frac\pi2 + 2\pi n}$$

are equivalent for $n \in \mathbb{Z} $.

Answer 2

Since $i^2 = -1$, we will need to multiply our result by $-1$...to wit:

$$e^{i (\ln |-i| + i \arg (-i))}$$ $$e^{i (\ln 1 + \frac {3 \pi i}{2} \pm 2\pi i k)}$$ $$e^{i (0 + \frac {3 \pi i}{2} \pm 2\pi i k))}$$ $$e^{i (\frac {3 \pi i}{2} \pm 2\pi i k)}$$ $$e^{(-\frac {3 \pi}{2} \pm 2\pi k)}$$

The last answer is fine on its own, but using Quanto's answer above, each of the coterminal angles can be taken either clockwise or counterclockwise. Some books imply the exponential is positive, so in this case, we can use the coterminal angle to express this (i.e. $-\frac {3 \pi}{2} = \frac {\pi}{2})$; thus $$e^{(-\frac {3 \pi}{2} \pm 2\pi k)}\equiv e^{(\frac {\pi}{2} \pm 2\pi k)} $$

Answer 3

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.