# Lower bound for the modulus of general complex polynomial of degree $n$?

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!

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}