I'm trying to find all complex numbers $z$ for which $e^{z^2}=-1$. I found this related question:
Finding the set $\{z: e^z=-1\}$
While this did help me to develop some ideas, I could not solve my question, because of the $z$ squared term was still causing me trouble.
What I guess I need is, that $-1=e^{\pi i}$ and $e^{z}=e^{x}e^{(2k+1)\pi i}$, where $e^x=1$ and $e^{(2k+1)\pi i}=y$ but since we have $e^{z^2}$ which would be $e^{x^2}e^{-y^2}e^{2xyi}$, I'm a little confused as to how to proceed. Any suggestions would be appreciated.
Since$$e^z=-1\iff z=\pi i+2n\pi i\text{ for some }n\in\mathbb Z,$$and since the number $i$ has two square roots: $\pm\frac1{\sqrt2}(1+i)$, then$$e^{z^2}=-1\iff z=\pm\frac{\sqrt{(2n+1)\pi}}{\sqrt2}(1+i)\text{ for some }n\in\mathbb Z.$$