Question

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

• A. ab
• B.
• C.
• D.

Answer 1

The correct option is B

$Leta+ib=r{e}^{i\theta }a-ib=r{e}^{-i\theta }\therefore ilog\frac{a-ib}{a+ib}=ilog\left(e^{-2\theta i}\right)=i(-2\theta i)loge=2\theta \therefore cos\left(ilog\frac{a-ib}{a+ib}\right)=cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}=\frac{a^2-b^2}{a^2+b^2}$