Let $\tan^{-1} \big ( e^{i \theta} \big) = \alpha + i\beta$

$\therefore e^{i\theta} = \tan (\alpha + i \beta)$

$\therefore \cos \theta + i \sin \theta = \tan (\alpha + i \beta ) \ \cdots \ \cdots (1)$

Hence we can write $\cos \theta - i \sin \theta = \tan (\alpha - i \beta) \ \cdots \ \cdots (2)$

$\therefore \tan 2\alpha = \tan \big [ (\alpha + i\beta) + (\alpha - i\beta ) \big ]$

Using $\tan (A+B) = \dfrac {\tan A + \tan B}{1-\tan A \tan B}$

$\therefore \tan (2\alpha ) = \dfrac {\tan (\alpha + i \beta) + \tan (\alpha - i\beta) }{1-\tan (\alpha + i \beta) \tan (\alpha - i \beta)}$

Using values of (1) and (2)

$\therefore \tan (2\alpha) = \dfrac {(\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta)}{1- (\cos \theta + i \sin \theta) (\cos \theta - i \sin \theta)}$

$\therefore \tan (2 \alpha) = \dfrac {2 \cos \theta}{1-(\cos ^2 \theta + \sin ^2 \theta)}$

$\therefore \tan (2\alpha) = \dfrac {\cos \theta} {0}$

$\therefore 2\alpha = \dfrac {\pi}{2} $ and $2\alpha = n \pi + \dfrac {\pi } {2}$ (general value)

$\therefore \alpha = \dfrac {n\pi} {2} + \dfrac {\pi}{4} \ \cdots \ \cdots (A)$

Also, $\tan 2 i\beta = \tan \big [ (\alpha + i \beta ) - (\alpha - i \beta ) \big ]$

$\therefore \tan (2i\beta ) = \dfrac {\tan (\alpha + i\beta) - \tan (\alpha - i \beta )}{1 \mp \tan (\alpha - i \beta)}$

Using values of (1) and (2)

$\therefore \tan (2 i\beta) = \dfrac {(\cos \theta + i \sin \theta) - (\cos \theta - i \sin \theta)}{1+ (\cos \theta + i \sin \theta) (\cos \theta - i \sin \theta)}$

$\therefore \tan (2i\beta) = \dfrac {2i \sin \theta}{1+1} = i \sin \theta$

$\therefore i \tanh 2 \beta = i \sin \theta$

$\therefore \tanh 2\beta = \sin \theta$

$\therefore 2\beta = \tanh^{-1}\sin \theta = \dfrac {1}{2} \log \left [ \dfrac {1+ \sin \theta} { 1- \sin \theta} \right ] = \dfrac {1}{2} \log \left [ \dfrac {1+\sin \theta}{1- \sin \theta} \right ]$

$= \dfrac {1}{2} \log \left [ \dfrac {\cos \frac {\theta}{2} + \sin \frac {\theta}{2}}{\cos \frac {\theta}{2} - \sin \frac {\theta}{2}} \right ]^2 = \log \left [ \dfrac {\cos \frac {\theta}{2}+ \sin \frac {\theta}{2}}{\cos \frac {\theta}{2} - \sin \frac {\theta}{2}} \right ]$

$\beta = \dfrac {1}{2} \log \left [ \dfrac {1+ \tan \frac {\theta}{2}} { 1- \tan \frac {\theta}{2}} \right ]$

$\beta = \dfrac {1}{2} \log \tan \left ( \dfrac {\pi}{4} + \dfrac {\theta}{1} \right ) \ \cdots \ \cdots (B)$

Hence,

real part $= \alpha = \dfrac {n \pi}{2} + \dfrac {\pi}{4}$ and

imaginary part $= \beta = \dfrac {1}{2} \log \tan \left ( \dfrac {\pi}{4} + \dfrac {\theta}{2} \right ) $