$|z|=z^5$ How many solutions does this equation have?

Question

Multiple choice question : How many solutions does the equation $|z|=z^5$ have?

A - $1$ solution

B - $2$ solutions

C - $5$ solutions

D - $6$ solutions

These were the $4$ possible answers.

I started by allowing $z^5\in \Bbb R$.

$|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.

I found $2$ obvious solutions, $z=1$ and $z=0$ However I couldn't find any other complex solutions; and the B answer was not correct.

I have also tried to expand with $z=a+ib$ but couldn't find anything, still.

The correct answer was D but I couldn't understand why.

Excuse my English, I am not used to doing maths in English.

Answer 1

Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.

Answer 2

It is indeed true that$$\lvert z\rvert=z^5\implies\lvert z\rvert=0\vee\lvert z\rvert=1.$$However, I don't understand your proof. You can prove it as follows:\begin{align}\lvert z\rvert=z^5\implies&\bigl\lvert\lvert z\rvert\bigl\lvert=\lvert z^5\rvert\\\iff&\lvert z\rvert=\lvert z\rvert^5\\\iff&\lvert z\rvert=0\vee\lvert z\rvert^4=1\\\iff&\lvert z\rvert=0\vee\lvert z\rvert=1.\end{align}If $\lvert z\rvert=0$, then $z=0$ and $0$ is indeed a solution. And if $\lvert z\rvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.

Answer 3

$$ \eqalign{ & \left| z \right| = z^{\,5} \,\quad \mathop \Rightarrow \limits^{z = Ae^{\,i\,\alpha } } \quad A = A^{\,5} e^{\,i\,5\alpha } \quad \Rightarrow \cr & \Rightarrow \quad \left\{ {\matrix{ {A = A^{\,5} } \cr {e^{\,i\,5\alpha } = 1} \cr } } \right.\quad \Rightarrow \quad \left\{ {\matrix{ {A = 0,1} \cr {\alpha = 2k\pi /5} \cr } } \right.\quad \Rightarrow \cr & \Rightarrow \quad z = 0,e^{\,2k\pi /5} \;\left| {\;k = 0, \cdots ,4} \right.\quad \Rightarrow \quad 6\quad sol. \cr} $$