# Modulus of Reciprocal of Polynomial Bounded From Above When Exterior to Circle

I am reading the book Complex Variables and Applications, (Brown and Churchill, 9th edition). I was reading the following example:

If n is a positive integer and if $$a_0, a_1, a_2, . . . ,a_n$$ are complex constants, where $$a_n\neq 0$$, the quantity $$P(z) = a_0 (a) + a_1 z + a_2 z^2 + · · · + a_n z^n$$ is a polynomial of degree n. We shall show here that for some positive number R, the reciprocal $$1/P(z)$$ satisfies the inequality $$\left \lvert \frac{1}{P(z)} \right \rvert<\frac{2}{|a_n|R^n} \qquad \text{whenever }|z| > R$$

As part of the solution (attached below), they define: $$w= \frac{a_0}{z^n} +\frac{a_1}{z^{n-1}}+\frac{a_2}{z^{n-2}}+..+\frac{a_{n-1}}{z}$$ which leads to the inequality: $$(9) \qquad |w|\leq \frac{|a_0|}{|z|^n} +\frac{|a_1|}{|z|^{n-1}}+\frac{|a_2|}{|z|^{n-2}}+..+\frac{|a_{n-1}|}{|z|}$$

In regards to $$(9)$$, they say: "Now that a sufficiently large positive number R can be found such that each of the quotients on the right in inequality $$(9)$$ is less than the number $$|a_n|/(2n)$$ when $$|z| > R$$.

Why is this last statement true? How can I say what the relationship is between the quotients on the right of $$(9)$$ and the value of $$|a_n|$$?

Excerpt from the book's solution

$$n$$ and all the $$a_i$$ are fixed.
Since we want $$\frac {|a_i|}{|z|^{n-i}}<\frac {|a_n|}{2n}$$ for every $$i$$, a reasonable $$R$$ to pick is $$\max_i\sqrt[n-i]{2n\frac{|a_n|}{|a_i|}}$$
the quotients and $$|a_n|$$ need not be related.
• Can you please explain how you arrived at this particular $R$? Commented Oct 15, 2021 at 10:51