If tanh^{-1}(x+iy) = frac{1}{2} tanh^{-1}left | vedclass

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(C) We know the relationship between the inverse hyperbolic tangent and the inverse trigonometric tangent function: $	anh^{-1}(z) = \frac{1}{i} 	an^

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$i 	an^{-1}(y)$

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(C) We know the relationship between the inverse hyperbolic tangent and the inverse trigonometric tangent function: $	anh^{-1}(z) = \frac{1}{i} 	an^{-1}(iz)$.To find $	anh^{-1}(iy)$,we substitute $z = iy$ into the formula:$	anh^{-1}(iy) = \frac{1}{i} 	an^{-1}(i(iy))$Since $i^2 = -1$,this simplifies to:$	anh^{-1}(iy) = \frac{1}{i} 	an^{-1}(-y)$Using the property $	an^{-1}(-y) = -	an^{-1}(y)$,we get:$	anh^{-1}(iy) = \frac{1}{i} (-	an^{-1}(y)) = -\frac{1}{i} 	an^{-1}(y)$Since $\frac{1}{i} = -i$,we have $-\frac{1}{i} = i$.Therefore,$	anh^{-1}(iy) = i 	an^{-1}(y)$.Thus,option $C$ is correct.

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

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