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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?

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  • $\begingroup$ $y_k = 0 + \lambda_k + _ \lambda_k + \cdots$ $\endgroup$ – AHusain Mar 22 '19 at 0:22
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    $\begingroup$ This is correct (and trivial). When all $\lambda_j$ are small the LHS is small everywhere except small neighborhoods of $x_k$. To each of these neighborhoods apply Rouche's theorem. $\endgroup$ – Alexandre Eremenko Mar 22 '19 at 1:59
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    $\begingroup$ @AlexandreEremenko sorry, but for the sake of understanding. How is this theorem applied now? What are the holomorphic functions $f,g$ one would consider?- If you think it is off-topic I will close the question also, but I would appreciate an answer very much. Thank you for your time. $\endgroup$ – user121558 Mar 22 '19 at 2:44
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    $\begingroup$ @AlexandreEremenko I now received two answers and both of them seem to rely not only on distances $x_j-x_k$ but on all the parameters $\lambda_j$ as well. That's nice, but not what I asked for. Was the way you intended to apply Rouch\'e's theorem able to avoid that problem? $\endgroup$ – user121558 Mar 23 '19 at 22:24
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    $\begingroup$ @J.Doe: You should mark Alexandre Eremenko's answer as the accepted one then. $\endgroup$ – Willie Wong Mar 28 '19 at 1:30
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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 $x_1$.

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    $\begingroup$ Smallness bound of your $\eta$ depends on $x_j$ and the rest of $\lambda_j$. Was not the question about $\eta$ depending on $x_j$ only (but not on the rest of $\lambda_j$? $\endgroup$ – Alexandre Eremenko Mar 22 '19 at 21:27
  • $\begingroup$ @AlexandreEremenko: I dunno. If the OP can clarify that'd be great! $\endgroup$ – Willie Wong Mar 25 '19 at 7:42
  • $\begingroup$ see my answer. I made the estimate independent on the rest of $\lambda_j$. $\endgroup$ – Alexandre Eremenko Mar 25 '19 at 12:03
  • $\begingroup$ @AlexandreEremenko: thanks! That's a very nice answer. I +1'd it. $\endgroup$ – Willie Wong Mar 28 '19 at 1:29
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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.

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  • $\begingroup$ This is not wgat the question asks for, it only assumes $\lambda_k$ to be small. $\endgroup$ – user69109 Mar 22 '19 at 10:32
  • $\begingroup$ @Tokoyo good point, I've updated the answer to handle the case when $\lambda_j$ is small only for $j=k$. $\endgroup$ – Skeeve Mar 22 '19 at 15:06
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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.

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