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.
 A: 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.
