# Number of zeros outside the disk $\{ z : |z| \leq 2 \}$

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.

• @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. May 20, 2021 at 3:31
• The Theorem applies to zeroes inside the disk, where $\ 9z^4 + 3 \$ dominates. How many zeroes does that leave outside the disk?
– user882145
May 20, 2021 at 4:20
• @boojum, three ! May 20, 2021 at 13:53
• @boojum, Wait! I think it does not dominate, because $|9z^4 +3| \geq 9|z|^4-|3| = 9 \times 2^4 -3 =141$ and $|z^7-7z| \leq |z|^7 +7|z| = 2^7 +7\times2 =128+14 =144$. May 20, 2021 at 14:36
• 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.
– user882145
May 20, 2021 at 20:36

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