On solving the equation $\sin ^{2}\left ( z \right )= i\pi$ One obvious approach is to rewrite $\sin\left ( z \right )$ as $\frac{e^{iz}-e^{-iz}}{2i}$ and then proceed applying an appropriate substitution. However that leads to a quartic complex equation, which gets a bit(actually a lot) messy. Is there another(maybe obscure) approach suitable for this problem?

You can replace $\sin z = w$ and notice that $w$ solve the equaiton $w^2=i\pi.$
This means that $\sin z$ can either be $\sqrt\pi e^{i \frac\pi4}$ or $\sqrt\pi e^{\frac{5\pi}{4}}$.
Then, you are solving the easier equation of $\sin z = w$ for some known $w$...
