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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $x =$

Similar Questions

The locus of $z$ which lies in the shaded region is best represented by

Let ${z_1}$ and ${z_2}$ be two roots of the equation ${z^2 + az + b = 0}$,where ${z}$ is a complex number. Further,assume that the origin,${z_1}$,and ${z_2}$ form an equilateral triangle. Then:

$A$ point $z$ moves on the Argand diagram in such a way that $|z - 3i| = 2$. Then its locus will be:

The equation $z\overline{z} + (2 - 3i)z + (2 + 3i)\overline{z} + 4 = 0$ represents a circle of radius

Let the curve $z(1+i)+\bar{z}(1-i)=4, z \in \mathbb{C}$,divide the region $|z-3| \leq 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha-\beta|$ equals :

Vedclass Test Series

Exam Paper Generator

Online Exam Module