# Cauchy's integral formula or Bounds for analytic functions

The problem is here.

Consider the polynomial $P(z)=a_0+a_1z+...+a_{n-1}z^{n-1}+z^n$. Show that, for all sufficiently large values of $r$, there must exist points $z_0, z_1$ on the circle $|z|=r$ such that $|e^{P(z_0)}|=e^{(1/2)r^n}, |e^{P(z_1)}|=e^{-(1/2)r^n}$.

I tried to show that there exist $z_0, z_1$ with same modulus $r$ s.t $Re(P(z_0))= \frac{1}{2}r^n =-Re(P(z_1))$, but failed. I don't know why the statement 'sufficiently large values of $r$' is given. Also I thought Fundamental theorem of algebra, but I can't use the theorem to solve this problem. I learned Liouville's thm, maximal modulus property, Cauchy's integral formula and so on... but I can not use any of them! Please help me! Thank you for reading.

## 1 Answer

For $t\in U=\{t; |t|=1\}$ put $f(t)=P(rt)+\overline{P}(r/t)-r^n$, where $\overline{P}(z)=\overline{a_0}+\cdots+\overline{a_{n-1}}z^{n-1}+z^n$ is the polynomial with its coefficients conjugate to those of $P$. Then your problem is to show that the function $f$ (from $U$ to $\mathbb{R}$) takes the value $0$. Now, as $U$ is connected, and $f$ is clearly continuous, the image $I=f(U)$ of $U$ by $f$ is an interval $I$. We have $f(1)=P(r)+\overline{P}(r)-r^n=r^n+Q(r)$, with $Q$ polynomial with coefficients in $\mathbb{R}$ of degree $\leq n-1$. Hence for $r$ sufficiently large, it takes values $>0$. Do the same for $f(\exp(i\pi/n))$, and you show that for $r$ sufficiently large, it takes values $<0$. Hence for $r$ sufficiently large, as $I$ contain positive and negative numbers, it contain $0$, and there exists $t\in U$ with $f(t)=0$.

I have not done the computations for the second case, but I suppose that it has a similar proof.