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Equation of Conics in Complex Form

If z=x+iy and...

Question

If $z = x + i y$ and $| z - 1 |^{2} + | z + 1 |^{2} = 4$ , then the locus of $z$ is

x^{2} - y^{2} = 1

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x^{2} + y^{2} = 1

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x^{2} - y^{2} = 4

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x^{2} + y^{2} = 2

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Solution

The correct option is B $x^{2} + y^{2} = 1$
Let $z = x + i y$
$\Rightarrow | x + i y - 1 |^{2} + | x + i y + 1 |^{2} = 4 \Rightarrow | x - 1 + i y |^{2} + | x + 1 + i y |^{2} = 4 \Rightarrow (x - 1)^{2} + y^{2} + (x + 1)^{2} + y^{2} = 4 \Rightarrow 2 x^{2} + 2 y^{2} = 2$

Hence, the required locus is $x^{2} + y^{2} = 1$

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