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Category:Utilities for WindowsQ:
Finding the domain and range of all the complex solutions of $z^2=2i$
I started with the constraint $r=\sqrt{2}$ and so I was making a rotation about z axis to get the desired origin. Then since $x=\cos\theta,y=\sin\theta$ we have
$$z=r(\cos\theta+i\sin\theta)$$
$$\iff r(\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta)=2\cos\theta+2i\sin\theta$$
which gives $\cos\theta=0$ and $\sin\theta=0$.
Then we are done. The answer is $|z|=r\ge0$ and $z=0$ for $z\in\mathbb{C}$. This goes by the name of “Gershgorin circle theorem” from differential equations. But it turned out that my attempt is wrong.
Thanks in advance!
A:
If you make a rotation about the $z$-axis, you can see that
$$z^2=2i \iff z=r\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$$
for $r=\sqrt{2}$. This is some solution to the given equation.
In order to see that this is the only solution, we can find all of the real solutions to the given equation, and then every other solution must be complex conjugate. For example, a general solution is
$$z_p=re^{i\frac{\pi}{4p}}$$
for any integer $p$. As $p$ ranges from $1$ to $4$, this gives us $4$ solutions. Notice, though, that the only $4$th root of $2i$ is $0$, so no other solution can be complex-conjugate to any of those roots. If there were more roots, there would be some solution, say $z_1$ and $z_2$, such that $z_1+\bar{z_2}$ (the “conjugate sum”) were another root, but this is impossible since $ http://www.propertymajalengka.com/wp-content/uploads/2022/06/yulyvan.pdf
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