Here's the problem I'm trying to proof:
Let functions $f(z)$ and $g(z)$ be both analytic in the disc $|z| ≤ 2$, let $|f(z)| > |g(z)|$ on the circle $|z| = 2 $and let $|g(z)| > |f(z)|$ on the circle $|z| = 1$. Prove that the number of zeros of the function f(z) in the disc $|z| < 2$ is greater or equal the number of zeros of the function $g(z)$ inside the disc $|z| < 1$.
I'm wondering how I may want to use Rouche's Theorem. From the given conditions, we can find out that in the disc $|z| ≤ 2$, $f(z)$ and $f(z)+g(z)$ have the same number of zeros, and in the disc $|z| ≤ 1$, $g(z)$ and $f(z)+g(z)$ have the same number of zeros and counting multiplicities. What should I do next? How can I show that the number of zeros of the function f(z) in the disc $|z| < 2$ is greater or equal the number of zeros of the function $g(z)$ inside the disc $|z| < 1$? Is there anything other than Rouche’s theorem I should consider? Thanks for the help!
Let $N(r, f)$ denote the number of zeros of the holomorphic function $f$ in the disk $|z| < r$, counted with multiplicity.
You correctly demonstrated that $N(2, f) = N(2,f+g)$ and $N(1, g) = N(1, f+g)$.
It only remains to observe that $N(2, f+g) \ge N(1, f+g)$, which is obvious from the definition. Then $$ N(2, f) = N(2,f+g) \ge N(1, f+g) = N(1, g) $$ is the desired conclusion.