Complex Numbers And Quadratic Equations question 305
Question: If $ {{\tan }^{-1}}(\alpha +i\beta )=x+iy, $ then x = [RPET 2002]
Options:
A) $ \frac{1}{2}{{\tan }^{-1}}( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} ) $
B) $ \frac{1}{2}{{\tan }^{-1}}( \frac{2\alpha }{1+{{\alpha }^{2}}+{{\beta }^{2}}} ) $
C) $ {{\tan }^{-1}}( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} ) $
D) None of these
Show Answer
Answer:
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}}( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} ) $ .
BETA
