Consider the integral $$N(\lambda) = {1 \over 2 \pi i} \oint_C {f'(z) + \lambda g'(z) \over f(z) + \lambda g(z)} dz$$ where $0 \le \lambda \le 1$ . By the hypotheses, the function $f + \lambda g$ is non-singular on $C$ or on the interior of $C$ and has no zeros on $C$ . Hence, by the argument principle, $N(\lambda)$ equals the number of zeros (counted with multiplicity) of $f + \lambda g$ contained inside $C$ . Note that this means that $N(\lambda)$ must be an integer.

Since $C$ is compact, both $|f|$ and $|g|$ attain minima and maxima on $C$ . Hence there exist positive real constants $a$ and $b$ such that $$|f(z)| > a > b > |g(z)|$$ for all $z$ on $C$ . By the triangle inequality, this implies that $|f(z) + \lambda g(z)| > a - b$ on $C$ . Hence $1/(f + \lambda g)$ is a continuous function of $\lambda$ when $0 \le \lambda \le 1$ and $z \in C$ . Therefore, the integrand is a continuous function of $C$ and $\lambda$ . Since $C$ is compact, it follows that $N(\lambda)$ is a continuous function of $\lambda$ .

Now there is only one way for a continuous function of a real variable to assume only integer values - that function must be constant. In particular, this means that the number of zeros of $f + \lambda g$ inside $C$ is the same for all $\lambda$ . Taking the extreme cases $\lambda = 0$ and $\lambda = 1$ , this means that $f$ and $f+g$ have the same number of zeros inside $C$ .