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Each of the polynomial of the form $p(z)=a_0+\dots+a_{n-1}z^{n-1}+z^n$ satisfies the inequality $\sup\left\{\,|p(z)|\,\big\vert\,|z|\le 1\,\right\}\ge 1$

Is this statement true or false that we have to find. well MMP says that sup will be attained at $|z|=1$ so when $|z|=1$ we have $|p(z)|=|a_0+a_1\dots+ a_{n-1}+1|\le |a_0|+\dots+|a_{n-1}|+|1|$ I can not conclude more.plz help.

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up vote 2 down vote accepted

Using Rouche theorem if $$| z^n+a_0+\cdots +a_{n-1}z^{n-1}| < |z^n|+|a_0+\cdots +a_{n-1}z^{n-1}|$$ for every $z$ with $|z|=1$ then these $z^n$ and $a_0+\cdots +a_{n-1}z^{n-1}$ would have the same amount of roots inside the circe, so this cannot be happening cause of the degrees. so there is a point of equality and you get what you want.

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there is a point of quality meaning? – Un Chien Andalou Jun 14 '12 at 9:56
a point $|z_0|=1$ such that $|z_0^{n} + a_0+ \cdots +a_{n-1}z_0^{n-1}|= |z_0^n| +|a_0+ \cdots +a_{n-1}z_0^{n-1}| \geq 1$ if the the inequality is strict Rouche theorem can be applied – clark Jun 14 '12 at 9:59


  • Compute $I=\displaystyle\oint_S p(z)\dfrac{\mathrm dz}{z^{n+1}}$, where $S=\{z\in\mathbb C\mid |z|=1\}$.
  • Prove that $|I|\leqslant2\pi M$ with $M=\max\{|p(z)|;z\in S\}$.
  • Deduce a lower bound on $M$.
  • Finally, compare $M$ and $\max\{|p(z)|;|z|\leqslant1\}$.
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