# Find all the points which satisfy $z^n = z$, where $z$ is a complex number. How many different solutions are there altogether?

Question: Find all the points which satisfy $z^n = z$, where $z$ is a complex number, and $n$ is a whole number where $2 ≤ n ≤ 9$. How many different solutions are there altogether?

My attempt:

Usually with these root questions using complex numbers , I like to convert into polar form.

$$z^n = z$$

$$z = r(cis(\theta))$$

$$z^n = r^n(cis(n\theta))$$

$$z = r^{n * \frac{1}{n}} (cis \frac{1}{n}(n\theta + 2\pi \times k))$$

where k is any integer

$$z = r^1 (cis (\theta+\frac{2\pi k}{n}))$$

If this is correct where do I go from here? Or if this is wrong please show me the correct way :)

Either $z=0$ or $z \neq 0$, in which case you can divide by $z$ (note that $z=0$ is always a solution to the equation). If you divide the equation by $z$, you get $z^{n-1} = 1$ ; so the solutions are the $(n-1)^{\text{th}}$ roots of units. Therefore your solution set for a fixed $n$ is $$\{0 \} \cup \{ e^{2\pi i k/(n-1)}, k \in \{0,\cdots,n-2\} \}.$$ Afterwards you look at all your solutions and look at the distinct ones ; in other words, you need to check if a solution to $z^3=z$ is not a solution to $z^9=z$ for instance. But if $z^3 = z$, then $$z^9 = (z^3)^3 = z^3 = z$$ so you can remove these cases in your counting.
P.S. : If you converted to polar form, you would have to divide by $z$ eventually to solve for the radius and angle anyway. This is the only reasonable way to proceed.