A) \[2\text{ }tan{{h}^{-1}}(y)\]
B) \[-2\text{ }tan{{h}^{-1}}(y)\]
C) \[\text{i }ta{{n}^{-1}}y\]
D) \[\text{-i }ta{{n}^{-1}}(y)\]
Correct Answer: C
Solution :
Given, \[{{\tanh }^{-1}}(x+iy)=\frac{1}{2}{{\tanh }^{-1}}\left( \frac{2x}{1+{{x}^{2}}+{{y}^{2}}} \right)\] \[+\frac{i}{2}{{\tan }^{-1}}\left( \frac{2y}{1-{{x}^{2}}-{{y}^{2}}} \right);x,y\in R\] Put \[x=0,\] \[{{\tanh }^{-1}}(iy)=\frac{1}{2}{{\tanh }^{-1}}(0)+\frac{i}{2}{{\tan }^{-1}}\left( \frac{2y}{1-{{y}^{2}}} \right)\] \[=0+\frac{i}{2}{{\tan }^{-1}}\left( \frac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)\] (Put \[y=tan\,\theta\] ) \[=\frac{i}{2}{{\tan }^{-1}}(\tan 2\theta )=\frac{i}{2}2\theta\] \[=i{{\tan }^{-1}}[y]\]
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