<s>[INST] $\exp(z^2)=i$ 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.
 [/INST] If $\exp(z^2) = i$ then $z^2 = \log i = \ln|i| + i\arg i = 0 + i(\frac{\pi}{2} + 2\pi n) = \frac{4n+1}{2}\pi i$ where $n$ is an integer. These are the numbers $\ldots, -\frac{7\pi i}{2}, -\frac{3 \pi i}{2}, \frac{\pi i}{2}, \frac{5 \pi i}{2}, \frac{9\pi i}{2}, \ldots$. Taking square roots (using that $\sqrt{i} = \pm\frac{1+i}{\sqrt{2}}$) 
we get $$z = \pm \frac{1+i}{2} \sqrt{(4n+1)\pi}$$
for $n \geqslant 0$ and 
$$z = \pm\frac{-1+i}{2} \sqrt{(-4n-1)\pi}$$
for $n < 0$.
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