JEE 2023 — Mathematics PYQ

We know that $z+\bar{z}=2\text{Re}(z)$

$\therefore (a{z}^2+bz)+(a{\bar{z}}^2+b\bar{z})=2a$

$\Rightarrow a(z^2+{\bar{z}}^2)+b(z+\bar{z})=2a\dots (i)$

Add $(b{z}^2+az)+(b{\bar{z}}^2+a\bar{z})=2b$

$\Rightarrow b(z^2+{\bar{z}}^2)+a(z+\bar{z})=2b\dots (ii)$

From $(i)\times b-(ii)\times a$

$(b^2-a^2)(z+\bar{z})=0$

$z+\bar{z}=0(\because a\ne b)$

From $(i)\times a-(ii)\times b$

$(a^2-b^2)(z^2+{\bar{z}}^2)=2(a^2-b^2)[a^2\ne b^2]$

$z^2+{\bar{z}}^2=2\Rightarrow (z+\bar{z}{)}^2-2z\bar{z}=2$

$\Rightarrow z\bar{z}=-1\Rightarrow 1+1^2=1\text{(No solution)}$

But when $a=-b$

$\text{Re}(a{z}^2-az)=a$

Put $z=x+iy$

$\therefore x^2-x-1=y^2$

For any real value of y there two values of x, hence infinite complex number are possible.