Let a neq b be two non-zero real numbers. The | vedclass

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(C) Given $\operatorname{Re}(a z^2 + bz) = a$ and $\operatorname{Re}(b z^2 + az) = b$.Let $z = x + iy$. Then $z^2 = x^2 - y^2 + 2ixy$.The condition $\

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$0$

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(C) Given $\operatorname{Re}(a z^2 + bz) = a$ and $\operatorname{Re}(b z^2 + az) = b$.Let $z = x + iy$. Then $z^2 = x^2 - y^2 + 2ixy$.The condition $\operatorname{Re}(a z^2 + bz) = a$ implies $a(x^2 - y^2) + bx = a$ $(1)$.The condition $\operatorname{Re}(b z^2 + az) = b$ implies $b(x^2 - y^2) + ax = b$ $(2)$.Multiply $(1)$ by $b$ and $(2)$ by $a$:$ab(x^2 - y^2) + b^2x = ab$ $(3)$$ab(x^2 - y^2) + a^2x = ab$ $(4)$Subtract $(4)$ from $(3)$:$(b^2 - a^2)x = 0$.Since $a 
eq b$, if $a 
eq -b$, then $b^2 - a^2 
eq 0$, which implies $x = 0$.Substituting $x = 0$ into $(1)$ gives $a(-y^2) = a$. Since $a 
eq 0$, we get $y^2 = -1$, which has no real solution for $y$.Thus, for $a 
eq \pm b$, there are no solutions, so the number of elements is $0$.

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

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