Complex Polynomial Inequality Verification


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


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

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.


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

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


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

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

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


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

And thus the inequality is verified.