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

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Question

$\operatorname{Re}\left(a z^2+b z\right)=a$

$a z^2+b z+a \bar{z}^2+b \bar{z}=2 a$

$a\left(z^2+\bar{z}^2\right)+ b ( z +\overline{ z })=2 a$

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Vedclass · Accepted Answer

$0$

Vedclass · Answer

$\operatorname{Re}\left(a z^2+b zight)=a$

$a z^2+b z+a \bar{z}^2+b \bar{z}=2 a$

$a\left(z^2+\bar{z}^2ight)+ b ( z +\overline{ z })=2 a$

$\operatorname{Re}\left( bz z^2+ az ight)= b$

$b z^2+a z+b \bar{z}^2+ az =2 b$

$b \left( z ^2+\overline{ z }^2ight)+ a ( z +\overline{ z })=2 b$

$(1) 	imes b-(2) 	imes(a)$

$\Rightarrow\left(b^2-a^2ight)(z+\bar{z})=0$

$\Rightarrow \quad(z+\bar{z})=0 \quad\left(a^2 
eq b^2ight)$

$(1) 	imes a-(2) 	imes(b)$

$\Rightarrow \quad\left(a^2-b^2ight)(z+\bar{z})=2\left(a^2-b^2ight) \quad\left(a^2 
eq b^2ight)$

$z^2+\bar{z}^2=2$

$\Rightarrow(z+\bar{z})^2-2 z \bar{z}=2$

$z \bar{z}=-1$

$\Rightarrow 1+1^2=-1$

$\Rightarrow 	ext { No solution }$

$	ext { But when } a=-b$

$\operatorname{Re}\left(a z^2-a zight)=a$

$\Rightarrow \quad \operatorname{Re}\left(a\left(x^2-y^2+i 2 x yight)-a(x+i y)ight)=a$

$\Rightarrow a\left(x^2-y^2ight)-a x=a$

$\Rightarrow x^2-y^2-x=1$

$\Rightarrow x^2-x-1=y^2$

For any real values of $y$ there two values of $x$, hence infinite complex numbers are possible.

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

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