There is a theorem in complex analysis which states that if $\gamma$ is a closed simple curve in a domain $G$ of the complex plane, and $a_1,…,a_n$ are its zeros (repeated according to multiplicity) then $\frac{1}{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}dz$ equal the number of zeros of $f$ inside of $\gamma$.
Suppose $p(z)$ is a polynomial of degree $n$. I would like to use the above theorem to prove that $p$ must have exactly $n$ zeros. My idea is the following. I would like to show that $\frac{1}{2\pi i}\int_{|z|=R}\frac{p'(z)}{p(z)}dz=n$ for large $R$. It think there might be a way to use the M-L inequality, however I'm stuck how to proceed. What should I try from here?
Best Answer
Hint: Show that $p'(z)/p(z) = n z^{-1} + O(|z|^{-2})$ as $|z|\to \infty$ (maybe figure out a more precise form of(/an upper bound for) the $O(...)$ term), and then use this to estimate second term in the following identity $$ \frac{1}{2\pi i} \int_{|z|=R} \frac{p'(z)}{p(z)} \, dz = \frac{1}{2\pi i} \int_{|z|=R} \frac{n}{z} \, dz + \frac{1}{2\pi i} \int_{|z|=R} O(|z|^{-2}) \, dz. $$ Can you see that the last term should decay to zero for large $R$? To show the result, use the fact that the expression $ \frac{1}{2\pi i} \int_{|z|=R} \frac{p'(z)}{p(z)} \, dz$ can only take integer values.