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.


Your conjecture is correct. We can assume that the largest critical point occurs at a positive $z=R>0$. If we denote the zeros and poles by $a_n$ and $b_n$, respectively, then the condition for a critical point is $$ \sum \frac{1}{R-a_n} = \sum \frac{1}{R-b_n} . $$ In particular, the real parts of both sides must be equal to one another, and if $R>1$, then, given an $|a|\le 1$, we can find a real $-1\le\alpha\le 1$ such that $\textrm{Re}\: 1/(R-a) = 1/(R-\alpha)$. (To do this, I just confirmed that $1/(R+1)\le\textrm{Re}\: 1/(R-a)\le 1/(R-1)$ by a slightly tedious, but elementary calculation, but maybe there's a better way to see this.)

This means that our search for optimal functions can be restricted to those with real zeros and poles, but then it's clear that your example gives the largest possible $R$, by just looking at the effect of changing a single $a_n$ or $b_n$; notice that $F(R)=LHS-RHS$ will have the same sign as $j-k$ for $R$ larger than the largest zero.

For $j=k$, there is no bound: consider $\frac{z^2}{(z-a)(z-b)}$. This has its critical point at $z=2ab/(a+b)$, which we can make arbitrarily large.

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  • $\begingroup$ Christian Remling, are you aware of a text or article which contains this result? $\endgroup$ – Trevor J Richards Sep 17 '17 at 18:11
  • $\begingroup$ @TrevorRichards: No, I'm not. $\endgroup$ – Christian Remling Sep 17 '17 at 22:06

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