$\exp(z^2)=i$ solution check Find all complex solutions to the equation $\exp(z^2)=i$.
Attempt
We have $z^2 = x^2-y^2+2ixy$ where $z=x+iy$. Then
$$exp(x^2-y^2)=1 \text{ and } 2xy=\frac{\pi}{2}$$
So $x^2=y^2$ and $xy = \frac{\pi}{4}$ Thus $x = y$ and so $x=y= \pm \frac{\sqrt{\pi}}{2}$
I just don't feel very confident about this for some reason.

Your answer is fine except at the step where you say $2xy=\frac{\pi}{2}.$ In fact, you can have $2xy=\frac{\pi}{2}+2\pi k$ for any integer $k$, or $xy=\frac{\pi}{4}+\pi k.$
If $k\geq 0$ then you get $x=\pm\sqrt{\frac{\pi}{4}+\pi k}=\pm\frac{\sqrt{\pi}}{2}\sqrt{4k+1}$ and $y=x.$
If $k<0$ then you get $x=\pm\sqrt{-\pi k -\frac{\pi}{4}}=\pm\frac{\sqrt{\pi}}{2}\sqrt{-4k-1}$ and $y=-x.$

One neat thing you can do to simplify: 
If $j$ is non-negative integer, then we can write:
$$\begin{align}x&=\pm\frac{\sqrt{\pi}}{2}\sqrt{2j+1}\\y=&(-1)^jx\end{align}$$
and this gets all solutions.
