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!

  • 2
    $\begingroup$ 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! $\endgroup$ Commented Jan 24, 2020 at 22:52
  • $\begingroup$ Well I went ahead and did it anyway! ;) $\endgroup$ Commented Jan 25, 2020 at 1:40

1 Answer 1



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


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

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

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


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

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.


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