IMC 1995 · Day 2 · P9

It is enough to consider only polynomials with leading coefficient 1. Let $P(z) = (z - \alpha_1)(z - \alpha_2) \dots (z - \alpha_n)$ with $|\alpha_j| = 1$, where the complex numbers $\alpha_1, \alpha_2, \dots, \alpha_n$ may coincide.

We have $$\begin{align*} \widetilde{P}(z) \equiv 2 z P'(z) - n P(z) &= (z + \alpha_1)(z - \alpha_2) \dots (z - \alpha_n) \\ &\quad + (z - \alpha_1)(z + \alpha_2) \dots (z - \alpha_n) + \cdots + (z - \alpha_1)(z - \alpha_2) \dots (z + \alpha_n). \end{align*}$$ Hence, $\dfrac{\widetilde{P}(z)}{P(z)} = \sum\limits_{k=1}^{n} \dfrac{z + \alpha_k}{z - \alpha_k}$. Since $\operatorname{Re} \dfrac{z + \alpha}{z - \alpha} = \dfrac{|z|^2 - |\alpha|^2}{|z - \alpha|^2}$ for all complex $z$, $\alpha$, $z \ne \alpha$, we deduce that in our case $\operatorname{Re} \dfrac{\widetilde{P}(z)}{P(z)} = \sum\limits_{k=1}^{n} \dfrac{|z|^2 - 1}{|z - \alpha_k|^2}$. From $|z| \ne 1$ it follows that $\operatorname{Re} \dfrac{\widetilde{P}(z)}{P(z)} \ne 0$. Hence $\widetilde{P}(z) = 0$ implies $|z| = 1$.