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
Correct Answer: A
Solution :
\[{{\tan }^{-1}}(\alpha +i\beta )=x+iy\] \[\Rightarrow \] \[\alpha +i\beta =\tan (x+iy)\] ?..(i) Taking conjugate, \[\Rightarrow \] \[(\alpha -i\beta )=\tan (x-iy)\] ?..(ii) \[\Rightarrow \] \[\tan 2x=\tan [(x+iy)+(x-iy)]\] \[\therefore \,\,\tan 2x=\frac{(\alpha +i\beta )+(\alpha -i\beta )}{1-(\alpha +i\beta )\,\,(\alpha -i\beta )}=\frac{2\alpha }{1-({{\alpha }^{2}}+{{\beta }^{2}})}\] \[\therefore \,x=\frac{1}{2}{{\tan }^{-1}}\left( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} \right)\].
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