This question was asked in my quiz and I was unable to solve it.
If $p$ is a polynomial and if maximum of $|p|$ on a region $X$ is attained at an interior point, then without using the maximum modulus principle, show that $p$ is constant.
Examiner also gave a hint: Compute the integral of $\frac{p(z)}{z-a}$ over the circle with centre has contained in $X$.
I computed the integral using residue theorem. It comes out 2$\pi$ i $p(a)$. But I have no idea on how can I use it to prove $p$ to be constant.
Kindly give some hints.
Best Answer
The value of the integral is $2\pi ip(a)$ not $p(a)$. Also you don't need Residue Theorem . You can explicitly evaluate the integral by integrating each term in the polynomial. The idea of this exercise is to show that you can avoid all big theorems.
The integral of $\frac {p(z)} {z-a}$ over the circle with center $a$ and radius $r$ is $i \int_0^{2\pi} p(a+re^{it}) dt$. Hence we get $p(a)= \frac1 {2\pi}\int_0^{2\pi} p(a+re^{it}) dt$ and $|p(a)| \leq \frac1 {2\pi}\int_0^{2\pi} |p(a+re^{it})| dt$. But $|p(a)|$ is the maximum value of the integrand so this implies that $|p(a+re^{it})|=|p(a)|$ for all $t$. Varying the radius $r$ we conclude that $|p(z)|$ is a constant in some disk around $a$. Henec $p$ is a constant polynomial.