Q. If $\tan^{- 1}{(\alpha + i\beta) = x + iy}$, then $x$ is equal to
A) $\frac{1}{2}\tan^{- 1}\left( \frac{2\alpha}{1 - \alpha^{2} - \beta^{2}} \right)$
B) $\frac{1}{2}\tan^{- 1}\left( \frac{2\alpha}{1 + \alpha^{2} + \beta^{2}} \right)$
C) $\tan^{- 1}\left( \frac{2\alpha}{1 - \alpha^{2} - \beta^{2}} \right)$
D) None of these
Explanation
(a)
We have, $\tan^{- 1}{(\alpha + i\beta) = x + iy}$
$\Rightarrow \ \alpha + i\beta = \tan{(x + iy)}\ \ \ \$ …(i)
Taking conjugate,
$(\alpha + i\beta = \tan(x + iy)\$ …(ii)
$\therefore\ \tan{2x = \tan\left\lbrack (x + iy) + (x - iy) \right\rbrack}$
$\Rightarrow \ \tan{2x = \frac{(\alpha + i\beta) + (\alpha - i\beta)}{1 - (\alpha + i\beta) + (\alpha - i\beta)}}$
$= \ \frac{2\alpha}{1 - (\alpha^{2} + \beta^{2})}$
$\therefore\ \ x = \frac{1}{2}\tan^{- 1}\left( \frac{2\alpha}{1 - \alpha^{2} - \beta^{2}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.