I am having trouble proving the following identity:
$$\frac{\sinh \tau +\sinh i\sigma }{\cosh \tau +\cosh i\sigma }=-\coth \left(i \frac{\sigma +i\tau }{2}\right)$$
I have tried using identities and the definitions but haven't had much luck. This is a missing step in inverting the bipolar coordinates. Any assistance is appreciated.
Best Answer
The way I ended up solving it was by applying the same technique as in [ Hint to show $\tanh(z)=\frac{\sinh(2x)+i\sin(2y)}{\cosh(2x)+\cos(2y)}$? ]
To prove the identity: $$\coth z=\frac{\sinh 2x-i \sin 2y}{\cosh 2x-\cos 2y}$$\xo
Then I used the fact that $\sinh (-\tau)=-\sinh \tau $ and $\cosh (-\tau)=\cosh \tau$ to get: $$-\coth \left(\frac{-\tau +i \sigma }{2}\right)=\frac{\sinh \tau+i \sin \sigma}{\cosh \tau-\cos \sigma }$$
Then, use the identities $\sinh z=-i \sin (i z)$ and $\cosh z =\cos (i z)$ to get the identity into the desired form.