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

Given,

$X=\left\{z\in C:\mathrm{Re}\left(a{z}^2+bz\right)=a \mathrm{and} \mathrm{Re}\left(b{z}^2+az\right)=b\right\}$

Now, let $z=x+iy$

So, $\mathrm{Re}\left(a{z}^2+bz\right)=a$

$\Rightarrow \mathrm{Re}\left(a{\left(x+iy\right)}^2+b\left(x+iy\right)\right)=a$

$\Rightarrow \mathrm{Re}\left(a\left(x^2-y^2+2ixy\right)+b\left(x+iy\right)\right)=a$

$\Rightarrow a\left(x^2-y^2\right)+bx=a ... \left(\mathrm{i}\right)$

And, $\mathrm{Re}\left(b{z}^2+az\right)=b$

$\Rightarrow \mathrm{Re}\left(b{\left(x+iy\right)}^2+a\left(x+iy\right)\right)=b$

$\Rightarrow \mathrm{Re}\left(b\left(x^2-y^2+2ixy\right)+a\left(x+iy\right)\right)=b$

$\Rightarrow b\left(x^2-y^2\right)+ax=b ... \left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right), \left(\mathrm{i}\right)-\left(\mathrm{ii}\right)$ we get,

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

$\Rightarrow x^2-y^2-x=1 ... \left(\mathrm{iii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right), \left(\mathrm{i}\right)+\left(\mathrm{ii}\right)$

$\left(\left(x^2-y^2\right)+x-1\right)\left(a+b\right)=0$ (Note: here $a+b\ne 0$ is considered but it is not clear from the question)

$\Rightarrow x^2-y^2+x-1=0 ... \left(\mathrm{iv}\right)$

Now from $\left(\mathrm{iii}\right)$ and $\left(\mathrm{iv}\right)$ we get,

$x=0, y^2=-1$ (No solution)