Locating roots of a special class of polynomials in $\mathbb{Z}[X]$

Question

I am reading Prof. Ram Murty's Prime number and Irreducible polynomials. I am having problem in understanding a part of the following lemma:

Statement: Suppose that $\alpha$ is a complex root of a polynomial $$ f(x)=x^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0 $$ with coefficients $a_i$ equal to $0$ or $1$. If $|arg \ \alpha| \le \frac{\pi}{4}$ then $|\alpha| < \frac{3}{2}$. Otherwise $\mathfrak{R}(\alpha) < \frac{1+\sqrt{5}}{2\sqrt{2}}$.

Proof: The cases $m=1$ and $m=2$ can be verified directly. Assuming that $m \ge 3$, we compute for $z \ne 0$:

$$ \left| \frac{f(z)}{z^m}\right| \ge \left| 1+\frac{a_{m-1}}{z}+\frac{a_{m-2}}{z^2}\right| - \left(\frac{1}{|z|^3}+\cdots+\frac{1}{|z|^m}\right). $$

For $z$ satisfying $|arg \ z| \le \frac{\pi}{4}$ it is true that $\mathfrak{R}(\frac{1}{z^2}) \ge 0$, so for such $z$ we have

$$ \left|\frac{f(z)}{z^m}\right| > 1 - \frac{1}{|z|^2(|z|-1)} $$

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1. How to obtain the cases when $m=1,2$?

2. How are the above two inequalities obtained?

3. Where are we using the fact: For $z$ satisfying $|arg \ z| \le \frac{\pi}{4}$ it is true that $\mathfrak{R}(\frac{1}{z^2}) \ge 0$?

Answer 1

Here is a proof of the first inequality.

$$f(z) = z^m + a_{m-1}z^{m-1} + a_{m-2}z^{m-2} + a_{m-3}z^{m-3}+ \dots+a_0\\ \frac{f(z)}{z^m} = \left(1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right) + \left(\frac{a_{m-3}}{z^3}+ \dots+\frac{a_0}{z^m}\right)\\ \frac{f(z)}{z^m} - \left(\frac{a_{m-3}}{z^3}+ \dots+\frac{a_0}{z^m}\right)= \left(1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right)$$ $$\left|\frac{f(z)}{z^m}\right| + \left(\frac{|a_{m-3}|}{|z|^3}+ \dots+\frac{|a_0|}{|z|^m}\right)\ge \left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right|\tag{1}$$ $$\left|\frac{f(z)}{z^m}\right| + \left(\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m}\right)\ge \left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right|\tag{2}\\ \left|\frac{f(z)}{z^m}\right|\ge \left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right| - \left(\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m}\right)$$ Where $(1)$ is the triangle inequality, and $(2)$ is because $\forall i, |a_i| \le 1$.

[this only works for $|z| > 1$]For the second inequality, note that if $|z| < 1$, then $$\begin{align}\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m} &= \frac{1}{|z|^3}\left(1 + |z|^{-1} + \dots+|z|^{-(m-3)}\right)\\ &=\frac{1}{|z|^3}\left(\dfrac{1-|z|^{-(m-2)}}{1-|z|^{-1}}\right)\\ &\le\frac{1}{|z|^2(|z|-1)}\end{align}$$ and if $|z| > 1$, then $$\begin{align}\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m} &\le \frac{1}{|z|^3}\left(1 + |z|^{-1} + \dots\right)\\ &=\frac{1}{|z|^3}\left(\dfrac1{1-|z|^{-1}}\right)\\ &=\frac{1}{|z|^2(|z|-1)}\end{align}$$ So if we can show $\left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right| \ge 1$, the inequality you are after follows. This is where $\mathfrak R\left(\frac1{z^2}\right) \ge 0$ comes in. Though not mentioned, the argument restriction also implies $\mathfrak R\left(\frac1{z}\right) \ge 0$. Hence $$\mathfrak R\left(1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right) \ge 1 + 0 + 0 = 1$$ But the modulus of any number with real part $\ge 1$ must be $\ge 1$ as well. Thus $$\left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right| \ge 1$$ as desired.