# Proof of an inequality regarding polynomials with complex constants?

Page 12 from Brown and Churchill's Complex Variables and Applications 9th ed.

If $$n$$ is a positive integer and if $$a_0, a_1, a_2, ..., a_n$$ are complex constants, where $$a_n \not= 0$$, the quantity
$$P(z) = a_0 + a_1z + a_2z^2 + ... + a_nz^n$$
is a polynomial of degree $$n$$. We shall show here that for some positive number $$R$$, the reciprocal $$1/P(z)$$ satisfies the inequality
$$|\frac{1}{P(z)}| < \frac{2}{|a_n|R^n}$$ whenever $$|z| > R$$.

In class, my professor proved this, and I have been looking around to find another proof online (usually I like to look at different presentations of a proof to get a better understanding). Can anyone give me a hand? Much appreciated!

• Maybe you could add a few words about how your professor did it; I mean, I've done this problem before but am hesitant to write up my solution if it duplicates your prof's approach. Cheers! Commented Jan 24, 2020 at 22:52
• Well I went ahead and did it anyway! ;) Commented Jan 25, 2020 at 1:40

Given

$$P(z) = \displaystyle \sum_0^n a_i z^i \in \Bbb C[z], \tag 1$$

with

$$a_n \ne 0, \tag{1.5}$$

we may write $$P(z)$$ in the form

$$P(z) = z^n\displaystyle \sum_0^n a_i z^{i -n} = a_nz^n \sum_0^n \dfrac{a_i}{a_n}z^{i - n}; \tag 2$$

then we may estimate a lower bound for $$\vert P(z) \vert$$ as follows:

$$\vert P(z) \vert = \left \vert a_nz^n \displaystyle \sum_0^n \dfrac{a_i}{a_n}z^{i - n} \right \vert = \vert a_n z^n \vert \left \vert \displaystyle \sum_0^n \dfrac{a_i}{a_n}z^{i - n} \right \vert$$ $$= \vert a_n z^n \vert \left \vert 1 + \displaystyle\sum_0^{n - 1} \dfrac{a_i}{a_n} z^{i - n} \right \vert = \vert a_n z^n \vert \left \vert 1 - \left ( -\displaystyle\sum_0^{n - 1} \dfrac{a_i}{a_n} z^{i - n} \right ) \right \vert$$ $$\ge \vert a_n z^n \vert \left \vert \vert 1 \vert - \left \vert -\displaystyle\sum_0^{n - 1} \dfrac{a_i}{a_n} z^{i - n} \right \vert \right \vert = \vert a_n z^n \vert \left \vert \vert 1 \vert - \left \vert \displaystyle\sum_0^{n - 1} \dfrac{a_i}{a_n} z^{i - n} \right \vert \right \vert; \tag 3$$

now with

$$\vert z \vert > R \tag 4$$

(3) yields

$$\vert P(z) \vert \ge \vert a_n \vert R^n \left \vert \vert 1 \vert - \left \vert \displaystyle\sum_0^{n - 1} \dfrac{a_i}{a_n} z^{i - n} \right \vert \right \vert; \tag 5$$

turning now to the sum occurring on the right of this inequality, we have

$$\left \vert \displaystyle \sum_0^{n - 1} \dfrac{a_i}{a_n} z^{i - n} \right \vert \le \displaystyle \sum_0^{n - 1} \left \vert \dfrac{a_i}{a_n} \right \vert \vert z \vert^{i - n}; \tag 6$$

since every power of $$\vert z \vert$$ occurring in the sum on the right is negative, by choosing $$R$$ sufficiently large we may, in light of (4), force this sum to be arbitrarily small; in particular we may ensure that

$$\displaystyle \sum_0^{n - 1} \left \vert \dfrac{a_i}{a_n} \right \vert \vert z \vert^{i - n} < \dfrac{1}{2}; \tag 7$$

then

$$\left \vert \vert 1 \vert - \left \vert \displaystyle\sum_0^{n - 1} \dfrac{a_i}{a_n} z^{i - n} \right \vert \right \vert > \dfrac{1}{2}; \tag 8$$

hence, via (5)

$$\vert P(z) \vert > \dfrac{1}{2} \vert a_n \vert R^n ; \tag 9$$

reciprocating this relationship yields

$$\dfrac{1}{\vert P(z) \vert} < \dfrac{2}{\vert a_n \vert R^n}, \; \forall \vert z \vert > R, \tag{10}$$

the desired result.