Only words

Let $z=re^{\iota\theta}$ . Hence the statement is converted into the equation

$z=\bar{z}^3\\\Rightarrow re^{\iota\theta}=r^3e^{-3\iota\theta}$

Comparing absolute values, $r=r^3\Rightarrow r=0,1\quad (r\geq 0)$ .

For $r=0, z=0$

For $r=1,$ we comparing te arguments giving $\theta=2n\pi-3\theta\\\Rightarrow \theta=\frac{n\pi}{2}$

which gives 4 values of $z$ which are $\iota,-\iota,-1,1$ .

Therefore total of 5 values (including 0).

Using Elementary Algebra. The problem means "How many solution of $a+bi$ such that $a - bi = (a+bi)^3$

Case 1: If $a \neq 0$ and $b \neq 0$ this follow that $a- bi \neq (a+bi)^3$

Case 2: If $a = 0 \Rightarrow -bi = -ib^3$

Solving for $b$ . So $b = 1,-1, 0$ . Therefore, $i, -i, 0$ is the solution for case 2

Case 3: If $b = 0 \Rightarrow a = a^3$

Solving for $a$ . So $a = 1, -1, 0$ . Therefore, $1, -1, 0$ is the solution for case 3.

Therefore, $i, -i, 1, -1, 0$ are the solution. They have $5$ solution.