For $n\geq 3$, consider the rational function $\left(\sum_{k=1}^n \frac{\lambda_k z}{z-a_k}\right) -1$ where $0 < \lambda_k < 1$, $\sum_{k=1}^n \lambda_k =2$, $|a_k|=1$, and the $a_k$ are all distinct. The two things that I think are true about the zeros of this function are:

  1. that the zeros all lie on the unit circle,
  2. that the zeros are "interwoven" with the poles - i.e., if I traverse the unit circle I'll hit a zero, then a pole, then another zero, then another pole, etc.

I tried some test cases in Wolfram, and my "conjectures" seem to hold. It looks like there is some symmetry in the polynomial (the numerator) that could possibly be exploited to at least prove that the zeros have norm 1. Other than that, I really have no clue where to start.


closed as off-topic by Stefan4024, The Phenotype, Saad, Mohammad Riazi-Kermani, Claude Leibovici Mar 14 '18 at 8:52

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Pick $b_1,\ldots,b_n$ on the unit circle such that $\prod b_n = - \prod a_n$ and let $f(z) = \frac {\prod(z-b_k)}{\prod(z-a_k)}$.

Then, $f(z)$ can be written as $f(z) = (\sum_{k=1}^n \frac{\lambda_k z}{z-a_k} )-1 $ where, by evaluating $(z-a_k)f(z)$ at $a_k$, you get $\lambda_k a_k = \prod(a_k-b_i) / \prod_{i \neq k} (a_k-a_i)$, and so $\lambda_k = \frac {a_k-b_k}{a_k}\prod_{i \neq k} \frac{a_k-b_i}{a_k-a_i}$.

(you get $\sum \lambda_k = 2$ by evaluating $f$ at $\infty$)

We want to look at the argument of $\lambda_k$ (hoping to prove it is real).

If $u$ and $v$ are on the unit circle, you have $\arg(u-v) \equiv \pi/2 + (\arg(u)+ \arg(v))/2 \pmod \pi$.

And so, modulo $\pi$, we have : $\arg(\lambda_k) = (\pi + \arg(a_k)+\arg(b_k))/2 - \arg(a_k) + \sum_{i\neq k} (\arg(b_i)- \arg(a_i))/2 \\ = \pi/2 + (\sum \arg(b_i) - \sum \arg(a_i))/2$

Since we have $\prod b_i = - \prod a_i$, the two sums differ by $\pi$, and so $\arg(\lambda_k) = \pi/2 + \pi/2 = 0 \pmod \pi$.

So this proves that the functions obtained in this way have coefficients $\lambda_k$ that are real.

Moreover, they are a continuous function of the $b_k$ (and $a_k$ as long as they are distinct), and a coefficient $\lambda_i$ is $0$ if and only if one of the $b_k$ is $a_i$, so the signs of the $\lambda_i$ cannot change when you move the $a_i$ and $b_i$ as long as they stay alternating and distinct, and we can probably check with a simple example that there is a case with alternating poles and roots where all the coefficients are positive.

This is not a complete proof yet because I haven't shown that you can obtain all the possible $n$-uplets of sum $2$ while the $b_k$ vary over all $n$-uplets with the constrained products, but this is still some good evidence for the conjecture. Some careful study of limit cases should finish the proof.


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