Bounds on zeros of rational function 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 $N$!
 A: 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.
