$(ii) \ Let \ \dfrac {a-ib}{a+ib}= r \ e^{i\theta} $ $\dfrac {a-ib}{a+ib}= \dfrac {a-ib}{a+ib}\times \dfrac {a-ib}{a-ib} = \dfrac {(a-ib)^2} { a^2 -(ib)^2} = \dfrac {a^2 -b^2 - 2 iab}{a^2 +b^2} = r \ e^{i \theta} $ $r= \sqrt{\left ( \dfrac {a^2- b^2}{a^2+b^2} \right )^2+ \left ( - \dfrac {2ab}{a^2+b^2}\right)^2} $ $=\dfrac {\sqrt{a^4+b^4-2a^2b^2 + 4a^2b^2}}{a^2+b^2} = \dfrac {\sqrt{a^4+b^4+2a^2 b^2}}{a^2+b^2}$ $=\dfrac {\sqrt{(a^2+b^2)^2}}{a^2+b^2} $ $\therefore \ r=1$ $\theta = \tan^{-1}\left ( \dfrac {-2ab}{a^2-b^2} \right ) $ $\tan \theta = \dfrac {-2ab}{a^2-b^2} \ \cdots \ (A) $ $\tan \left ( i \log \left ( \dfrac {a-ib}{a+ib} \right ) \right ) $ $=\tan (i\log r \ e^{i\theta})$ $= \tan [i (i\theta \times \log e)] \ \cdots \ \cdots { \because r=1 }$ $=\tan [i^2 \theta] = \tan [-\theta]$ $= -\tan \theta$ $\therefore, \tan \left ( i \log \left ( \dfrac {a-ib}{a+ib} \right ) \right ) = \dfrac {2ab}{a^2-b^2} \ \cdots \ \cdots From \ (A)$