Question

$\tan i\left[{\log }_e\left(\frac{a-ib}{a+ib}\right)\right]$ is equal to

• A. $\frac{a^2-b^2}{2ab}$
• B. $\frac{2ab}{a^2+b^2}$
• C. ab
• D. $\frac{2ab}{a^2-b^2}$

Answer 1

The correct option is D $\frac{2ab}{a^2-b^2}$

$\tan i\left[{\log }_e\left(\frac{\cos \theta -i\sin \theta }{\cos \theta +i\sin \theta }\right)\right]$ By putting $a=r\cos \theta$ and $b=r\sin \theta$
$=\tan \left[i{\log }_e\right(\cos 2\theta -i\sin 2\theta \left)\right]$
$=\tan \left[i{\log }_e{e}^{-2i\theta }\right]$
$=\tan \left[i\right(-2i\theta \left)\right]$
$=\tan 2\theta$
$=\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2ab}{a^2-b^2}$