Limit of $\frac{1}{x+i y}$ for $y \rightarrow 0$ and distributional relations

Question

So I know for $y \rightarrow 0$ I have the following (distrubutional) relation:

$\frac{1}{x+i y} = \frac{x}{x^2+y^2} - i \frac{y}{x^2+y^2} = P(1/x) - i \pi \delta(x) $

where in the last expression the limit was taken, $P$ denotes Cauchy prinzipal value and $\delta$ the Dirac-delta distribution.

Now, I would like to take the limit $y \rightarrow 0$ of the expression $\frac{1}{x+\frac{1}{z+i y}}$ . Having not dealt with limits in a long time, can you guys estimate if this is possible? Maybe even ideas on how to go about this?

*edit

The original question in proper notation:

Let $x_1,x_2 \neq 0 \in \mathbb{R} $, $x\in \mathbb{R}$, $\varepsilon > 0$.

What happens to the expression $\frac{1}{x+x_1+\frac{1}{x+x_2 \pm i \varepsilon}}$ for $\varepsilon \rightarrow 0$ for a given $x$.

Answer 1

Much of the answer depends on the allowed values for the parameters involved.

If $x=0$, the expression can be rewritten as $z+iy$, which tends to $z$ as $y\to0$.

Suppose $x\neq0$ and let $z=u+iv$, $$ \frac{1}{x+\frac{1}{z+iy}}=\frac{u+i\left(y+v\right)}{\left(1+xu\right)+ix\left(y+v\right)}. $$ If $v\neq0$, as $y\to0$, we get $$ \frac{u+iy}{\left(1+xu\right)+ix} $$ If $v=0$, let also $y=\pm\varepsilon,$ for positive $\varepsilon$, $$ \frac{u\pm i\varepsilon}{\left(1+xu\right)\pm ix\varepsilon} \rightarrow u \left[PV\left(\frac{1}{1+xu}\right)\mp i\pi\delta(1+xu) \right]. $$

Answer 2

\begin{equation*} \frac{1}{x+\frac{1}{x+iy}}=\frac{x+iy}{1+x(x+iy)}=\frac{x+iy}{1+x^{2}+ixy} \rightarrow \frac{x}{1+x^{2}} \end{equation*} Edit

For the correct expression for general $z$ \begin{equation*} \frac{1}{x+\frac{1}{z+iy}}=\frac{z+iy}{1+x(z+iy)}\overset{y\rightarrow 0}{% \longrightarrow }\frac{z}{1+xz} \end{equation*} Distributions do not seem to come into play.