Prove $\coth\;2v = \frac{x^2 + y^2 + 1}{2y}$

Question

Q: Given that $x + jy =\tan (u + jv)$, prove that

$$\coth\;2v = \frac{x^2 + y^2 + 1}{2y}$$

I would like to ask this question, how can we prove it? I had tried to expand the equation $x + jy =\tan (u + jv)$ to the following equations

$$x = \frac{\tan\;u(1 +\tanh^2v)}{1-\tan^2u\;\tanh^2v}$$

and $$y = \frac{\tanh\;v(1 -\tan^2u)}{1-\tan^2u\;\tanh^2v}$$

I tried to relate to the question but lastly I stuck at here

$$\coth\;2v = \frac{x(1-\tan^2u)}{2y\;\tan\;u}$$

Hope somebody can give me some tips to solve it, thank you : )

Answer 1

Since $\frac{x}{y}=\frac{\tan u(1+\tanh^2v)}{\tanh v(1-\tan^2u)}$, you want to prove $\coth2v=\frac{1+\tanh^2v}{2\tanh v}$, which is well-known.