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!
 A: 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.
A: You are right that those are all the solutions, but I think there are easier ways to get there. For instance, I generally try to hold off on substituting $z = a+bi$ for as long as possible, since it rarely makes things nicer.
First off, begin by taking absolute values on both sides, giving $|z| = |z|^4$. This has the benefit of making the equation about a single, real number $|z|$ instead of one complex number $z$ and one real number $|z|$, which means it will be easier to extract useful information.
Since $|z|$ is a non-negative, real number, we must have either $|z| = 0$ or $|z| = 1$. Clearly $z = 0$ solves the original equation, so we look at what's left: $|z| = 1$.
If $|z| = 1$, the original equation reads $z^4 = 1$. This is well-known to have the four solutions $\pm 1, \pm i$. We insert back into the original equation to double-check that they are indeed solutions, and we're done.
A: If we define $z = re^{i\theta}$, we have $|z| = r$ and $z^4 = r^4e^{4i\theta}$. Notice that they are equal when $r^3e^{i(4\theta)} = 1$ or $r = 0$. From here, you can easily find $\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ with $r = 1$ and corresponding $z$ values will be $1, i, -1, -i$ and for $r = 0$, we have $z = 0$ so you have already found all the solutions.
A: The result is correct, but it faster with the exponential form of complesx numbers: if $z=r\,\mathrm e^{i\theta}$, the equation becomes
$$r=r^4\,\mathrm e^{4i\theta}\iff\begin{cases}r=0\quad\text{or}\\
r^3=1\:\wedge\:\mathrm e^{4i\theta}=1\end{cases}\iff \begin{cases}z=0\quad\text{or}\\ r=1\:\wedge\: \theta\equiv 0\mod\frac{2\pi}4=\frac\pi2. \end{cases}$$
A: Your first method is a bit unclear because of the use of the fourth root: it is not true, even for a real number, that $\sqrt {z^2}=z$. A good piece of advice is to avoid "taking roots" unless unavoidable. 
Your second method works by chance due to the fact that in this particular case all solutions have either real or imaginary part equal to zero. 
One "clean" way of doing this is to see the equation as 
$$
r^4e^{4i\theta}=r.
$$
This forces $r^4=r$, so $r=0$ or $r=1$; if $r=1$, we also have  $e^{4i\theta}=1$, which gives  the other four solutions $1,-1,i,-i$.
A: Yes you are correct, indeed note that
$$\begin{cases}\lvert z \lvert = z^4 \iff |z|=0 \quad \lor\quad |z|=1\\\\z^4 = \lvert z \lvert =\bar z^4 \iff \mathcal{Im}(z)=0\end{cases}$$
thus all the non trivial solutions $z\neq0$ are
$$z=e^{ik\frac{\pi}{2}} \quad \forall k \in \mathbb{Z}$$
