Let $z_0 = \left(a,b\right)$ find $z$ such that $z^2 = z_0$ where $z_0 , z$ are complex numbers.
Here is my attempt. taking absolute values on both sides, we obtain $\left|z\right|^2 = \left|z_0\right|^2$
we let $z = u +vi$ so that
$u^2 + v^2 = \sqrt{a^2 + b^2} $
Any help from here?
If $z_0=a+ib$ and $z=x+iy$ with $z^2=z_0$ then
$x^2-y^2+2ixy=a+ib$ hence
$x^2-y^2=a$ and $2xy=b$.
Your turn !
It would be better to write the complex number $z_0$ in rectangular form:
$$z_0 = a + bi = \sqrt{a^2 + b^2}\cdot\bigg({\frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{a^2 + b^2}}i}\bigg)$$
Then write the complex number $z_0$ in polar form:
$$z_0 = {r_0}\exp(i\arg{z_0})$$
where $r_0 = \sqrt{a^2 + b^2}$ and $\arg{z_0} = \arctan\frac{b}{a}$.
We now have to solve the equation $$z^2 = z_0$$ for $z$. But $z = r\exp(i\arg{z})$ means $z^2$ in polar form is $$z^2 = {r^2}\exp\bigg(i(2\arg{z})\bigg).$$
Equating moduli and arguments for $z^2$ and $z_0$, we get
$$r^2 = r_0 = \sqrt{a^2 + b^2} \iff r = \pm \sqrt{r_0} = \pm \sqrt{\sqrt{a^2 + b^2}} = \pm (a^2 + b^2)^{1/4}$$ and $$2\arg{z} = \arg{z_0} \iff \arg{z} = \frac{\arg{z_0}}{2}.$$
Thus, $z$ is given explicitly in polar form as: $$z = {\pm (a^2 + b^2)^{1/4}}\cdot{\exp\bigg(i\frac{\arg{z_0}}{2}\bigg)} = {\pm (a^2 + b^2)^{1/4}}\cdot{\exp\bigg(i\frac{\arctan\frac{b}{a}}{2}\bigg)}.$$
