Question

Prove that tan$\left(ilo{g}_e\left(\frac{a-ib}{a+ib}\right)\right)=\frac{2ab}{a^2-b^2}$ (where $a,b\in R^{+}$).

Answer 1

Let $a+ib=r{e}^{i\theta }$
$⟹a-ib=r{e}^{-i\theta }$
$⟹\frac{a-ib}{a+ib}=e^{-2i\theta }⟹{\log }_e\left(\frac{a-ib}{a+ib}\right)=-2i\theta$
$⟹\tan \left(i{\log }_e\left(\frac{a-ib}{a+ib}\right)\right)=\tan 2\theta$
$=\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{\frac{2b}{a}}{1-\frac{b^2}{a^2}}=\frac{2ab}{a^2-b^2}$