Question

$\cos \left(i\log \left(\frac{a-ib}{a+ib}\right)\right)$ is equal to

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

Answer 1

Answer: (B)

$\frac{a^2-b^2}{a^2+b^2}$

Let $a+ib=r{e}^{i\theta }$ and $a-ib=r{e}^{-i\theta }.$

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

$\therefore \cos \left(i\log \frac{a-ib}{a+ib}\right)=\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$=\frac{1-\left(\frac{b^2}{a^2}\right)}{1+\left(\frac{b^2}{a^2}\right)}=\frac{a^2-b^2}{a^2+b^2}$