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 $x_k.$

Here is why I think this should be true:

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?

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?

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 $x_k.$

Here is why I think this should be true:

Let $x=x_k - i y_k$ by multiplying the equation with $(x-x_k),$ we find

$$ \lambda_k+\sum_{j=1, j \neq k}^n \frac{\lambda_j}{x_k-x_j+i y_k} =y_k.$$

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?

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 $x_k.$

Here is why I think this should be true:

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?