How do I find zeros in D(0,2)

Question

$p(z) = z^8 - 20z^4 + 7z^3 + 1$. I know there is 4 real roots, but how do i figure out how many zeroes are there in $D(0,2)$?

Answer 1

Let $f(z) = -20z^4$, and $g(z) = z^8 + 7z^3 + 1$, then on the the boundary of $D(0,2)$, that is on $\Gamma: |z| = 2$, we have: $|f(z)| = 320 > 313 \geq |g(z)|$. So by Rouche's theorem, $f$ and $f+g$ have the same number of zeroes inside $\Gamma$. But $f(z) = -20z^4$ has $4$ zeros counted with muliplicity of $4$. So $f + g = z^8 -20z^4 + 7z^3 + 1$ has $4$ zeros inside $D(0,2)$.