Show that the equation, $\;\tan\left(i \log\dfrac{x−iy}{x+iy}\right)=2\;$ represents the rectangular hyperbola $\;x^2 − y^2 = xy\;$.
What I could do is to simplify the expression
$\tan\left(i\log\dfrac{x−iy}{x+iy}\right)=2\quad$ to the following one :$$\tan\left(i\log\dfrac{x^2-y^2-2ixy}{x^2+y^2}\right)\;.$$
I don’t know how to proceed further. Any hints will be much appreciated.
Hint. Consider the following identity:
$$\tan(i\ln(X)) \equiv \frac{i \left(X^2-1\right)}{X^2+1}$$
where in your case $X = \frac{x-iy}{x+iy}$.
Substitute, arrange and you're done.
Let $\ln\dfrac{x-iy}{x+iy}=2it$ where $t$ is real
$$\implies\dfrac{x-iy}{x+iy}=\dfrac{e^{it}}{e^{-it}}$$
Applying Componendo and Dividendo $$\dfrac x{iy}=\dfrac{\cos t}{-i\sin t}\implies\tan t=-\dfrac yx$$
$$2=\tan(i(2it))=\tan(-2t)=-\tan2t=\dfrac{2\tan t}{1-\tan^2t}=?$$
