If {tan ^{ - 1}}(alpha + ibeta ) = x + iy, th | vedclass

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(A) Given ${	an ^{ - 1}}(\alpha + i\beta ) = x + iy$. Taking the tangent of both sides,we get $\alpha + i\beta = 	an (x + iy) \dots (i)$. Taking the

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$\frac{1}{2}{	an ^{ - 1}}\left( {\frac{{2\alpha }}{{1 - {\alpha ^2} - {\beta ^2}}}} ight)$

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(A) Given ${	an ^{ - 1}}(\alpha + i\beta ) = x + iy$. Taking the tangent of both sides,we get $\alpha + i\beta = 	an (x + iy) \dots (i)$. Taking the complex conjugate of both sides,we get $\alpha - i\beta = 	an (x - iy) \dots (ii)$. Using the formula $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have: $	an 2x = 	an [(x + iy) + (x - iy)] = \frac{	an (x + iy) + 	an (x - iy)}{1 - 	an (x + iy) 	an (x - iy)}$. Substituting the values from $(i)$ and $(ii)$: $	an 2x = \frac{(\alpha + i\beta ) + (\alpha - i\beta )}{1 - (\alpha + i\beta )(\alpha - i\beta )} = \frac{2\alpha }{1 - (\alpha ^2 + \beta ^2)}$. Therefore,$2x = 	an ^{-1} \left( \frac{2\alpha }{1 - \alpha ^2 - \beta ^2} ight)$. Thus,$x = \frac{1}{2} 	an ^{-1} \left( \frac{2\alpha }{1 - \alpha ^2 - \beta ^2} ight)$.

If ${\tan ^{ - 1}}(\alpha + i\beta ) = x + iy,$ then $x =$

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