Number of solution of a complex equation with high degree I am asked to give the number of complex solution of the following equation:
$z^5 = |z|$
I tried to proceed by replacing $|z|^2$  by  $zz^*$
(where $z^*$ is the complex conjugate of $z$).
But then the equation becomes:
$z^9=z^*$
I really don't know how to solve either of the equations.
Thank you for your help.
 A: $$\rho^5e^{i5\theta}=\rho,$$
then $\rho=0$ or
$$\rho^4e^{i5\theta}=1.$$
As the RHS is real, $e^{i5\theta}$ as well and there are five additional solutions,
$$\rho=1,\\\theta=\frac{2k\pi}5.$$
A: 
According to OPs first step we use $|z|^2=z\overline{z}$ and we get
  \begin{align*}
z^{10}&=z\overline{z}\\
\end{align*}
We see $z=0$ is a solution. Excluding it from now on we can divide by $z$ and get
  \begin{align*}
z^9&=\overline{z}
\end{align*}

Now we follow the comment to OPs question.

It's convenient to use polar coordinates and represent $z=re^{i\varphi}$ with $r> 0$ and $0\leq \varphi < 2\pi$. ($r>0$, since we have excluded $z=0$). We obtain
  \begin{align*}
r^9e^{9i\varphi}&=re^{-i\varphi}\\
r^8e^{10i\varphi}&=1
\end{align*}
  Since $r>0$, $r=1$ follows and we get 
  \begin{align*}
e^{10i\varphi}=1
\end{align*}

The last equation implies since $e^{2k\pi i}=1$ with $k\in \mathbb{Z}$
\begin{align*}
10i\varphi&=2k\pi i\\
\varphi&=\frac{k}{5}\pi \qquad k=0,2,4,6,8
\end{align*}
Note: The selection of $k$ is according to $\arg(z^5)=\arg(|z|)=0$ and we have to respect that $0\leq \varphi < 2\pi$.
$$ $$

We conclude the number of solutions of the equation $z^5=|z|$ is $6$, with solutions
  \begin{align*}
z\in\{0\}\cup\left\{\left.e^{\frac{2k\pi i}{5}}\right|k=0,\ldots,4\right\}
\end{align*}

A: Somebody followed the comment's suggestion, but used the wrong $|z|=zz^\ast$, and when I pointed this out, deleted the answer. Let me do that with the correct $|z|^2=zz^\ast$. The equation becomes, by squaring:
$$z^{10}=zz^\ast.$$
$z=0$ is a solution, so dropping it we can divide by $z$ and we get:
$$z^9=z^\ast,$$
as OP says. Now we follow the suggestion, and set $z=\rho e^{i\theta}$. Substitute that and we get:
$$\rho^9e^{i9\theta}=\rho e^{-i\theta}\iff\rho^8e^{i10\theta}=1.$$
But then we have $\rho^8=1$, or $\rho=\pm1$ but $\rho\geq0$ so $\rho=1$. Then $e^{i10\theta}=1$, which means $\theta=\frac{2\pi+2k\pi}{10}=\frac{\pi}{5}+k\frac{\pi}{5}$, where $k=0,\dotsc,9$.
Then again, without using the expression for $|z|$ we just substitute and get:
$$\rho^5e^{i5\theta}=\rho\iff\rho^4e^{i5\theta}=1.$$
So again $\rho=1$ and $e^{i5\theta}=1$ so $\theta$ is as above. So using $|z|^2=zz^\ast$ yields a longer path in this case.
Edit
As the deleter of the now-undeleted A made me notice in comments, squaring the equation adds 5 extra solutions. Only the odd $k$s are actual solutions to the equation, whereas the even $k$s solve the squared equation but not the starting one. Indeed, taking an even $k$ shows an argument of $\frac{\pi+2n\pi}{5}$, which doesn't solve the original equation as that equation needs $e^{i5\theta}=1$, as I showed just above, and such an argument gives $e^{i5\theta}=-1$. Hence, the equation has 5 solutions ($k=0,2,4,6,8$ above) plus the null solution. Let me list them:
$$e^{i(\frac{\pi}{5}+\frac{\pi}{5})},e^{i(\frac{\pi}{5}+3\frac\pi5)},e^{i(\frac\pi5+5\frac\pi5)},e^{i(\frac\pi5+7\frac\pi5)},e^{i(\frac\pi5+9\frac\pi5)},0$$
Or:
$$0,e^{i\frac25\pi},e^{i\frac45\pi},e^{i\frac65\pi},e^{i\frac85\pi},e^{i\frac{10}{5}\pi}=1.$$
