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 $x^n=1$
Algebraically, $$(a+bi)^2 = (a^2-b^2) + 2abi$$ So suppose $(a+bi)^2 = c+di$, then you can solve the system of equations by comparing real and imaginary terms, beginning with $\frac{d}{2a} = b$ and substituting this into $a^2-b^2=c$. This will let you solve for $a$. Then you can use your solution for $a$ to solve for $b$.
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 $z= \pm \sqrt{r} (\cos \frac{\psi}{2} + i \sin \frac{\psi}{2})$
