Find the number of zeroes (counting multiplicity) of $p(z)=z^6+z^3+10z^2+4z+3$ inside the annulus $1<|z|<2$.
I think this can be solved using Rouché’s theorem. First consider $|z|\leq 1$ on the boundary $|z|=1$ and $|z^6+z^3|<|10z^2+4z+3|$ which can be shown using triangle inequalities, the left side is at most $2$ while right side is at least $3$. We could also consider $|z|\leq 1+ \epsilon$ and our proof still works. Thus $p(z)$ has 2 roots inside $|z|\leq 1$. Also for $|z|= 2$ we have $|z^6+z^3|>|10z^2+4z+3|$. Thus we have 6 zeroes of $p(z)$ inside $|z|<2$ and therefore 4 of them are in $1<|z|<2$. Is this correct? The only "tricky" part is ruling out that there are zeros on $|z|=1$.