Amann/Escher, Analysis I, Exercise I.11.8: Show that $|p(z)|>R$ for all $z \in \mathbb{C}$ such that $|z|>R$ I'm doing Exercise I.11.11 from textbook Analysis I by Amann/Escher.



The fact that my proof is too simple makes me wonder if I miss something. Could you please verify if my attempt contains logical gaps/errors?

My attempt:
Since $|z| > R \ge 1$, $|z|^k > 1$ for all $k \in \mathbb N^+$.
We have $$\begin{aligned} |p(z)| &= |z^{n}+a_{n-1} z^{n-1}+\cdots+a_{1} z+a_{0}| \\ &\ge |z|^{n}+|a_{n-1}| |z|^{n-1}+\cdots+|a_{1}| |z|+|a_{0}| \\ &> 1 + |a_{n-1}| +\cdots+|a_{1}| +|a_{0}| \\ &= R\end{aligned}$$
 A: As noted in the comments, you applied the triangle inequality incorrectly.
 
But assuming $n\ge 2$, ideas along the same lines can be made to work:

\begin{align*}
|p(z)|&=\left|z^{n}+\sum_{k=0}^{n-1}a_kz^k\right|\\[4pt]
&\ge |z|^{n}-\sum_{k=0}^{n-1}|a_k||z|^k
&&\text{[a version of the triangle inequality]}\\[4pt]
&\ge |z|^{n}-\sum_{k=0}^{n-1}|a_k||z|^{n-1}
&&\text{[since $|z| > 1$]}\\[4pt]
&=|z|^{n}-|z|^{n-1}\sum_{k=0}^{n-1}|a_k|\\[4pt]
&=|z|^{n}-|z|^{n-1}(R-1)\\[4pt]
&=|z|^{n-1}\bigl(|z|-(R-1)\bigr)\\[4pt]
&>|z|^{n-1}\bigl(R-(R-1)\bigr)
&&\text{[since $|z| > R$]}\\[4pt]
&=|z|^{n-1}\\[4pt]
&> R^{n-1}
&&\text{[since $|z| > R$ and $n\ge 2$]}\\[4pt]
&\ge R
&&\text{[since $R \ge 1$ and $n\ge 2$]}\\[4pt]
\end{align*}
hence, for the case $n\ge 2$, we have $|p(z)| > R$, as was to be shown.

However, for the case $n=1$, the claim does not always hold.

For a counterexample, let $p=x+1$, and let $z=-{\large{\frac{5}{2}}}$.

Then $R=1+1=2$, and $|z|={\large{\frac{5}{2}}} > R$, but $|p(z)|=\left|-{\large{\frac{5}{2}}}+1\right|={\large{\frac{3}{2}}} < R$.
A: I have figured out the proof and posted it here. It would be great if someone helps me verify it.

My attempt:
It follows from $|z| > R = 1 + \sum_{k=0}^{n-1} |a_k| \ge 1$ that $|z|^k \le |z|^{n-1}$ for all $k \le n-1$ and $|z| - \sum_{k=0}^{n-1} |a_k| > 1$. It follows that
$$\begin{aligned} |p(z)| & = \left | z^n +\sum_{k=0}^{n-1} a_k z^k \right | \\
& \ge |z^n| - \left | \sum_{k=0}^{n-1} a_k z^k \right |, \quad \text{Triangle Inequality}\\
& \ge |z^n| - \sum_{k=0}^{n-1} |a_k| |z|^k, \quad \text{Triangle Inequality} \\
& \ge |z^n| - \sum_{k=0}^{n-1} |a_k| |z|^{n-1}, \quad |z|^k \le |z|^{n-1} \\
& = |z^n| - |z|^{n-1} \sum_{k=0}^{n-1} |a_k| = |z|^{n-1} \left ( |z| - \sum_{k=0}^{n-1} |a_k| \right ) \\
& > |z|^{n-1} >|z|  > R\end{aligned}$$
This completes the proof.
