Why must c be a real number? Question: Let $c = a+bi$ be a fixed complex number, where $c\not=0, \pm1, \pm2,...,$ and note that $i^c$ is multiple-valued. What additional restriction must be placed on the constant so that the values of $|i^c|$ are all the same.
My Work:
Using $z^c = e^{c\log{z}}$, it follows that $i^c = e^{c\log{i}}$
$$\log(i) = \ln{|i|}+i\arg{i} \Rightarrow$$ $$i^c = e^{ai[\frac{\pi}{2}+2k\pi]}e^{-b{[\frac{\pi}{2}+2k\pi}]} \Rightarrow$$ $$|i^c| = e^{-b[\frac{\pi}{2}+2k\pi]}$$
After this I cannot see why the restriction must be that c is a real number. 
 A: You might notice that $e^{-b(\frac{\pi}{2} + 2k\pi)}$ is multivalued in $k$. As $k$ can be chosen to be any integer, this term has no single well-defined value. However, if $b$ is $0$, then that issue does not emerge.
A: What about using Euler's formula, so we've got
$$
e^{i(2\pi n+\frac{\pi}{2})}=i\implies i^c={(e^{i(2\pi n+\frac{\pi}{2})})}^c=e^{ci(2\pi n+\frac{\pi}{2})}
$$
which gives us with $c=a+bi$ the following setting for easier notation $n=0$ (the solution goes the same for $n\in\mathbf{Z}$)
$$
i^c={(e^{i\frac{\pi}{2}})}^c=e^{ci\frac{\pi}{2}}=e^{(a+bi)i\frac{\pi}{2}}=e^{(ai-b)\frac{\pi}{2}}=e^{ai\frac{\pi}{2}}e^{-b\frac{\pi}{2}}
$$
and therefore
$$
|i^c|=|e^{ai\frac{\pi}{2}}e^{-b\frac{\pi}{2}}|=|e^{ai\frac{\pi}{2}}||e^{-b\frac{\pi}{2}}|=1\cdot|e^{-b\frac{\pi}{2}}|
$$
so the $b\equiv0$ since otherwise the magnitude of $i^c$ will change all the time according to varying $b$. This means $c\in\mathbf{R}$.
A: Using $i=e^{i(1+4k)\pi/2}$,
$i^c=e^{ic(1+4k)\pi/2}=e^{i(a+ib)(1+4k)\pi/2}=e^{ia(1+4k)\pi/2}e^{-b(1+4k)\pi/2}$
As $|i|=1$, $|i^{ia(1+4k)\pi/2}|=1$, so:
$|i^c|=e^{-b(1+4k)\pi/2}$
Unless $b=0$, $|i^c|$ is also multi-valued.
