If $ anh^{- 1}{(x + iy)} = \frac{1}{2} anh^{- 1}\left( \frac{2x}{1 + x^{2} + y^{2}} ight) + \frac{i}{2} an^{- 1}\left( \frac{2y}{1 - x^{2} - y^{2}} ight),\ x,\ y \in R$, then $ anh^{- 1}{(iy)}$ is

Question

If $	anh^{- 1}{(x + iy)} = \frac{1}{2}	anh^{- 1}\left( \frac{2x}{1 + x^{2} + y^{2}} ight) + \frac{i}{2}	an^{- 1}\left( \frac{2y}{1 - x^{2} - y^{2}} ight),\ x,\ y \in R$, then $	anh^{- 1}{(iy)}$ is

Dr. Abubacker · Accepted Answer

(c)
Given, $	anh^{- 1}{(x + iy) = \frac{1}{2}	anh^{- 1}\left( \frac{2x}{1 + x^{2} + y^{2}} ight)} + \frac{i}{2}	an^{- 1}\left( \frac{2y}{1 - x^{2} - y^{2}} ight);x,\ y \in R$
Put $x = 0$,
$	anh^{- 1}{(iy) = \frac{1}{2}	anh^{- 1}{(0) + \frac{i}{2}	an^{- 1}\left( \frac{2y}{1 - y^{2}} ight)}}$
$= 0 + \frac{i}{2}	an^{- 1}\left( \frac{2	an	heta}{1 - 	an^{2}	heta} ight)\ \ \ (put\ y = 	an	heta)$
$= \frac{i}{2}	an^{- 1}{(	an{2	heta})}$
$= \frac{i}{2}\ 2	heta$
$= i	an^{- 1}y$

Q. If $\tanh^{- 1}{(x + iy)} = \frac{1}{2}\tanh^{- 1}\left( \frac{2x}{1 + x^{2} + y^{2}} \right) + \frac{i}{2}\tan^{- 1}\left( \frac{2y}{1 - x^{2} - y^{2}} \right),\ x,\ y \in R$, then $\tanh^{- 1}{(iy)}$ is

Explanation

Reviewed by Dr. Abubacker

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