Distinct roots of $z^n-z$

Question

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?

Answer 1

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.

Answer 2

Hint: $$ z^n-z = z \, (z^{n-1}-1) $$ and the $n-1$ roots of unity are different from each other (and different from $0$).

Answer 3

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.

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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 $r=s=0$.