Solve for all $z$: $z^3+4|z|=0$ in a better way

Question

I need to solve for all $z$

$$z^3+4|z|=0$$

It's quite straight forward, once you insert $a+bi$ instead of $z$.

the answers are:

$a=0$ and $b=0$

or $a=1$ and $b=\sqrt{3}$

or $a=1$ and $b=-\sqrt{3}$

From experience, these exercises usually have some better, shorter solutions than this, does somebody have an idea?

It's important to note that these are my answers and not a textbook's so I might have a mistake, please correct if I'm wrong

Answer 1

Since $$z^3=-4|z|\implies |z|^3 =4|z| \implies |z|\in\{0,2\}$$

we have two possibilities:

a) $|z|=0$ so $z=0$.

b) $|z|=2$ so $z^3+8=0$ so $$(z+2)(z^2-2z+4)=0$$ so $z=-2$ or $z=1\pm i\sqrt{3}$

Answer 2

If $z^3+4|z|=0$, then $z^3=-4|z|$ and therefore $|z^3|=\bigl|-4|z|\bigr|=4|z|$. So, $z=0$ or $|z|^2=4$. This last equation is equivalent to $z=2(a+bi)$, with $a^2+b^2=1,$ and\begin{align}\bigl(2(a+bi)\bigr)^3+4\bigl|2(a+bi)\bigr|&\iff8(a+bi)^3=-8\\&\iff(a+bi)^3=-1\\&\iff a+bi=\frac12\pm\frac{\sqrt3}2i\vee a+bi=-1\\&\iff z=1\pm\sqrt3i\vee z=-2.\end{align}

Answer 3

If we let $z=r\exp(i\theta)$, $r\ge 0$,$\theta \in \mathbb{R}$ then we have

$$r^3\exp(3i\theta)+4r=0$$

Hence we can have $r=0$, suppose not, then we have

$$r^2\exp(3i\theta)=-4=4\exp\left(i(\pi+2k\pi) \right)$$

that is we have $r=2$, and $\theta=\frac{(2k+1)\pi}3$

In summary, all the solutions are $r=0$ or $r=2\exp\left(\frac{i(2k+1)\pi}{3} \right), k=-1,0,1.$