A) \[\frac{1}{2}\tan \left( \frac{2\alpha }{1+{{\alpha }^{2}}+{{\beta }^{2}}} \right)\]
B) \[\frac{1}{2}{{\tan }^{-1}}\left( \frac{2\alpha }{1+{{\alpha }^{2}}+{{\beta }^{2}}} \right)\]
C) \[\frac{1}{2}{{\tanh }^{-1}}\left( \frac{2\beta }{1+{{\alpha }^{2}}+{{\beta }^{2}}} \right)\]
D) \[\frac{1}{2}{{\tan }^{-1}}\left( \frac{2\beta }{1-{{\alpha }^{2}}-{{\beta }^{2}}} \right)\]
Correct Answer: D
Solution :
\[{{\tan }^{-1}}(\alpha +i\beta )=x+iy\] \[\Rightarrow \] \[\tan (x+iy)=\alpha +i\beta \] \[\therefore \] \[\tan (x-iy)=\alpha -i\beta \] (conjugate) \[tan\text{ }2x=tan[x+iy+x-iy]\] \[=\frac{\tan (x+iy)+\tan (x-iy)}{1-\tan (x+iy)\tan (x-iy)}\] \[=\frac{\alpha +i\beta +\alpha -i\beta }{1-(\alpha +i\beta )(\alpha -i\beta )}\] \[=\frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}}\] \[\Rightarrow \] \[x=\frac{1}{2}{{\tan }^{-1}}\left( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} \right)\]
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