How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$? How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 &lt; |z| &lt; 2$?

I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| &lt; |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.
General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.

How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|&lt;|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.

I will let you do the same with the circle of radius $2$.
I think total zeros are 7-2=5 inside the annulus. Because, for |z|&lt;1: 2 roots and |z|&lt;2: 7 roots. Is it ok.