Roots of rational function

Question

Sorry, I asked a similar question yesterday which contained a mistake in the question posed, here is the real question.

Let $(x_n)_{n=1}^N$ be a sequence taking values in $[1,2]$ with the property that $x_1<x_2<...<x_N$ and $$\frac1N \gtrsim \vert x_j-x_{j-1} \vert \gtrsim \frac1N.$$

We then define a function

$$f(x) = \sum_{j=1}^{N} \frac{\alpha_j}{x-x_j},$$

where $\alpha_j$ are numbers with positive real part satisfying $1/N^2 \lesssim \Re(\alpha_i) \lesssim 1/N$ and $ 1 \lesssim \Im(\alpha_i) \lesssim 1.$

I would like to show that any solution to $f(x)=-i$, where $i$ is the imaginary unit, satisfies $\Im(x) \gtrsim 1/N^2.$

The tildes in the inequalities mean we have these estimates up to a constant independent of $j$ and $N$!

Answer 1

The conjecture appears to be very far from true.

The image of a Mathematica notebook below shows the imaginary parts (multiplied by $n^2$) of the roots of $f(x) = -i$ for $n:=N=10, 20, 30, 50, 100$ for $x_j=1+j/n$ ($j=1,\dots,n$), the real parts of the $\alpha_j$'s randomly selected between $1/n^2$ and $1/n$, and the imaginary parts of the $\alpha_j$'s randomly selected between $1$ and $3$.

The common, steady pattern is that there is just one root of $f(x) = -i$ with an imaginary part $\asymp1$ whereas the imaginary parts of all the other roots are very, very close to $0$, even when multiplied by $n^2$: