Is zero to the power of an imaginary number still zero? I just want to make sure that $0^i = 0$, but for some reason I couldn't find anything about this online.
Is this true?
--Background--
I'm trying to prove that some exponent is zero. I thought I'd raise each side to the power of $i$ so that I could use Euler's formula.
 A: Standard definition of a complex exponent $w^z$ is $$ w^z=e^{z\log(w)} = e^{a\log(w)}e^{bi\log(w)}$$
where $z = a +bi.$ For $w = re^{i\theta},$ this gives
$$
w^z = e^{a\ln(r)}e^{ai\theta}e^{bi\ln(r)}e^{-b\theta}.
$$
For $w = 0,$ we have $r=0$ and $\theta =$ whatever. $\ln(0)$ is undefined but we can take the limit $r\downarrow0$ so that $\ln(r)\downarrow-\infty.$ The limit of the expression for $w^z$ is only zero if $a >0,$ otherwise it blows up or oscillates. In other words if  $\Re(z)>0$ then $0^z=0.$ Otherwise it's infinite or undefined.
A: Let’s consider the general expression $x^i$.
We can say that:
$$
x^i = e^{i \ln(x)}
$$
and, according to Euler’s formula $e^{i \theta} = cos(\theta) + i \sin(\theta)$, we can say that:
$$
x^i = e^{i \ln(x)} = \cos(\ln(x)) + i \sin(\ln(x))
$$
Now, if $x = 0$, we have:
$$
0^i = \cos(\ln(0)) + i \sin(\ln(0))
$$
$ln(0)$ doesn’t exist, but, if we take the limit, we can say that $ln(0) \to - \infty$
Therefore:
$$
0^i = \cos(-\infty) + i \sin(-\infty)
$$
But the functions $\sin(x)$ and $\cos(x)$ don’t have limits when $x$ approaches to infinite, since they oscillate.
Therefore, $\cos(-\infty) + i \sin(-\infty)$ is undefined, hence $0^i$ is undefined and it’s not equal to 0.
For the same reason $\infty^i$ and $(-\infty)^i$ are undefined too.
