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 $$\begin{aligned} |p(z)| &= |z^{n}+a_{n-1} z^{n-1}+\cdots+a_{1} z+a_{0}| \\ &\ge |z|^{n}+|a_{n-1}| |z|^{n-1}+\cdots+|a_{1}| |z|+|a_{0}| \\ &> 1 + |a_{n-1}| +\cdots+|a_{1}| +|a_{0}| \\ &= R\end{aligned}$$
Best Answer
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 $|p(z)|=\left|-{\large{\frac{5}{2}}}+1\right|={\large{\frac{3}{2}}} < R$.