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  • 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)

best_answer

\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.

Posted by

Nehul

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