complex numbers - Find all roots of $z^2=3-4i$.

Find all roots of $z^2=3-4i$.

$z^2=3-4i$

$z^2+4i-3=0$

But how do I go on from here?

Answer

The way people will really solve it.

If we have a root $z_0$, then $-z_0$ is also a root. Since this is a quadratic, there can't be more than $2$ roots, so we just need to guess one root $z_0$. How do we do that?

Observe that $|z^2| = \sqrt{3^2 + 4^2} = 5$, therefore $|z|=\sqrt{5}$. Now, the only nice numbers with that absolute value are $\pm 1 \pm 2i$ and $\pm 2 \pm i$. Geometric intuition says that one of the roots must be found in the 4th quadrant. A quick check shows that indeed $z_0 = 2-i$ fits the bill:
$$
(2-i)^2 = 4 - 4i + i^2 = 3 - 4i.
$$

So the answer is $2-i$ and $-2 + i$.

This is how a lazy person like me solves a problem like this. It is perfectly legitimate to guess the answer (if you have a good argument why other answers are impossible, which in this case follows from the general knowledge that a quadratic has at most two roots). Only when guessing doesn't work right away do people start testing more powerful general approaches.