Consider the expression $$ \frac{r^2+z^2-a^2+i(2ar)}{r^2+z^2+a^2-2az}$$ for real-valued variables $r$, $a$, $z$. Show that this is equivalent to $$ \frac{r+i(z+a)}{r+i(z-a)}$$
I came across this problem while trying to show that two transformations (conformal mappings) are equivalent and I was surprised that I couldn't find any logical way forward from the first equation. Sage provides that they are, in fact, equivalent, which is good, and now I only want to show the derivation of the latter equation from the former (i.e. reduce the former equation to the latter, do not simply show that their difference is zero).
I feel in my gut that there's a way by factoring, and multiplying by a complex conjugate, which would introduce the imaginary part in the denominator, but I haven't found anything that works (completing squares, grouping terms and factoring, etc).
The expression is equivalent to $$ \frac {(r+ia)^2 - i^2 z^2} { r^2 - i^2 (z-a)^2 } $$ $$ \frac {(r+ia-iz)(r+ia+iz)}{(r+iz-ia)(r-iz+ia)} $$ $$ \frac {r+i(z+a)}{r+i(z-a)} $$