Complex Polynomial Inequality Verification

Definition

For $$z, a_i \in \mathbb C$$ and $$n \in \mathbb R^+$$

$P(z) = a_0 + a_1 z + a_2 z^2 + \dots + a_n z_n$

with $$a_n \neq 0$$ (i.e. has a $$n$$th term) is called a *polynomial* of degree $$n$$.

The inequality we wish to verify is

Theorem

$\left\vert\frac{1}{P(z)}\right\vert \lt \frac{2}{\vert a_n\vert R^n}$

whenever $$\vert z\vert > R$$ for some $$R \in \mathbb R^+$$. In other words, $$\left\vert\frac{1}{P(z)}\right\vert$$ is bounded above by $$\frac{2}{\vert a_n \vert R^n}$$ when $$z$$ is sufficiently large.

We proceed by abuse of algebra.

Proof

Let $$\displaystyle{w = \frac{a_0}{z^n} + \frac{a_1}{z^{n-1}} + \frac{a_2}{z^{n-2}} + \dots + \frac{a_{n-1}}{z}}$$, then $$P(z) = (a_n+ w)z^n$$. Also note that $$\vert P(z)\vert = \vert a_n + w \vert \cdot \vert z^n \vert$$. Since $$\vert z^n \vert = \vert z \vert ^n$$, we can write $$\vert P(z)\vert = \vert a_n + w \vert \cdot \vert z \vert ^n$$.

We see that

$wz^n = a_0 + a_1 z + a_2 z^2 + \dots + a_{n-1}z^{n-1}$

So by the general triangle inequality

$\vert w \vert \vert z \vert ^n \leq \vert a_0 \vert + \vert a_1 \vert \vert z \vert + \vert a_2 \vert \vert z \vert ^2 + \dots + \vert a_{n-1} \vert \vert z \vert ^{n-1}$

or,

$\vert w \vert \leq \frac{\vert a_0 \vert}{\vert z \vert ^ {n}} + \frac{\vert a_1 \vert}{\vert z \vert ^ {n - 1}} + \frac{\vert a_2 \vert}{\vert z \vert ^ {n - 2}} + \dots + \frac{\vert a_{n - 1}\vert}{\vert z \vert}$

Certainly, if we pick a large enough $$z$$, the following is true:

$\vert w \vert \lt \vert a_n \vert$

For some reason, the above will not be nearly as useful to us as the following, so that will be what we will work with.

$\vert w \vert \lt \frac{\vert a_n \vert}{2}$

That is, the $$a_n z^n$$ term asymptotically dominates the rest - for a large enough $$z$$. So by the triangle inequality,

$\vert a_n + w \vert \geq {\Large\vert} \vert a_n \vert - \vert w \vert {\Large\vert}$

Since $$\vert w \vert \lt \frac{\vert a_n \vert}{2}$$, then $${\Large\vert} \vert a_n \vert - \vert w \vert {\Large\vert}> \frac{\vert a_n \vert}{2}$$ — that is, taking away less than half of $$\vert a_n \vert$$ leaves more than half of $$\vert a_n \vert$$. So we then have

$\vert a_n + w \vert \geq {\Large\vert} \vert a_n \vert - \vert w \vert {\Large\vert} > \frac{\vert a_n \vert}{2}$

All provided that $$z$$ is of sufficient size. Since $$\vert a_n + w \vert \cdot \vert z \vert^n > \frac{\vert a_n \vert}{2} \cdot \vert z \vert ^n$$, we have

$\vert P(z) \vert = \vert a_n + w \vert \cdot \vert z \vert^n > \frac{\vert a_n \vert}{2} \cdot \vert z \vert^n$

Then, if $$\vert z \vert > R$$, we have

$\vert P(z) \vert > \frac{\vert a_n \vert}{2} \cdot \vert z \vert ^n > \frac{\vert a_n \vert}{2} \cdot R^n$

or,

$\left\vert \frac{1}{P(z)}\right\vert \lt \frac{2}{\vert a_n \vert R^n}$

And thus the inequality is verified.