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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 inadvance! 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, wecan 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

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$p$ 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 besome solution, say $z_1$ and $z_2$, such that$z_1+\bar{z_2}$ (the "conjugate sum") were anotherroot, but this is impossible since $Download

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