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The locus of $z$ such that ${| z - 1 |}^{2} + {| z + 1 |}^{2} = 4$ is

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

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

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

$x^{2} + y^{2} = 5$

answer is B.

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Detailed Solution

${| z - 1 |}^{2} + {| z + 1 |}^{2} = 4 \Rightarrow {| (x - 1) + i y |}^{2} + {| (x + 1) + i y |}^{2} = 4 \Rightarrow {(x - 1)}^{2} + y^{2} + {(x + 1)}^{2} + y^{2} = 4 \Rightarrow 2 x^{2} + 2 y^{2} + 2 = 4 \Rightarrow x^{2} + y^{2} = 1$

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