Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X=\left\{z \in \mathbb{C}: \operatorname{Re}\left(a z^{2}+b z\right)=a\right.$ and $\left.\operatorname{Re}\left(b z^{2}+a z\right)=b\right\}$ is equal to :

Let $z=x+i y$ <br/><br/>$$ \begin{array}{ll} &\Rightarrow z^2=x^2-y^2+2 i x y \\\\ &\therefore a z^2+b z \\\\ & =a\left(x^2-y^2+2 i x y\right)+b(x+i y) \\\\ & =a\left(x^2-y^2\right)+b x+2 a i x y+b i y \end{array} $$ <br/><br/>$\begin{array}{ll}\operatorname{Re}\left(a z^2+b z\right)=b \\\\ \Rightarrow a\left(x^2-y^2\right)+b x=a \\\\ \Rightarrow x^2-y^2+\frac{b}{a} x=1 .........(i)\end{array}$ <br/><br/>$\begin{aligned} & \text { and } b z^2+a z \\\\ & =b\left(x^2-y^2+2 i x y\right)+a(x+i y) \\\\ & =b\left(x^2-y^2\right)+a x+2 b i x y+a i y \\\\ & \operatorname{Re}\left(b z^2+a z\right)=b \\\\ & \Rightarrow b\left(x^2-y^2\right)+a x=b \\\\ & \Rightarrow x^2-y^2+\frac{a}{b} x=1 ..........(ii)\end{aligned}$ <br/><br/>On subtracting Equation (ii) from Equation (i), we get <br/><br/>$$ \begin{aligned} & \frac{b}{a} x-\frac{a}{b} x=0 \\\\ & \Rightarrow \left(\frac{b}{a}-\frac{a}{b}\right) x=0 \\\\ & \Rightarrow x=0 \text { or } \frac{b}{a}-\frac{a}{b}=0 \\\\ & \Rightarrow b^2-a^2=0 \\\\ & \Rightarrow a= \pm b \\\\ & \Rightarrow a=-b(\text { since } a \neq b) \end{aligned} $$ <br/><br/>From (i), when $x=0$, then <br/><br/>$$ \begin{aligned} & 0-y^2=1 \Rightarrow y^2=-1 \\\\ & \Rightarrow y \in \phi \Rightarrow z \in \phi \text { has no solution. } \end{aligned} $$ <br/><br/>When, $a=-b$, then $x^2-y^2-x=1$ has infinitely many solutions.