Rouché's theorem in annulus $1<|z|<2$ I have to find number of roots of the polynomial $p(z)=z^4-8z+10$ in the annulus $1<|z|<2$ .
I'm trying to do this using Rouché's theorem.And, by this theorem , I have that $p$ does not have zeros in $|z|<1$ , which means that number of zeros in the given annulus is the same as the number of zeros on $|z|<2$ .
Then I try to see how does this polynomial 'behave' in $|z|=2$ and I see that the coefficients  $2^4$ and $|-8*2|$ are the same ,and in that way I cannot decide what should I choose for $g$ so that $|p-g|<|g|$ and so to apply Rouché's theorem .
Can someone help me do this?
Any advice is appreciated.
Thank you in advance!

We plan to use the version of Rouché's theorem which says that if $|g(z)| < |f(z)|$ on the boundary of our region then $f$ and $f+g$ have the same number of zeros with multiplicty in the region.
Let's first set a goal for ourselves. If we can show that $|p(z)| > 3$ for $|z| = 2$, then Rouché's theorem tells the number of zeros of $p(z)$ and $q(z) = z^4 - 8z + 7$ agree. We see that $z = 1$ is a zero of $q(z)$, so using polynomial long division we get that $q(z)= (z-1)(z^3 + z^2 + z - 7)$. Note this also rules out zeros in $|z| = 2$, and ruling out zeros in $|z| = 1$ is already handled by analysis you mention having done in this question.
The factor $r(z) = z^3 + z^2 + z - 7$ then yields to ad-hoc analysis. By checking $1$ and $3/2$, the intermediate value theorem finds a root $x_0$ of $r(z)$ in $(1, 3/2)$. Furthermore, $r'(z) = 3z^2 + 2z + 1$, which the quadratic formula tells us has only imaginary roots, hence $r(z)$ is monotone on $\mathbb{R}$. We conclude that $x_0$ is the only real root of $r(z)$, and so the other roots of $r(z)$ are conjugate. Call them $\alpha$ and $\overline{\alpha}$. Then factoring shows $-7 = -x_0 |\alpha|^2$, and our bounds on $x_0$ force $|\alpha|^2 > 4$, hence $|\alpha| > 2$. Together with what you've already shown, this proves that $p(z)$ has exactly $2$ zeroes in the annulus.
So we just need to make this estimate. It suffices to show that $|p(z)|^2 > 9$ for $|z| = 2$. For $|z| = 2$, doing out the multiplication gives $$|p(z)|^2 = (z^4 - 8z + 10)(\bar{z}^4 - 8\bar{z} + 10) = 612 + 10\text{Re}(z^4) - 16 \text{Re}(z^3) - 80 \text{Re}(z),$$ where $612$ arises as $2^8 + 8 * 8 * 2^2 + 100$. We bound $\text{Re}(z^4)$ by $-16$, bound $\text{Re}(z^3)$ by $8$, and $\text{Re}(z)$ by $2$ to see that $$|p(z)|^2 \geq 612 - 10 * 16 - 16 * 8 - 80*2 = 612 - 448 > 9,$$ and this completes our proof.
