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I'm trying to solve this question but I am having trouble connecting the dots. The question reads:

Assume that we have a complex polynomial: $$P(z) = a_0+a_1z+...+a_nz^n$$ Satisfies $|P(z)|\leq 1$ whenever $|z|=1$. Show that $|a_n|\leq 1 \>\>\forall n$.

So, I have simplified $|P(z)|\geq|a_n|\left|1-\frac{|a_{n-1}|}{|a_n|} -...-\frac{|a_0|}{|a_n|}\right|$, using the fact that $|z|=1$.

Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?

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    $\begingroup$ Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle. $\endgroup$ Commented Nov 12, 2018 at 4:55

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This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+\cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(\frac 1 z)$. $q$ is a polynomial and $|q(z)| \leq 1$ if $|z|=1$. By MMP $|q(0)| \leq 1$ which is what we need.

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  • $\begingroup$ Ah, I misunderstood the question – I thought it was asking for $|a_k|\le1$ for $0\le k\le n$. $\endgroup$ Commented Nov 13, 2018 at 10:06
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One way to show this is via Cauchy estimates. From Cauchy integral formula, for the $n$-th derivative of $P$ we have $$ P^{(n)}(z) = \frac{n!}{2\pi i } \int_{|\xi| = 1} \frac{P(\xi)}{(z - \xi)^{n+1}}d\xi. $$ Hence, using the fact that $|P|\leq 1$ on $|\xi| = 1$, we obtain $$ n! |a_n| \leq \frac{n!}{2\pi } \int_{|\xi| = 1} \frac{|d\xi| }{|z - \xi|^{n+1}} = \frac{n!}{2\pi} \int\limits_0^{2\pi} \left|\frac{ie^{i\theta}} {(e^{i\theta})^{n+1}}\right| d\theta = n!, $$ which gives $|a_n|\leq 1$.

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