Distinct roots of $z^n-z$ How would we prove that for any positive integer $n$ the complex roots of $z^n-z$ are all distinct?
In the case that $n=1,2,3$ I have factored it directly but how can we do it in general?
 A: 
The polynomial $f(z)$ of degree $n>0$ has $n$ distinct roots if and only if it has no common root with the derived polynomial $f'(z)$.

If we assume this result, we have to check $f(z)=z^n-z$ and $f'(z)=nz^{n-1}-1$. Let $f'(\alpha)=0$, so $\alpha^{n-1}=\frac{1}{n}$. Since
$$
f(\alpha)=\alpha(\alpha^{n-1}-1)=\alpha\left(\frac{1}{n}-1\right)\ne0
$$
(unless $n=1$), we are done. The case $n=1$ should be excluded, because you get the zero polynomial.
You find the proof of the above criterion in every textbook.

Here's a sketch of the proof. Suppose $(z-\alpha)^2$ divides $f(z)$. Then
$$
f(z)=(z-\alpha)^2g(z)
$$
so
$$
f'(z)=2(z-\alpha)g(z)+(z-\alpha)^2g'(z)
$$
and so $\alpha$ is a common root of $f(z)$ and $f'(z)$.
Conversely, suppose $f(\alpha)=f'(\alpha)=0$. Then
$$
f(z)=(z-\alpha)^2q(z)+rz+s
$$
for some $r,s$. Then $f'(z)=2(z-\alpha)q(z)+(z-\alpha)^2q'(z)+r$ and from the hypotheses we get
$$
r\alpha+s=0,\qquad r=0
$$
Thus $r=s=0$.
A: From the fundamental theorem
of algebra (FTA),
you know that a
polynomial of degree $n$
with complex coefficients
has at most $n$ complex roots.
Writing your equation as
$z(z^{n-1}-1) = 0$,
we see that
$0$ is a root and
the $n-1$ roots of
$1$ are roots
(these,
by DeMoivre,
are
$e^{2\pi i k/(n-1)}$
for $k = 0 $ to $n-2$).
There are a total of $n$
distinct roots.
By the FTA,
there can be no more roots.
Therefore,
the polynomial has
exactly these
$n$ distinct roots.
A: Hint:
$$
z^n-z = z \, (z^{n-1}-1)
$$
and the $n-1$ roots of unity are different from each other (and different from $0$).
