Let $f(z)=(z-e^i)(z-e^{2i})(z-e^{3i}) \cdots (z-e^{50i})$. Prove that there exist $z_1$ such that $|z_1|=1$ and $|f(z_1)|\ge 1$.
What I have already done is sadly not too much... I know that I can take any $z_1=e^{\alpha i}$ that holds $|z_1|=1$ obviously. But how to pick such $z_1$ that $|f(z_1)|\ge 1$... I'm lost.
$f(z)$ is a holomorphic function over the unit disk and $\left|f(0)\right|=1$. Since $f$ is non-constant, by the maximum modulus principle there is some $z\in S^1$ such that $\left|f(z)\right|>1$.
Numerically we have $$ \frac{1}{2\pi}\int_{0}^{2\pi}\left|f(e^{i\theta})\right|^2\,d\theta\approx 4.3834$$ hence the previous statement can be strenghtened into for some $z\in S^1$, $|f(z)|>\color{red}{2}$.