TITLE: Derivation of Inequality in Fundamental Theorem of Algebra
QUESTION [1 upvotes]: I've been reading through a proof of the Fundamental Theorem of Algebra and I am trying to understand how an inequality was derived. The actual proof is located here https://mathscholar.org/2018/09/simple-proofs-the-fundamental-theorem-of-algebra/. 

In the proof, it's given that
$$p(z)=p_0+p_1z+p_2z^2+\cdots +p_nz^n$$
where the coefficients $p_i$ are complex numbers with neither $p_0$ nor $p_n$ equal to zero.

The line I'm confused about is this...

First note that for large 
  $z$
  , say 
  $|
z
|
>
2
\max_
i
\left|
\large{
\frac{p_i}{p_n}
}
\right|$, the 
  $z^
n$
   term of 
  $p
(
z
)$
   is greater in absolute value than the sum of all the other terms.

How was the inequality derived?

REPLY [0 votes]: Suppose $|z| > 2\max_i\left|{\large{\frac{p_i}{p_n}}}\right|$.

In particular, using $i=n$, we get $|z| > 2$, hence $2(|z|-1) > |z|$.

Letting $r=\max_i |p_i|$, we have $|z| > {\Large{\frac{2r}{|p_n|}}}$, hence
\begin{align*}
&\left|\sum_{i=0}^{n-1}p_iz^i\right|\\[4pt]
\le\;&\sum_{i=0}^{n-1}|p_iz^i|\\[4pt]
=\;&\sum_{i=0}^{n-1}|p_i||z|^i\\[4pt]
\le\;&r\sum_{i=0}^{n-1}|z|^i\\[4pt]
=\;&r\left(\frac{|z|^n-1}{|z|-1}\right)\\[4pt]
=\;&2r\left(\frac{|z|^n-1}{2(|z|-1)}\right)\\[4pt]
 < \;&2r\left(\frac{|z|^n}{|z|}\right)\\[4pt]
=\;&2r|z|^{n-1}\\[4pt]
=\;&|p_n|\cdot \frac{2r}{|p_n|}\cdot |z|^{n-1}\\[4pt]
 < \;&|p_n|\cdot |z|\cdot |z|^{n-1}\\[4pt]
=\;&|p_nz^n|\\[4pt]
\end{align*}
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