We consider the equation


$$ \sum\_{j=1}^n \frac{\lambda\_j}{x-x\_j} =i$$


where $\lambda\_j>0$ and $x\_j$ are real distinct numbers.


I want to show that if $\lambda\_k$ is small compared to the distance of all $x\_j$ from $x\_k$ then there exists a solution $x\approx x\_k- i y\_k$ to this equation in the neighbourhood of $x\_k.$


**Heuristic argument:**


Let $x=x\_k - i y\_k$ by multiplying the equation with $(x-x\_k),$ we find


$$ \lambda\_k - \sum\_{j=1, j \neq k}^n \frac{i y\_k\lambda\_j}{x\_k-x\_j-i y\_k} =y\_k.$$


Now, if $x\_k-x\_j$ is large, then the sum is small and we can choose $y\_k\approx \lambda\_k.$


However, this argument is (obviously) non-rigorous. 


Can we make it rigorous?



Existence of solution to linear fractional equation
Here's a "real" method proof:


Without loss of generality, assume $k = 1$. You are equivalently looking for the roots of 


$$ P(x, \eta) = \eta \prod\_{j > 1} (x - x\_j) + \sum\_{\ell > 1} \lambda\_\ell \prod\_{j \neq \ell} (x - x\_j) - i \prod\_{j} (x - x\_j) $$


where we wrote $\eta$ for $\lambda\_1$. (We consider $\lambda\_\ell$ and all $x\_j$ to be fixed.)


Now allow $\eta$ to be a parameter in $\mathbb{C} \cong \mathbb{R}^2$. 


For $\eta = 0$, by inspection $P(x\_1, 0) = 0$. The derivative 


$$ \partial\_\eta P(x\_1,\eta) = \prod\_{j > 1} (x\_1-x\_j) \cdot \mathrm{Id}$$


is non-critical. So by implicit function theorem for all sufficiently small $\eta$ there exists a solution close to $x\_1$. 


