$\exp(z^2)=i$ 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.
 A: 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.
A: 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 $n < 0$.
A: $\displaystyle\exp(z^2)=i\iff z^2=\ln(i)=i\frac{\pi}2+2ik\pi=i\frac\pi2(4k+1)\quad$ with $k\in\mathbb Z$.
This $k$ comes from the fact that the exponential fonction is $2i\pi$ periodic on the complex numbers.
Depending on the sign of $k$ you then need to solve $w^2=\pm i\ $ which has $4$ solutions.


*

*$w=\pm(1+i)/\sqrt{2}\quad$ for $w^2=i$ and $k\ge 0$

*$w=\pm(1-i)/\sqrt{2}\quad$ for $w^2=-i$ and $k<0$


Finally $$z=\pm(1+i\,\sigma(k))\sqrt{|4k+1|\frac{\pi}4}$$ where $k\in\mathbb Z$ and $\sigma(k)=1_{\{k\ge 0\}}-1_{\{k<0\}}$ (same as signum function with $\sigma(0)=+1$)
