Inf of the Sup of complex polynomials Consider $\mu\left(P\right)=\underset{\left|z\right|=1}{\text{sup}}\left|P\left(z\right)\right|$ where $P$ is a complex polynomial. Then let $C_n$ be the set of complex polynomials of degree $n$ such that $P=X^n+\sum_{k=0}^{n-1}a_kX^k$. I've proved that
$$
\forall P \in C_n, \ \underset{\left|z\right|=1}{\text{sup}}\left|\prod_{k=1}^{n}\left(z-a_k^2\right)\right|\leq \mu\left(P\right)^2
$$
I'm asked to deduce from this the value of $\mu_n=\underset{P \in C_n}{\text{inf}}\mu\left(P\right)$ for $n \geq 1$.
I know that we have a minoration that is satisfied by $\mu_n$ but I can't find its value, any hint ?
 A: Actually using the inequality above the result follows directly (without Bernstein's inequality which is a non-trivial result), as $P(z)=z^n$ shows that $\mu_n \le 1$ while the inequality above applied to a sequence $P_n, \mu(P_n) \to \mu_n$, shows that $\mu_n \le LHS \le \ \mu(P_n)^2 \to \mu_n^2$, so $\mu_n \le \mu_n^2 \le \mu_n$ since $\mu_n \le 1$, so $\mu_n =1$ or $\mu_n =0$. We show below that $\mu_n >0$, hence $\mu_n =1$ for all $n \ge 1$
We can assume $n \ge 2$ as by elementary geometry we see that given arbitrary complex $a$ there is at least an angle $\theta$ s.t. $|e^{i\theta}-a| \ge 1$. If $a=0$ all angles will do, otherwise join $a$ with $0$ by a line and take $\theta$ the angle where the line crosses the unit circle away from $a$, so $e^{i\theta}, 0, a$ are collinear and in this order.
Now assume $n \ge 2$ and then if $P(z)=(z-z_1)...(z-z_n)$ there is an argument $|\theta - \arg{z_k}| \ge \frac{\pi}{n}$ for all $k=1,..n$ (as the largest distance between consecutive arguments on the unit circle is at least $\frac{2\pi}{n}$, so one can choose $\theta$ in the middle there, and then obviosuly $|e^{i\theta}-z_k| \ge \sin{\frac{\pi}{n}}$ for all $k$ by elementary geometry, so $|P(e^{i\theta})|\ge (\sin{\frac{\pi}{n}})^n$, hence $\mu_n \ge (\sin{\frac{\pi}{n}})^n >0$
