1
$\begingroup$

Let $p(z) = a_nz^n + \cdots + a_0$ be a complex polynomial of degree $n$. Show that for large enough $|z|$, there exists $c > 0$ such that $c|z|^n < |p(z)|$.

It is easy to see that the statement is true, but I am having a hard time proving it completely rigorously. For example, I don't want to say things like "the $a_nz^n$ term eventually dominates". I am sure it can work if I take $c = |a_n| / 2$, but what would be the threshold for $|z|$?

1 Answers 1

1

Reverse engineering: observe that

$$c|z|^n\le|p(z)|\iff c|z|^n<|a_nz^n+\ldots+a_0|\iff c<\left|a_n+\frac{a_{n-1}}z+\ldots+\frac{a_0}{z^n}\right|$$

We are assuming that $\;a_n\neq0\;$ , thus

$$\left|a_n+\frac{a_{n-1}}z+\ldots+\frac{a_0}{z^n}\right|\le\left|a_n\right|+\left|\frac{a_{n-1}}z\right|+\ldots+\left|\frac{a_0}{z^n}\right|\xrightarrow[|z|\to\infty]{}|a_n|$$

so you can indeed take $\;c:=\frac{|a_n|}2\;$ or anything smaller than $\;|a_n|\;$

  • 0
    Thanks! Why does the first equation alternate between strict and non strict inequality? – 2017-02-05
  • 0
    @ihmth Just a typo. Editing. – 2017-02-05
  • 0
    Actually, as for the second equation, why does this ensure that we can take any $c < |a_n|$? I thought it was clear to me but could you expand a little more? I could see how everything would work out if we could direcly say that $\left|a_n+\frac{a_{n-1}}z+\ldots+\frac{a_0}{z^n}\right|\xrightarrow[|z|\to\infty]{}|a_n|$, but is this true for complex numbers? – 2017-02-05