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

1. Equations: Let $$z = x + iy$$. The real parts give:
   • $$a(x^2 - y^2) + bx = a$$
   • $$b(x^2 - y^2) + ax = b$$
2. Subtracting: $$(a - b)(x^2 - y^2) + (b - a)x = a - b \implies \mathbf{x^2 - y^2 - x = 1}$$
3. Adding: $$(a + b)(x^2 - y^2) + (a + b)x = a + b \implies \mathbf{x^2 - y^2 + x = 1}$$
4. Solve: Subtracting these two results gives $$2x = 0 \implies \mathbf{x = 0}$$.
5. Check $$y$$: Substitute $$x=0$$ into $$x^2 - y^2 + x = 1 \implies -y^2 = 1 \implies \mathbf{y^2 = -1}$$.
6. Conclusion: No real value of $$y$$ exists.

Number of elements = 0.