Why does $1^{-i}$ equal 1?

Question

At one point, I found an equation that works with complex logarithms, but I lost the book that contains the equation. If I feed this to Wolfram|Alpha, it states that $1^{-i}$ is equal to 1. Why is $1^{-i} = 1$?

Answer 1

Since $a^b = e^{b \log a}$, we have

$$1^{-i} = e^{-i \log 1} = e^{-i \cdot 2k\pi i} = e^{2k\pi}$$

Note that in the complex numbers $\log^\mathbb C z = \log^\mathbb R |z| + (2k \pi + \arg z)i$, so there are infinite choices for its value.

One of the values that $1^{-i}$ takes is $1$, but it is not always $1$. If we set $k = 0$, the resulting number it's called principal value (http://en.wikipedia.org/wiki/Principal_value), and that is what wolfram reports.

With $k=0$ you get $e^0 = 1$

Answer 2

Given any $a,b\in \mathbb{C}$ we define $a^b=e^{b\ \log\ (a)}$. Thus in this case $1^{-i}$ = $e^{-i\log (1)}$ = $1$.