Solve $\lvert z \lvert = z^4, z \in \mathbb{C}$ $$z^4 =\lvert z \lvert ,    z \in \mathbb{C}$$
Applying the formula to calculate $ \sqrt[4]{z} $, I find that solutions have to have this form:
$$z=\sqrt[4]{\lvert z \lvert}$$
$$z=\sqrt[4]{\lvert z \lvert} \ e^{i \frac{\pi}{2}}=i \ \sqrt[4]{\lvert z \lvert}$$
$$z=\sqrt[4]{\lvert z \lvert} \ e^{i \frac{3 \pi}{2}}=-i \ \sqrt[4]{\lvert z \lvert}$$
$$z=\sqrt[4]{\lvert z \lvert} \ e^{i \pi}=-\sqrt[4]{\lvert z \lvert}$$
 
Using the Cartesian form:
$$(a+i b)^4=\sqrt{a^2+b^2}$$

$z=0$ is a solution

If $a=0$ : $$(i b)^4=\lvert b \lvert $$
$$b^4=\lvert b \lvert$$

$-i$ and $i$ are solutions

If $b=0$ : $$a^4=\lvert a \lvert $$

$-1$ and $1$ are solutions. 

Finally, these are the solutions of $z^4=\lvert z \lvert$: $$0,-i,i,1,-1$$ 
Is it correct? Thanks!

From $|z|=z^4$ we get $|z|=|z|^4$ so $|z|=1$ or $|z|=0$ (so $z=0$.)
In the first case $z^4 =1$ so $$(z+i)(z-i)(z+1)(z-1)=0$$ 
so yes you find all the solutions.
