Question 142
Let $n$ be a positive integer. The polynomial with complex coefficients
$$
P(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0 \quad (a_n \neq 0)
$$
satisfies: for any complex number $z$ with $|z| = 1$, we have $|P(z)| \le 1$.
Prove that for any $k \in \{0, 1, \dots, n-1\}$, we have $|a_k| \le 1 - |a_n|^2$.
$$
P(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0 \quad (a_n \neq 0)
$$
satisfies: for any complex number $z$ with $|z| = 1$, we have $|P(z)| \le 1$.
Prove that for any $k \in \{0, 1, \dots, n-1\}$, we have $|a_k| \le 1 - |a_n|^2$.