Why this $$ \Im \frac{-2}{1+e^{-s + i a }} $$
equals to this expression:
$$\\\ \frac{\sin(a)}{\cos(a)+\cosh(s)} $$
I was trying to evaluate the Fourier transform of a hyperbolic function and my textbook and the other sources say this equality holds on. I only got to:
$$ \Im \frac{e^{-ia}}{e^{-ia}+e^{-s}} $$
Well, generally the imaginary part of $e^{-ia}$ is $\sin(a)$, but $e^{-s}$ is real. So I don't understand the $\cosh (s)$ part. I am just so confused.
The fourier tranform has been done on this function:
$$f(x) = \frac{\sinh(ax)}{\sinh(\pi x)}$$