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

Question

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?

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My attempt:

Since $|z| > R \ge 1$, $|z|^k > 1$ for all $k \in \mathbb N^+$.

We have $$\begin{aligned} |p(z)| &= |z^{n}+a_{n-1} z^{n-1}+\cdots+a_{1} z+a_{0}| \\ &\ge |z|^{n}+|a_{n-1}| |z|^{n-1}+\cdots+|a_{1}| |z|+|a_{0}| \\ &> 1 + |a_{n-1}| +\cdots+|a_{1}| +|a_{0}| \\ &= R\end{aligned}$$

Answer 1

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 $|p(z)|=\left|-{\large{\frac{5}{2}}}+1\right|={\large{\frac{3}{2}}} < R$.

Answer 2

I have figured out the proof and posted it here. It would be great if someone helps me verify it.

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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.