QJEEAdv22Ma4015: JEE Advanced 2022 Mathematics - Class 11 JEEAdv22Ma4015 Solution: Let denote the complex conjugate of a complex number and let .…

Let, $z = x + iy$

$\therefore$ $\overline z = x - iy$

Given, $\overline z - {z^2} = i(\overline z + {z^2})$

$\Rightarrow (x - iy) - {(x + iy)^2} = i\left[ {(x - iy) + {{(x + iy)}^2}} \right]$

$\Rightarrow (x - iy) - ({x^2} - {y^2} + 2ixy) = i[x - iy + {x^2} - {y^2} + 2ixy]$

$\Rightarrow (x - {x^2} + {y^2}) - iy(1 + 2x) = xi - {i^2}y + {x^2}i - i{y^2} + 2{i^2}xy$

$\Rightarrow (x - {x^2} + {y^2}) - iy(1 + 2x) = xi + y + i{x^2} - i{y^2} - 2xy$

$\Rightarrow (x - {x^2} + {y^2}) - iy(1 + 2x) = y(1 - 2x) + i(x + {x^2} - {y^2})$

Comparing both sides real part we get,

$x - {x^2} + {y^2} = y - 2xy$

$\Rightarrow x - {x^2} + {y^2} - y + 2xy = 0$ ..... (1)

And comparing both sides imaginary part we get,

$- y(1 + 2x) = x + {x^2} - {y^2}$

$\Rightarrow - y - 2xy = x + {x^2} - {y^2}$

$\Rightarrow x + {x^2} - {y^2} + y + 2xy = 0$ ...... (2)

Adding equation (1) and (2) we get,

$x - {x^2} + {y^2} - y + 2xy + x + {x^2} - {y^2} + y + 2xy = 0$

$\Rightarrow 2x + 4xy = 0$

$\Rightarrow 2x(1 + 2y) = 0$

$\therefore$ $x = 0$ or $1 + 2y = 0 \Rightarrow y = - {1 \over 2}$

Case 1 : When $x = 0$ :

Put $x = 0$ at equation (1), we get

${y^2} - y = 0$

$\Rightarrow y(y - 1) = 0$

$\Rightarrow y = 0,1$

$\therefore$ $z = 0 + 0i$ or $0 + i$

Case 2 : When $y = - {1 \over 2}$ :

Put $y = - {1 \over 2}$ in equation (1), we get

$x - {x^2} + {1 \over 4} + {1 \over 2} - x = 0$

$\Rightarrow {x^2} = {1 \over 4} + {1 \over 2}$

$\Rightarrow {x^2} = {3 \over 4}$

$\Rightarrow x = \, \pm \,{{\sqrt 3 } \over 2}$

$\therefore$ $z = {{\sqrt 3 } \over 2} - {i \over 2}$ or $z = - {{\sqrt 3 } \over 2} - {i \over 2}$

$\therefore$ Number of distinct $z = 4$