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

Let z = x + iy
$\begin{array}{l}\mathrm{a}\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)+\mathrm{bx}=\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to \left(1\right)\\ \mathrm{b}\left({\mathrm{x}}^2-y^2\right)+\mathrm{ax}=\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to \left(2\right)\end{array}$
(1) - (2)
$\begin{array}{l}\Rightarrow (\mathrm{a}-\mathrm{b})\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)+(\mathrm{b}-\mathrm{a})\mathrm{x}=\mathrm{a}-\mathrm{b}\\ \Rightarrow {\mathrm{x}}^2-{\mathrm{y}}^2-\mathrm{x}=1\to \left(3\right)\end{array}$
(1) + (2)
$\begin{array}{l}\Rightarrow (\mathrm{a}+\mathrm{b})\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)+\mathrm{x}(\mathrm{a}+\mathrm{b})=(\mathrm{a}+\mathrm{b})\\ \Rightarrow {\mathrm{x}}^2-{\mathrm{y}}^2+\mathrm{x}=1\left(\text{ or }\right)\mathrm{a}+\mathrm{b}=0\end{array}$
From (3) & (4) x = 0
$\Rightarrow$ y² = -1
$\therefore$ not possible [$\because$ y$\in$R]
No of complex number = 0
If a + b = 0 from (3) we will have infinite solutions.