I am studying for an exam in complex analysis and I have the following practice problem:
Let $P(z)=a_n z^n+\dots+a_1 z+a_0$ with $a_n,a_0\neq 0$, and assume $P(z)\neq 0$ on $|z|=1$. Let $P^*(z)=\overline{a_0}z^n+\overline{a_1}z^{n-1}+\dots+\overline{a_n}$. Prove that if $|a_0/a_n|>1$, then $P(z)$ and $\overline{a_0}P(z)-a_n P^*(z)$ have the same number of zeros in $|z|<1$.
I believe this problem should be solved with Rouche's theorem: if $|P(z)|=|P^*(z)|$ on $|z|=1$, then we may take $f(z)=\overline{a_0}P(z)$, $g(z)=-a_n P^*(z)$, and we have $|f(z)|>|g(z)|$ on $|z|=1$ and the result follows.
However, I am unable to show $|P(z)|=|P^*(z)|$ on $|z|=1$. I am unsure if I am approaching this problem the right way. Any help is very appreciated, thank you.