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Question: If \(\tan^{- 1}(\alpha + i\beta) = x + iy,\) then \(x =\)...

If tan⁡−1(α+iβ)=x+iy,\tan^{- 1}(\alpha + i\beta) = x + iy, then x=x =

A

12tan⁡−1(2α1−α2−β2)\frac{1}{2}\tan^{- 1}\left( \frac{2\alpha}{1 - \alpha^{2} - \beta^{2}} \right)

B

12tan⁡−1(2α1+α2+β2)\frac{1}{2}\tan^{- 1}\left( \frac{2\alpha}{1 + \alpha^{2} + \beta^{2}} \right)

C

tan⁡−1(2α1−α2−β2)\tan^{- 1}\left( \frac{2\alpha}{1 - \alpha^{2} - \beta^{2}} \right)

D

None of these

Answer

12tan⁡−1(2α1−α2−β2)\frac{1}{2}\tan^{- 1}\left( \frac{2\alpha}{1 - \alpha^{2} - \beta^{2}} \right)

Explanation

Solution

tan⁡−1(α+iβ)=x+iy\tan^{- 1}(\alpha + i\beta) = x + iy

tan⁡−1(α−iβ)=x−iy\tan^{- 1}(\alpha - i\beta) = x - iy

2x=x+iy+x−iy2x = x + iy + x - iy = tan⁡−1(α+iβ)+tan⁡−1(α−iβ)\tan^{- 1}(\alpha + i\beta) + \tan^{- 1}(\alpha - i\beta)

∴\therefore x=12tan⁡−12α1−α2−β2x = \frac{1}{2}\tan^{- 1}\frac{2\alpha}{1 - \alpha^{2} - \beta^{2}}= 12tan⁡−1α+iβ+α−iβ1−(α+iβ)(α−iβ)\frac{1}{2}\tan^{- 1}\frac{\alpha + i\beta + \alpha - i\beta}{1 - (\alpha + i\beta)(\alpha - i\beta)}

= 12tan⁡−1(2α1−α2−β2)\frac{1}{2}\tan^{- 1}\left( \frac{2\alpha}{1 - \alpha^{2} - \beta^{2}} \right)