real and imaginary part of square root of any complex number

let $k\in\mathbb{C}$. What are the real and imaginary parts of any complex number $x$ so that $x^2=k$ ?

My first idea was writing $k$ and $x$ in polar form: $x=r(\cos{\phi}+i\sin{\phi})$;$k=r'(\cos{\psi}+i\sin{\psi})$. Then use De Moivre's formula such that: $x^2=r^2(\cos{2\phi}+i\sin{2\phi})=r'(\cos{\psi}+i\sin{\psi})$.

Any hints how to go on ?

Another idea could be using roots of unity: We know how $x$ looks like when $x^n=1$

Answer

Well, you just answered yourself.

If $r_1 e^{i\theta_1} = r_2 e^{i \theta_2}$ then $r_1=r_2 , \theta_1 = \theta_2 + 2\pi n$. That means in this case that

$$r^2 = r' \Rightarrow r=\sqrt{r'}$$
$$2\phi = \psi + 2\pi n \Rightarrow \phi = \frac{\psi}{2} , \frac{\psi}{2} + \pi$$
Meaning the solution will be $z= \pm \sqrt{r} (\cos \frac{\psi}{2} + i \sin \frac{\psi}{2})$