Let a b be two non-zero real numbers. Then the number of elements in the set X = { z : Re ( a z 2 + b z ) = a and Re ( b z 2 + a z ) = b } is equal to [2023]

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Solution

Q.
Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X = {z \in ℂ : Re (a z^{2} + b z) = a and Re (b z^{2} + a z) = b}$ is equal to [2023]

1 2

2 0

3 3

4 1

Ans.
(2)

We have, $Re (a z^{2} + b z) = a$

$Re (a (x^{2} - y^{2} + 2 i x y) + b (x + i y)) = a$

$a (x^{2} - y^{2}) + b x = a ...(i)$

Also, $Re (b z^{2} + a z) = b$

$b (x^{2} - y^{2}) + a x = b ...(ii)$

From (i) and (ii), we get $(a - b) (x^{2} - y^{2}) - (a - b) x = a - b$

$\Rightarrow x^{2} - y^{2} - x = 1 ...(iii)$

Also, adding (i) and (ii), we get $(a + b) (x^{2} - y^{2}) + (a + b) x = a + b$

Case 1: If $a + b \neq 0$ , then $x^{2} - y^{2} + x = 1$ $...(iv)$

From (iii) and (iv), we get $x = 0, y^{2} = - 1$ , which is not possible.

Case 2: If $a + b = 0$ , then infinite number of solutions exist.

So, set X has infinite number of elements.

So, number of elements in the set X is 0.

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