Question

The expression $tan(ilog(\frac{a-ib}{a+ib}))$ reduces to:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $a+bi=r{e}^{i\theta }$. Then $tan\theta =\frac{b}{a}$
$tan(ilog\frac{a-ib}{a+ib}=tan\left(ilog{e}^{-2i\theta }\right)$
=$tan(-2i^2\theta )=\frac{2tan\theta }{1-ta{n}^{\theta }}=\frac{\frac{2b}{a}}{1-\left(\frac{b^2}{a^2}\right)}=\frac{2ab}{a^2-b^2}$