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?
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*}