Proving $\frac{\sinh\tau+\sinh i\sigma}{\cosh\tau+\cosh i\sigma }=-\coth\left(i\frac{\sigma+i\tau }{2}\right)$ for bipolar coordinates $(\sigma,\tau)$

Question

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.

Answer 1

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.

Answer 2

This appears to be wrong; for example, the case $\sigma=0$ reduces to $\tanh\tau=\coth\frac{\tau}{2}$. One approach is to use $\sinh A+\sinh B =2\sinh\frac{A+B}{2}\cosh\frac{A-B}{2},\,\cosh A+\cosh B=2\cosh\frac{A+B}{2}\cosh\frac{A-B}{2}$ with $A=\tau,\,B=i\sigma$, so the fraction reduces to $\tanh\frac{A+B}{2}=\tanh i\frac{\sigma-i\tau}{2}$. I suspect you may have meant to evaluate$$\frac{\sinh\tau+\sinh i\sigma}{\cosh\tau\mathbf{\color{orange}{-}}\cosh i\sigma}=\frac{2\sinh\frac{A+B}{2}\cosh\frac{A-B}{2}}{2\sinh\frac{A+B}{2}\sinh\frac{A-B}{2}}=\coth\frac{A-B}{2}=-\coth i\frac{\sigma+i\tau}{2}.$$