3
$\begingroup$

I need to count (including the multiplicities of the zeros) number of the zeros outside the disk $\{ z : |z| \leq 2 \}$ for the polynomial

$f(z) = z^7 +9z^4 -7z +3$.

I know this should be direct application for Rouche's theorem, but I tried all choices for the two functions to get the required inequality $ |p(z)| < |q(z)|$ for $|z|=2 $, but none of them works. Should I consider a different curve or what terms I should consider to get the required inequality?

I think $z^7 +3$ should work but couldn't confirm that.

$\endgroup$
7
  • $\begingroup$ @JitendraSingh, I actually read this post before I wrote my question, and it gives helpful hints, but the terms in my functions differ by 1 in the reverse directions. $\endgroup$ May 20, 2021 at 3:31
  • $\begingroup$ The Theorem applies to zeroes inside the disk, where $ \ 9z^4 + 3 \ $ dominates. How many zeroes does that leave outside the disk? $\endgroup$
    – user882145
    May 20, 2021 at 4:20
  • $\begingroup$ @boojum, three ! $\endgroup$ May 20, 2021 at 13:53
  • $\begingroup$ @boojum, Wait! I think it does not dominate, because $|9z^4 +3| \geq 9|z|^4-|3| = 9 \times 2^4 -3 =141$ and $|z^7-7z| \leq |z|^7 +7|z| = 2^7 +7\times2 =128+14 =144$. $\endgroup$ May 20, 2021 at 14:36
  • $\begingroup$ The argument will require a little refinement, then, since $ \ 9z^4 + 3 \ $ does dominate even at $ \ |z| = \frac74 \ $ and $ \ |z^7 - 7z| \ $ only starts to "catch up" when $ \ |z| \ $ gets to about $ \ 1.85 \ $ . The two are also close around $ \ |z| \approx 0.6 \ $ . Four of the zeroes are inside the unit disc and the moduli of the other three are all around $ \ 2.1 - 2.2 \ , $ which seems to be what makes the analysis a bit tricky. $\endgroup$
    – user882145
    May 20, 2021 at 20:36

1 Answer 1

1
$\begingroup$

We can apply Rouché's theorem to $$ f(z) = z^7 +9z^4 -7z +3 $$ and $$ g(z) = z^7 +9z^4 -10z = z(z^3+10)(z^3-1) \, . $$ For $|z| = 2$ is $$ |f(z)-g(z)| = |3z+3| \le 9 $$ and $$ |g(z)| \ge |z| (10 - |z^3|) (|z|^3-1) = 28 \, . $$ It follows that $f$ and $g$ have the same number of zeros inside ($4$) and outside ($3$) of the circle $|z|=2$.


How did I find the comparison polynomial $g$? Trial and error, essentially, but here is a possible approach: Numerical approximations (e.g. with WolframAlpha) indicate that $f$ has three roots with absolute value greater than two: $$ \begin{align} z_1 &\approx -2.14379 \\ z_2 &\approx 1.05711 - 1.84842 i \approx 2.12935 \cdot e^{-0.3346 i \pi }\\ z_3 &\approx 1.05711 + 1.84842 i \approx 2.12935 \cdot e^{0.3346 i \pi } \end{align} $$ These are – very roughly – the roots of $z^3+9.72$. This suggest to choose $g$ as $(z^3 + 10)$, multiplied with a fourth-degree polynomial having zeros only inside $|z|=2$. I chose the factor $z (z^3 - 1)$ so that in the difference $f-g$ both the $z^7$ and the $z^4$ terms vanish.

$\endgroup$
1
  • $\begingroup$ +1 Finding $g$ was only part of the solution, the way you lower bounded $|g(z)|$ a significant element. $\endgroup$
    – copper.hat
    May 22, 2021 at 16:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .