Prove: a $\mathbb{C}\to\mathbb{C}$-polynomial of degree $>1$ is not injective

Question

Show that a polynomial of degree $>1$ is not injective (as $\mathbb{C}\to\mathbb{C}$-map).

My attempt:

Consider $z_0\in\mathbb{C}$, then the polynomial $f$ is holomorphic in $z_0$. Suppose that $f$ is injective. In particular, $f$ is injective in a neighborhood of $z_0$, implying that $f'(z_0)\ne 0$. Thus $\operatorname{mult}_{z=z_0}f(z)=1$, so that $f(z)=(z-z_0)g(z)$ for $g:\mathbb{C}\to\mathbb{C}$ holomorphic in $z_0$ and $g(z_0)\ne 0$.

I want to show that $g$ has to be a constant function, because this would imply that $\deg f = 1$, yielding a contradiction. How can I do this?

Thanks.

Answer 1

Since $f$ is a polynomial function, $g$ is a polynomial function too. But $g$ has no zeros. Therefore, by the Fundamental Theorem of Algebra, it is a constant function.

Answer 2

It's not restrictive to assume the polynomial is monic. If it has two distinct roots $a$ and $b$, then the function is not injective as $f(a)=0=f(b)$.

Otherwise $f$ has a single root, hence it is of the form $f(z)=(z-a)^n$, where $n>1$ is the degree. Since there are $n$ distinct $n$-th roots of unity, the function is not injective.