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 poles at $\infty$ when $j\neq k$). What is the smallest number $R>0$ such that all the (finite) 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 $j\neq k$).

When $j=k$, I am not sure what I expect the answer to be.

NOTE: This question was originally posted on MSE, without receiving any answers.