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Let $p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_0$.

My question is: Is there an elementary way to show that for all $r > 0$ $$ \max \limits _{|z| = r} |p(z)| \ge |a_n|r^n$$ without using complex analysis machinery that falls out of the Cauchy integral formula?

One way to do it with the Cauchy integral formula:

$$ |a_n| \le \frac{1}{2 \pi} \Bigg|\int \limits_{|z| = r} \frac{f(\zeta)}{\zeta ^{n+1}}\ \text{ }d\zeta \Bigg| \le \frac{1}{2 \pi} 2 \pi r \max \limits _{|z| = r} \Bigg| \frac{p(z)}{z^{n+1}} \Bigg | = \max \limits _{|z| = r} \Bigg| \frac{p(z)}{z^{n}} \Bigg | $$

(I was looking at a previous post Domination of complex-value polynomial by highest power)

Thanks!

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  • $\begingroup$ you can do it using Rouche's theorem instead. Of course, it is still Cauchy implicitly... $\endgroup$
    – leshik
    Jul 4, 2013 at 3:07
  • $\begingroup$ Thanks - Rouche is verboten. ;) $\endgroup$
    – bryanj
    Jul 4, 2013 at 3:08

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You can actually almost avoid complex analysis (or rather Cauchy integral formula in the full generality) by averaging over the circle. Namely, you can check that $$\frac{1}{2\pi}\int_{0}^{2\pi}|P(re^{i\theta})|^2d\theta=|a_n|^2r^{2n}+|a_{n-1}|^2r^{2n-2}+....|a_0|^2.$$

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    $\begingroup$ well, you just need to evaluate an integral of the function $e^{ik\theta}$ over $[0,2\pi],$ that can be done without Cauchy. $\endgroup$
    – leshik
    Jul 4, 2013 at 3:20
  • $\begingroup$ Very nifty indeed. Thanks! $\endgroup$
    – bryanj
    Jul 4, 2013 at 3:27
  • $\begingroup$ @bryanj: Thanks! $\endgroup$
    – leshik
    Jul 4, 2013 at 3:29

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