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 positive numbers satisfying $1/N^2 \lesssim \alpha\_i \lesssim 1/N^2.$


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 $N$!



Bounds on zeros of rational function
Let us drop the assumption $x\_j\in[1,2]$, it is not needed.
Proving the result by contradiction, denote our function by $f\_N$, suppose that $f\_N(z\_N)=-i$, and $\mathrm{Im}\ z\_N= 1/(N^2R\_N)$ where $R\_N\to\infty$. Since nothing depends on a shift in horizontal direction, one may assume without loss of generality that
$z\_N=i/(N^2R\_N)$, so our equation is
$$\sum\_{j=1}^N\frac{\alpha\_{j,N}N^2R\_N}{i-x\_{j,N}N^2R\_N}=-i.$$
Now consider what happens when $N\to\infty$.
Let us order $x\_j$ by increasing modulus.
Your condition
about $|x\_j-x\_{j+1}|>c/N$ implies that $|x\_{j,N}|\geq c\_1(j-1)/N$ for all $j$ except $j=1$.


Then the sum of all terms for $j\geq 2$ tends to $0$
as $N\to\infty$.


Indeed,
$$\sum\_{j=2}^N\left|\frac{\alpha\_{j,N}N^2R\_N}{i-x\_{j,N}N^2R\_N}\right|=
\sum\_{j=2}^N\frac{\alpha\_{j,N}N^2R\_N}{\sqrt{1+x\_{j,N}^2N^4R\_n^2}}$$
$$\leq
(c\_3/N)\sum\_{j=2}^N1/(j-1)\to 0.$$
So we conclude that the term with $j=1$ must tend to $-i$. But this is impossible, since
$$\frac{\alpha\_{1,N}R\_N N^2}{i-R\_N N^2x\_{1,N}}$$


$$=\frac{-iN^2\alpha\_{1,N}R\_N-\alpha\_{1,N}R\_N^2N^4x\_{1,N}}{1+R\_N^2N^4x\_{1,N}^2},$$
so by comparing the imaginary parts and using that $\alpha\_{1,N}N^2$ is bounded from above and below,
we obtain
$$R\_N\sim R\_N^2x\_{1,N}^2N^4\to\infty,$$
so $x\_{1,N}N^2\to 0$, while comparing imaginary parts gives $N^2x\_{1,N}\to\infty$. This contradiction proves the result.


