For a very large parameter $d$, variable $x$ and constants $a$ and $b$ (with $a$ real and $b$ generally complex), I am trying to prove that the following expression
$$ \left( \frac{ia-xb}{ia+xb} \right)e^{-x d}=\pm 1, $$
will reduce perturbatively to
$$ x\approx \frac{-ia}{b}(1\pm 2 e^{iad/b}), $$ but I am unable to reach this solution.
Specifically, I am not sure whether I should treat the term $e^{-xd}$ as a whole and call it $\epsilon$ (small), or whether I should also expand the $x$ in this exponent as a series. I am not sure how the latter method can be done as it will have mixed powers of expansion parameter, but I tried the former method and re-wrote the problem as
$$\left( \frac{ia-xb}{ia+xb} \right) \epsilon=\pm 1,$$ convincing myself that the actual "smallness" here is due to $d$ being large (not $x$), making the exponent as a whole always small. Then I expanded $x$ to first order in $\epsilon$ as $$ x=x_{0}+\epsilon x_{1}+\cdots$$
This gives $x_{0}=-ia/b$ and $x_{1}=\pm a(1-i/b)$, but this is not the sought answer. I am not sure how to handle this problem using perturbative expansions.