I'm stuck on the following steps:
$$ \frac{m\sinh\phi}{\sinh\tau}\,\Re[e^{ims\cosh(\phi-\tau)}] = \frac{m\sinh(\phi+\tau)}{\sinh\tau}\,\Re[e^{ims\cosh(\phi)}] $$ I'm not quite sure how we can get the right-hand side from the left. I know $\cosh(a-b) = \cosh a\cosh b-\sinh a\sinh b$, but how can we take out the '$-\tau$' from the real part of this exponent?