Question

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

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

Answer 1

Answer: (B)

We have, $tan \left[i\,log\left(\frac{a-ib}{a+ib}\right)\right]$
Let $a=r cos\,\theta$, $b=r sin\,\theta$
$\therefore \, \frac{b}{a}=tan\,\theta \quad\ldots\left(i\right)$
$\therefore \, tan\left[i\,log\left(e^{-i\theta} \times e^{-i\theta}\right)\right]$
$=tan \left[i\, log\,e^{-2i\theta}\right]=tan\,2\theta\quad \left(\because\,log\,e=1\right)$
$=\frac{2\, tan\,\theta}{1-tan^{2}\,\theta}=\frac{2\frac{a}{b}}{1-\frac{b^{2}}{a^{2}}}$
$=\frac{2ab}{a^{2}-b^{2}}$ (using $(i)$)