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I got stuck on this problem:

Given a polynomial on complex plane $P(z) = z^n + a_{n-1}z^{n-1} + ... + a_1 z + a_0$ for $z \in \mathbb{C}$. Prove that $\max_{|z| = 1} |P(z)| \ge 1$

What I tried so far: For this problem, I thought about the Maximum Modulus Principle, so I achieved that the maximum of $|P(z)|$ on the disk $\overline{B}(0,1)$ is gained on the boundary $|z| = 1$. So suppose $\max_{|z| = 1} |P(z)| \lt 1$, then we must have that $|P(z)| \lt 1$ for all $z \in \overline{B}(0,1)$. So, I tried to choose some special value to get the contradiction, like choose $z = 0$, I have $|a_0| < 1$, $z = 1, -1, i, -i...$. But I still haven't come to the conclusion. Can anyone give me some hint to go on? Thanks a lot.

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I think I found the solution. Continuing with my reasoning, applying the Cauchy formula, we have: $$1 = a_n = {1 \over 2\pi i}\oint_{|z| = 1} {P(t) \over t^{n+1}} dt$$ Therefore, we have: $$1 = |a_n| = |{1 \over 2\pi i}\oint_{|z| = 1} {P(t) \over t^{n+1}} dt| \le {1 \over 2\pi}\max_{|z| = 1}|P(z)| l(|z| = 1) = \max_{|z| = 1}|P(z)|$$ Here $l(\gamma)$ is the length of the curve $\gamma$. And the inequality is from Estimation lemma

Therefore we proved the problem.

Please let me know if you see any issue with my solution. Thanks!

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    $\begingroup$ The $P(0)$'s should be $P(t)$'s, and since you've found $1 \le \max_{\lvert z\rvert = 1} \lvert P(z)\rvert$ as desired, there's no need to argue by contradiction. $\endgroup$ – kobe Apr 19 '16 at 3:22
  • $\begingroup$ Yeah, that's right. Thanks @kobe $\endgroup$ – le duc quang Apr 19 '16 at 3:58
  • $\begingroup$ Yeah, I edited it. A consequence from copy-paste the expression. Thanks $\endgroup$ – le duc quang Apr 19 '16 at 4:03
  • $\begingroup$ No problem. Your proof looks good now. :) $\endgroup$ – kobe Apr 19 '16 at 4:04
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Your method is probably having difficulty since a function like $z^n$ is in fact bounded by $1$ on the interior of the disk. Thus, choosing any "special" values on the interior - or any set of special values - cannot suffice as proof. It's perhaps possible that this exercise could be completed purely algebraically, as you are trying to do, but I wouldn't say that's a hopeful strategy.

A better method would be to use Cauchy's integral formula to relate the values on the boundary to the values and derivatives inside. In particular, to really get out of the algebraic mind set, try thinking about the following statement, true of functions on $\mathbb C$:

If $f$ is a holomorphic function and $f^{(n)}(0)=n!$, then $\max_{|z|=1}|f(z)|\geq 1$.

The idea to solving this is to use the formula $$f^{(n)}(0)=\frac{n!}{2\pi i}\oint \frac{f(z)}{z^{n+1}}\,dz$$ where the integral is taken over the boundary unit disk $|z|=1$, oriented counter clockwise. However, if you take absolute values, you get $$\left|f^{(n)}(0)\right|=\left|\frac{n!}{2\pi}\oint \frac{f(z)}{z^{n+1}}\,dz\right|\leq \frac{n!}{2\pi i}\oint\left|\frac{f(z)}{z^{n+1}}\right|=\frac{n!}{2\pi}\oint |f(z)| \leq n!\max_{|z|=1}|f(z)|$$ and you are essentially done.

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  • $\begingroup$ Oh, I just saw your answer after posting my solution. Actually, they're very similar, right? Thanks for your help! $\endgroup$ – le duc quang Apr 19 '16 at 3:15
  • $\begingroup$ @kobe Thanks; I fixed the issues. $\endgroup$ – Milo Brandt Apr 19 '16 at 13:29

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