Rouché for Polynomial $p(z)=z^7+z(z-3)^3+1$ around non-centered annulus. > Let $p(z)=z^7+z(z-3)^3+1$. Find the numbers of zeros (including multiplicities) in $B_{1}(3)$. I want to apply Rouché's Theorem for $p(z)$ and $g(z)=-z(z-3)^3$, but I don't know how to confine on $|z-3|=1$. The problem should be solved without using a calculator. Advice appreciated. Thanks

It's better to work with unit circle $|z|\leqslant1$, For this purpose let $w=z-3$ then $p(z)=(w+3)^7+w^4+3w^3+1$ and for $|w|=1$ we have $$|f(w)|=|w^4+3w^3+1|\leqslant|w|^4+3|w|^3+1=5<|w+3|^7=|g(w)|$$ therefore the number of zeros $p(w)$ and $g(w)$ in $|w|\leqslant1$ are the same, but $g(w)$ has no zeros there.