VITEEE 2017 Paper

Consider the equation $\overline{zz}$ + (2 - 3i) z + (2 + 3i) $\overline{z}$ + 4 = 0 ... (1) Let z = x + iy and $\overline{z}$ = x - iy , $\overline{zz}$ = $x^2+y^2$ Put value of z, $\overline{z}$ and $\overline{zz}$ in equation (1), we get $(x^2+y^2)$ + (2 - 3i) (x + iy) + (2 + 3i) (x - iy) + 4 = 0 ⇒ 4x + 6y + 4 + $x^2+y^2$ = 0 Now, we make it perfect square ⇒ $x^2+y^2$ + 4x + 6y + 4 + 4 + 9 = 4 + 9 ⇒ $(x+2)^2$ + $(y+3)^2$ = 9 This represents a circle of radius 3.