How can we prove that, in the limit of $|1/y| \leq |c|$, where $c,y$ are generally complex, the following approximation holds?
$$ \frac{1-yc}{1+yc} \rightarrow e^{-2/cy} $$
My attempt: if we call $cy=x$, with $|x|\geq 1$ then we have
$$ \frac{1-yc}{1+yc}=\frac{1-x}{1+x}=\frac{-x(1-1/x)}{+x(1+1/x)}=-\frac{(1-1/x)}{(1+1/x)}= -(1-1/x)(1+1/x)^{-1}\approx -(1-1/x)^{2}, $$
to first order in $1/x$. I then say that $(1-1/x)$ is also the first order approximation of $e^{-1/x}$, therefore we end up with
$$ \frac{1-yc}{1+yc} \approx -(1-1/x)^{2} \approx -e^{-2/x} = -e^{-2/cy}. $$
Is my method accurate/correct?