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I've seen many questions that regard a monic polynomial, where the main trick is to use Cauchy's integral formula to deduce about $P(z)$.
Well, this question is concerned about a general $P(z)$.
If $P(z)=a_nz^n+\ldots+a_0$ is a polynomial of degree n, and $max_{|z|=1}|P(z)|\leq1$, then $|P(z)|\leq|z|^n$ when $|z|>1$.

I've managed to achieve $|a_n|\leq1, |a_0|\leq1, |a_n+\ldots+a_0|\leq1$, but I need to have a bound on $|a_n|+\ldots+|a_0|$, no?

Lior Pollak
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Let $w=1/z$ for $|z| \ge 1$ and note that $Q(w)=\frac{P(z)}{z^n}=w^n\sum_{k=0}^n a_k/w^k=\sum a_{k}w^{n-k}$ is a polynomial in the closed unit disc that satisfies $|Q(w)|=|P(z)| \le 1$ for $|w|=1$ since then $|z|=1$ also.

By the maximum modulus one has $|Q(w)| \le 1$ in the unit disc or $|P(z)| \le |z|^n$ outside it

Conrad
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  • Why is that $|Q(\frac{1}{z})|=|P(z)|$ in the unit circle? – Lior Pollak Jan 23 '22 at 16:54
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    because $|z|=1$ so $|\frac{P(z)}{z^n}|=|P(z)|$ – Conrad Jan 23 '22 at 16:57
  • We have that for $z$ in the interior of the disc $Q(z)=\int_{\partial D}\frac{Q(w)}{w-z}dw$. How do we bound $Q(z)$ using the value on the boundary? It doesn't seem completely clear to me. I thought the maximum modulus principle was for bounds on a connected open set... – Kadmos Mar 25 '24 at 16:12
  • Maximum modulus applies on any connected open set and bounded holomorphic functions inside (if the functions are continuous on the boundary then one can use the maximum there, if not one uses $\limsup |f|$ at the boundary) – Conrad Mar 25 '24 at 16:16