If $f(z)$ has a repeated root, then the curves ${C_{\theta}}(f)$ are all singular.

If $f'(z) = 0$ and $f(z) \neq 0$, then $z$ lies on ${C_{\theta}}(f)$ for exactly one value of $\theta \in {{\mathbb R}}/\pi{{\mathbb Z}}$, namely $\theta \equiv \arg(f(z)) \pmod{\pi}$. On the other hand, if $f'(z)=0$ and $f(z)=0$, then $z$ is a singular point on all ${C_{\theta}}(f)$.

We can now prove:

Every polynomial $f(z)$ of degree $n$ has exactly $n$ roots.

We may suppose that $f(z)$ is monic. If $f(z)$ has a repeated root, then it certainly has a root; by induction on the degree, $f(z)$ has exactly $n$ roots.

Assume, then, that $f(z)$ has no repeated roots. Choose any two different angles $\alpha,\beta \in {{\mathbb R}}/\pi{{\mathbb Z}}$ such that $C_\alpha(f)$ and $C_\beta(f)$ are nonsingular; this is possible by Corollary \[cor:repeated\]. Applying the argument which proved Proposition \[prop:nonsing\] to the curves $C_\alpha(f)$ and $C_\beta(f)$, the theorem follows.

The idea at the heart of Gauss’s proof is that when $r$ is sufficiently large, the $2n$ points of $C_\alpha(f) \cap S_r$ are paired via components of $C_\alpha(f)$. This associates to $f$ a purely combinatorial object: a *matching*, or partition into $n$ subsets of size two, of the set $C_\alpha(f) \cap S_r$. 