Proof. Let $n$ denote the degree of $f$ . Without loss of generality, the assumption can be made that the leading coefficient of $f$ is $1$ . Thus, $\displaystyle f(z)=z^n+\sum_{m=0}^{n-1} c_m z^m$ .

Let $\displaystyle R=1+\sum_{m=0}^{n-1}|c_m|$ . Note that, by choice of $R$ , whenever $|z|>R$ , $f(z) \neq 0$ . Suppose that $|z| \ge R$ . Since $R \ge 1$ , $|z^a| \le |z^b|$ whenever $0<a<b$ . Hence, we have the following string of inequalities:

$$\left| \sum_{m=0}^{n-1} c_m z^m \right| \le 1 + \sum_{m=0}^{n-1} |c_m| |z^m| \le |z^{n-1}| + \sum_{m=0}^{n-1} |c_m| |z^{n-1}| \le R |z^{n-1}| \le |z^n|$$

Since polynomials in $z$ are entire, they are certainly analytic functions in the disk $|z| \le R$ . Thus, Rouché's theorem can be applied to them. Since $\displaystyle \left| \sum_{m=0}^{n-1} c_m z^m \right| \le |z^n|$ for $|z| \ge R$ , Rouché's theorem yields that $z^n$ and $f(z)$ have the same number of zeroes in the disk $|z| \le R$ . Since $z^n$ has a single zero of multiplicity $n$ at $z=0$ , which counts as $n$ zeroes, $f(z)$ must also have $n$ zeroes counted according to multiplicity in the disk $|z| \le R$ . By choice of $R$ , it follows that $f$ has exactly $n$ zeroes in the complex plane.