# How would I prove that if an integer $n>2$, then $\bar{z}=z^{n-1}$ has $n+1$ solutions

If $n$ is an integer and $n>2$, then $$\bar{z}=z^{n-1}$$ has $n+1$ solutions. Does this have something to do with the rational root theorem?

• $|z|=0$ or $|z|=1$. If $|z|=1$, we have $\overline{z}=\frac{1}{z}$, so that $z^{n}=1$, it gives $n$ solutions and $|z|=0$ gives the $(n+1)$-th solution. – MoebiusCorzer Nov 26 '15 at 16:20

Since $|z|=|\overline{z}|$, taking absolute values gives $$|z|=|z|^{n-1}$$ which, as $|z|$ is a non-negative real number, means $|z|=0$ or $|z|=1$.
The equation is obviously satisfied when $z=0$. If $|z|=1$, then $\overline{z}=\frac{1}{z}$, which means the equation reduces to $$\frac{1}{z}=z^{n-1}$$ whose solutions are the $n$th roots of unity.
Let $z=re^{it}$. We then have $$re^{-it} = r^{n-1}e^{(n-1)it}$$ This means $$r=0 \text{ or }r^{n-2} e^{nit} = 1$$ Now $r^{n-2} e^{nit} = 1$ gives us $r=1$ and $t = \dfrac{2 \pi k}n$, where $k \in \{0,1,\ldots,n-1\}$, which gives us a total of $n+1$ solutions (including the solution $z=0$).