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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?

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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 !

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  • $\begingroup$ Thanks a bunch, I actually wanted to try that but that horrendous calculations I saw coming turned me off. Lol $\endgroup$ – Icosahedron Jun 28 '17 at 9:46
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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: $$z = {\pm (a^2 + b^2)^{1/4}}\cdot{\exp\bigg(i\frac{\arg{z_0}}{2}\bigg)} = {\pm (a^2 + b^2)^{1/4}}\cdot{\exp\bigg(i\frac{\arctan\frac{b}{a}}{2}\bigg)}.$$

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