Let $f$ be a rational function with $j$ zeros and $k$ poles, all of which reside in the closed unit disk (excepting of course the zeros or pole at $\infty$ when $j\neq k$). What is the smallest number $R>0$ such that all the critical points of $f$ must lie in the closed disk centered at the origin with radius $R$?
When $j\neq k$, a lower bound for the answer is $\dfrac{j+k}{|j-k|}$, as can be seen by inspecting the example $f(z)=\dfrac{(z-1)^j}{(z+1)^k}$. I suspect that $R=\dfrac{j+k}{|j-k|}$ is the answer in general (again assuming $j\neq k$).