Question

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

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

Answer 1

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

Let $a=r\cos \theta$ and $b=r\sin \theta$

$\therefore \tan \theta =\frac{b}{a}$

Now, $\frac{a-ib}{a+ib}=\frac{r(\cos \theta -i\sin \theta )}{r(\cos \theta +i\sin \theta )}={(\cos \theta -i\sin \theta )}^2$

$=\cos 2\theta -i\sin 2\theta =e^{-2i\theta }$

$\therefore i\log \frac{a-ib}{a+ib}=i\log e^{-2i\theta }=i(-2i\theta )=2\theta$

$\therefore \tan \left[i\log \frac{a-ib}{a+ib}\right]=\tan 2\theta$

$=\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2\frac{b}{a}}{1-\frac{b^2}{a^2}}=\frac{2ab}{a^2-b^2}$

Hence option $C$ is the answer