Let be two-zero real numbers. Then the number of elements in the set <img alt="X=\left\{z \in \mathbb{C}:

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

Option: 1
0

Option: 2
2

Option: 3
1

Option: 4
3

Answers (1)

$\because z+\bar{z}=2 \operatorname{Re}(\mathrm{z}) \quad \text { If } z=x+i y$
$\Rightarrow z+\bar{z}=2 \mathrm{x}$
$\left(a z^2+b z\right)+\left(a \bar{z}^2+b \bar{z}\right)=2 a$
$\left(b z^2+a z\right)+\left(b \bar{z}^2+a \bar{z}\right)=2 b$
add (1) and (2)
$(\mathrm{a}+\mathrm{b}) \mathrm{z}^2+(\mathrm{a}+\mathrm{b}) \mathrm{z}+(\mathrm{a}+\mathrm{b}) \overline{\mathrm{z}}^2+(\mathrm{a}+\mathrm{b}) \overline{\mathrm{z}}=2(\mathrm{a}+\mathrm{b})$
$(\mathrm{a}+\mathrm{b})\left[\mathrm{z}^2+\mathrm{z}+(\overline{\mathrm{z}})^2+\overline{\mathrm{z}}\right]=2(\mathrm{a}+\mathrm{b})$
sub. (1) and (2)
$(a-b)\left[z^2-z+\bar{z}^2-\bar{z}\right]=2(a-b)$
$z^2+\bar{z}^2-z-\bar{z}=2$
$2 \mathrm{x}+2\left(\mathrm{x}^2-\mathrm{y}^2\right)=2 \quad \Rightarrow \mathrm{x}^2-\mathrm{y}^2+\mathrm{x}=1$
$\mathrm{2\left(x^2-y^2\right)-2 x=2 \quad \Rightarrow x^2-y^2-x=1 }$
Case I: If $\mathrm{a}+\mathrm{b} \neq 0$
From (3) & (4)
$\mathrm{\begin{array}{ll} 2 x+2\left(x^2-y^2\right)=2 & \Rightarrow x^2-y^2+x=1 \\ 2\left(x^2-y^2\right)-2 x=2 & \Rightarrow x^2-y^2-x=1 \end{array} }$
$(5)-(6)$
$\begin{array}{ll} 2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0 & \\ \text { from (5) } \quad y^2=-1 & \Rightarrow \text { not possbible } \\ & \therefore \text { Ans }=0 \end{array}$
Case II: If $\mathrm{a}+\mathrm{b}=0$ then infinite number of solution.
So, the set $\mathrm{X}$ have infinite number of elements.