Let and be the sum and the product of all the non-zero solutions of the equation ( z ¯ ) 2 + | z | = 0 , z C . Then is equal to [2024]

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Solution

Q.
Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation ${(\bar{z})}^{2} + | z | = 0, z \in C .$ Then $4 (α^{2} + β^{2})$ is equal to [2024]

1 2

2 6

3 4

4 8

Ans.
(3)

We have, ${(\bar{z})}^{2} + | z | = 0$ ...(i)

Let $z = x + i y \Rightarrow \bar{z} = x - i y$

From (i), ${(x - i y)}^{2} + | x + i y | = 0$

$\Rightarrow x^{2} - y^{2} - 2 i x y + \sqrt{x^{2} + y^{2}} = 0$

   $\Rightarrow (x^{2} - y^{2} + \sqrt{x^{2} + y^{2}}) - i (2 x y) = 0$

   $\Rightarrow x^{2} - y^{2} + \sqrt{x^{2} + y^{2}} = 0$ and $2 x y = 0$

Case I: $x = 0$ and $y \neq 0$

   $\Rightarrow - y^{2} + | y | = 0 \Rightarrow | y | = y^{2} \Rightarrow y = \pm 1$

Case II: $x \neq 0$ and $y = 0$

   $\Rightarrow x^{2} + | x | = 0$

   $\Rightarrow x = 0$ only $(∵ x \in R)$

   $\Rightarrow x = 0, y = 0$ (Rejected) $\Rightarrow z = i, - i$ are solutions.

Hence, $α = i + (- i) = 0$ and $β = i (- i) = - i^{2} = 1$

Hence, $4 (α^{2} + β^{2}) = 4 (0 + 1) = 4$

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