$\operatorname{Re}\left(a z^2+b z\right)=a$

$a z^2+b z+a \bar{z}^2+b \bar{z}=2 a$

$a\left(z^2+\bar{z}^2\right)+ b ( z +\overline{ z })=2 a$

$\operatorname{Re}\left( bz z^2+ az \right)= b$

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

$b \left( z ^2+\overline{ z }^2\right)+ a ( z +\overline{ z })=2 b$

$(1) \times b-(2) \times(a)$

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

$\Rightarrow \quad(z+\bar{z})=0 \quad\left(a^2 \neq b^2\right)$

$(1) \times a-(2) \times(b)$

$\Rightarrow \quad\left(a^2-b^2\right)(z+\bar{z})=2\left(a^2-b^2\right) \quad\left(a^2 \neq b^2\right)$

$z^2+\bar{z}^2=2$

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

$z \bar{z}=-1$

$\Rightarrow 1+1^2=-1$

$\Rightarrow \text { No solution }$

$\text { But when } a=-b$

$\operatorname{Re}\left(a z^2-a z\right)=a$

$\Rightarrow \quad \operatorname{Re}\left(a\left(x^2-y^2+i 2 x y\right)-a(x+i y)\right)=a$

$\Rightarrow a\left(x^2-y^2\right)-a x=a$

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

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

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