0 to the complex power!

Let's first compute $0^i$ $x=e^{\ln(x)}$ $x^i=e^{i\ln(x)}$ We know that $e^i\theta=\cos(\theta)+i\sin(\theta)$ $\Rightarrow x^i=e^{i\ln(x)}=\cos(\ln(x))+i\sin(\ln(x))$ In the case $x=0$ $\Rightarrow 0^i=\cos(\ln(0))+i\sin(\ln(0))$ And as $\ln(0)=\red{undefined}$ $\Rightarrow 0^i=\red{undefined}$

Now let's consider $z=a+bi;b\neq0$ $0^z=0^a\times0^{bi}=0\times(0^i)^b=0\times(\red{undefined})^b=\red{undefined}$ $\Rightarrow \Im(z)\cancel{\in}(-\infty,0)\cup(0,\infty)$

Now let's consider $z=a+0\times i=a$ $\Rightarrow \Im(z)=0,\space z\in\mathbb{R}$ $\Rightarrow 0^z=0$

Therefore, if $0^z$ is not a undefined value then $\Im(z)=0$