Solve $\sqrt{5-12i}$ by square root definition I KNOW it can be solved by the trig formula, but I want to solve it by the square root definition, so please don't just post an alternative way to do it.
By the square root definition:
$$z = 5-12i$$
$$\sqrt{z} = w\implies w^2 = z$$
So if I suppose $w = a+bi$ we have:
$$w^2 = z \implies (a+bi)^2 = 5-12i\implies\\a^2-b^2+2abi = 5-12i\implies \\5 = a^2-b^2\\-12 = 2ab\implies$$
$$b^4-5b^2-36 = 0\implies b = \pm \sqrt{-4} = \pm 2i, b = \pm 3$$
Then I get $4$ solutions:
For $b = \pm 2i$ I get $a = \pm 3i$ then $\sqrt{z}$ is 
$$\pm 2 + 3i$$
For $b = \pm 3$ I get $a = \pm 2$ and $\sqrt{z}$ is
$$2\pm 3i$$ 
But two of these, when squared, aren't $z$. What am I doing wrong?

An easier way is to solve:
$$\begin{cases}a^2-b^2=5\ \ (1)\\ 2ab=-12\ \ (2)\\a^2+b^2=|5-12i|=13\ \ (3),\end{cases}$$
because you don't have to solve a equation of degree $4$.
Therefore, by $(1)$ 
$$a^2=b^2+5$$
By $(3)$
$$2b^2=8\implies b^2=4\implies b=\pm2$$
and by $(2)$
$$ab=-6\implies a=\mp3$$
Therefore $$\sqrt{5-12i}=\mp3\pm 2i$$
