Existence of solution to linear fractional equation 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?
 A: Expanding the comment by Alexandre Eremenko (and assuming $\lambda_j$ small for $j=k$ only).
Let us rewrite the equation in the form
$$
\underbrace{\prod_{j=1}^n (x-x_j) + i \sum_{j\in\{1,\ldots,n\} \setminus\{k\}} \lambda_j \prod_{s\in\{1,\ldots,n\} \setminus \{j\}} (x-x_s)}_{f(x)} + \underbrace{i \lambda_k \prod_{s\in\{1,\ldots,n\} \setminus \{k\}} (x-x_s)}_{g(x)} = 0.
$$
Choose $r>0$ sufficiently small so that $f(x) \ne 0$ for $x\in\partial B_r(x_k)$. Then $\min_{x\in \partial B_r(x_k)} |f(x)| = m > 0$. On the other hand if $\lambda_k$ is sufficiently small then $|g(x)| < m$ for all $x \in \partial B_r(x_k)$. Since $f(x_k)=0$ it remains to apply Rouché's theorem.
A: 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 $x_1$. 
A: Here is an estimate that does not depend on $\lambda_j,j\geq 2$.
Lemma. Consider an equation $$\epsilon+az+(b+\phi(z))z^2=0,$$
 where $\epsilon>0$, $|a|\geq 1$ and  $b$, are complex constants, and
$\phi$ is analytic, $|\phi|<|b|/10.$ Then there exists a solution $|z|\leq 10\epsilon$.
Proof. If $|b|\leq |a|/(5\epsilon)$, apply Rouche to $|z|\leq2\epsilon/|a|$. The dominant term is $az$. If $b>|a|/(5\epsilon)$, apply Rouche to
$|z|\leq10\epsilon$. The dominant term is $(b+\phi)|z^2|$ now.
To apply this lemma, write your equation
$$\frac{\epsilon}{z}+\sum_2^n\frac{\lambda_k}{z-z_k}=i$$
which we bring to the form given in the lemma by
writing the sum minus $i$ in the form $a+(b+\phi(z))z$ and mulpiplying on $z$. So
$$a=-\sum_2^n\frac{\lambda_k}{z_k}-i,$$
(we have $|a|\geq 1$ because the sum is real).
$$b=-\sum_2^n\frac{\lambda_k}{z_k^2},$$
and $\phi$ is what remains. I used the identity
$$\frac{1}{z-w}=-\frac{1}{w}-\frac{z}{w^2}-\frac{z^2/w^3}{1-z/w},$$
in which I set $w=z_k$ and then add.
Now estimate $\phi$, assuming that $|z|/|z_k|<1/20$
for all $k\geq 2$.
$$|\phi(z)|=\sum_2^n\lambda_k\left|\frac{z/z_k^3}{1-z/z_k}\right|< \frac{1}{10}\sum_2^n\lambda_k/|z_k^2|<|b|/10.$$
It is important here that all $\lambda_j$ are positive.
