How to find the square roots of $z = 5-12i$ I am asked the following question:

Find the square roots of $z = 5-12i$

I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rightarrow \quad z_1 = -3+2i \quad z_2 = 3-2i
$$
My question is: can the following method (below) be used to solve the problem above? Motivation for this question: if I were to find the cubic roots of the number given, I couldn't use the first method.
I will use this "other method" it in a different problem.

Find the square roots of $ z = 2i $

The method:
Since $ \rho = 2 $ and $ \theta = \frac{\pi}{2} $ we have to find a complex number such that
\begin{align*}
z_k^2 &= 2i\\
z_k^2 &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\rho_k^2 \left( \cos 2\theta_k + i \sin 2 \theta_k \right) &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\end{align*} 
\begin{cases}
\rho^2 &= 2\\
2\theta_k &= \frac{\pi}{2} + 2k\pi
\end{cases}
\begin{cases}
\rho &= \sqrt{2}\\
\theta_k &= \frac{\pi}{4} + 2k\pi
\end{cases}
\begin{align*}
z_0 &= \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 1+i\\
z_1 &= \sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = -1-i
\end{align*}
Using this method for the first example seems like a dead end, specially because of the fact that the angle of $z$ is not as straightforward as the angle of the second example.
Thank you.
 A: All you need to do is to convert your $z$ to polar form:
$$z=5-12i=\sqrt{12^2+5^2}e^{i(\arctan(-\frac{12}{5})+k\pi)}=13e^{i(-1.176+k\pi)}$$
Hence, one answer of $z^{\frac{1}{2}}$ is
$$z_1=13^{\frac{1}{2}}e^{i\frac{-1.176}{2}}=\sqrt{13}e^{i(-0.588)}$$
Converting this to Cartesian will give:
$$z_1=\sqrt{13}\left(\cos(-0.588)+i\sin(-0.588)\right)=3-2i$$
which is one of your original results.
A: If you let $\theta$ be the angle in the first case, then using the fact that $\tan \theta = -12/5$, you can find $\cos\theta$ and $\cos\frac{\theta}{2}$ indirectly. One of the roots would be: 
$\sqrt{13}(-\frac{3}{\sqrt{13}} + i \frac{2}{\sqrt{13}}) = z_1.$
A: You can apply de Moivre's theorem (second example) on the first example without doubt. Though you need a calculator or a trigonometrical table handy to check the sin and cos of a certain angle which feels kinda painful, at least for me.
A: Your method should work fine in both cases (though the trigonometry is a bit more complicated), but I point out that your first method can still work on the square roots of $2i$: We see that
$$
a^2-b^2 = 0
$$
so $a = \pm b$, and then, secondly, $2ab = 2$.  This gives us $a = b = \pm 1$ as the solutions, so the square roots of $2i$ are $1+i$ and $-1-i$.
