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?

$$
sin(z) = \pm \sqrt{i \pi} \\
e^{iz} - e^{-iz} = \pm 2i \sqrt{i \pi}
$$
Let $e^{iz} = x$, then $x^2 \pm 2i \sqrt{i \pi} x - 1 = 0$.
Hence:
$e^{iz} = x = \frac{\pm 2i \sqrt{i \pi} \pm \sqrt{-4i \pi +4}}{2} = \pm i \sqrt{i \pi} \pm  \sqrt{1-i \pi}$
Thus, $z = -i \log(\pm i \sqrt{i \pi} \pm  \sqrt{1-i \pi})$.
Or:
$$
z_1 = -i \log(i \sqrt{i \pi} +  \sqrt{1-i \pi}) \\
z_2 = -i \log(i \sqrt{i \pi} -  \sqrt{1-i \pi}) \\
z_3 = -i \log(-i \sqrt{i \pi} +  \sqrt{1-i \pi}) \\
z_4 = -i \log(-i \sqrt{i \pi} -  \sqrt{1-i \pi})
$$
As $\log(-1) = i (1 + 2k) \pi$, $z_3 = z_2 + i(1+2k) \pi$, and $z_4 = z_1 + i(1+2k) \pi$, for $k$ integer.
