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Today I tried to show that $\mid P(z)\mid>\mid a_n\mid\mid z\mid ^n(\frac{\mid z\mid -R}{\mid z\mid -1})$. I already thought that I got it, but while reading trough my 'proof' I noticed a mistake, marked below with $*$. I'm trying to prove the result to show that all the roots of the complex polynomial $P(z)$ of degree $n$ lie in the disc $B(0,R)$, where $R=1+M, M=max\{\mid\frac{a_j}{a_n}\mid\},\,0\leq j\leq a_{n-1}$. Additional assumptions are $\mid z\mid\geq R\geq 1$ and $a_n\neq 0$.

$P(z)=a_nz^n+a_{n-1}z^{n-1}+...+a_0$. So I wrote (triangle inequality and geometric sum, $R\geq M$)

$\mid P(z)\mid>\mid a_n\mid(\mid z\mid^n-M(\mid z\mid^{n-1}+..+1))=\mid a_n\mid(\mid z\mid^n-M\frac{\mid z\mid^n-1}{\mid z\mid-1})>\mid a_n\mid(\mid z\mid^n-R\frac{\mid z\mid^n-1}{\mid z\mid-1})\stackrel{*}{>} \mid a_n\mid(\mid z\mid^n-R\frac{\mid z\mid-R}{\mid z\mid-1})>\mid a_n\mid(\mid z\mid^n-\mid z\mid^n\frac{\mid z\mid-R}{\mid z\mid-1})=\,\mid a_n\mid\mid z\mid^n(1-\frac{\mid z\mid-R}{\mid z\mid-1})>\mid a_n\mid\mid z\mid^n(\frac{\mid z\mid-R}{\mid z\mid-1}).$

I was thinking backwards at the $*$: $R\geq 1$ and $\mid z\mid^n\geq\mid z\mid$ so the estimate goes in wrong direction at $*$ because now the numerator is actually smaller and not greater than the nominator in the previous step. The rest of the proof is correct, $\mid z\mid^n\geq R$ and the last inequality can be computed:

$1-\frac{\mid z\mid-R}{\mid z\mid-1}>\frac{\mid z\mid-R}{\mid z\mid-1}\implies \frac{1}{2}>\frac{\mid z\mid-R}{\mid z\mid-1}\implies...\implies R>1$, which is correct by the definitions.

Oh, and there seems to be the special case when $P(z)=a_nz^n\implies M=0\implies R=1$. So maybe the estimate holds only with $\geq$ and not stricktly with $>$. I'm quite lost with this and I'd appreciate any advice a lot!

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Now I got it. $\mid P(z)\mid\,=\,\mid a_nz^n+a_{n-1}z^{n-1}+...+a_0\mid\,\geq\,\mid a_n\mid\mid z\mid^n-...-\mid a_1\mid\mid z\mid-\mid a_0\mid\,\geq\,\mid a_n\mid(\mid z\mid^n-M(\mid z\mid^{n-1}+...+1))=\,\mid a_n\mid(\mid z\mid^n-M\frac{\mid z\mid^n-1}{\mid z\mid-1})=\,\mid a_n\mid(\frac{\mid z\mid^n(\mid z\mid-1)-M(\mid z\mid^n-1)}{\mid z\mid-1})=\,\mid a_n\mid(\frac{\mid z\mid^{n+1}-\mid z\mid^n-M\mid z\mid^n+M}{\mid z\mid-1})=\, \mid a_n\mid\mid z\mid^n(\frac{\mid z\mid-1-M+\frac{M}{\mid z\mid^n}}{\mid z\mid-1})>\,\mid a_n\mid\mid z\mid^n(\frac{\mid z\mid-R}{\mid z\mid-1}).$

$M=max\{\mid\frac{a_j}{a_n}\mid\},\,0\leq j\leq n-1,$ $\mid z\mid\geq R=M+1$ and $1+M-\frac{M}{\mid z\mid^n}<R\iff M-\frac{M}{\mid z\mid^n}<R-1\iff M(1-\frac{1}{\mid z\mid^n})<R-1\iff M<\frac{R-1}{1-\frac{1}{\mid z\mid^n}}<\frac{R-1}{1-\frac{1}{R}}=\frac{R-1}{\frac{R-1}{R}}=\frac{R(R-1)}{R-1}=R=M+1\iff 0<1.$

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