All complex numbers ‘z’ which satisfy the relation $| z - | z + 1 | | = | z + | z - 1 ∣$ on the complex plane lie on the

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Line y = 0

Line x = 0

Line x = 0 or on a line segment joining $(- 1, 0) to (1, 0)$

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

answer is D.

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Detailed Solution

Given $| z - | z + {1|}^{2} = | z + | z - 1 | |^{2}$
$\begin{array}{l} ∴ (z - | z + 1 |) (\bar{z} - | z + 1 |) = (z + | z - 1 |) (\bar{z} + | z - 1 |) \\ z \bar{z} - z | z + 1 | - \bar{z} | z + 1 | + | z + 1 |^{2} \\ = z \bar{z} + z | z - 1 | + \bar{z} | z - 1 | + | z - 1 |^{2} \\ | z + 1 |^{2} - | z - 1 |^{2} = (z + \bar{z}) [| z - 1 | + | z + 1 |] \\ (z + 1) (\bar{z} + 1) - (z - 1) (\bar{z} - 1) = (z + \bar{z}) [| z - 1 | + | z + 1 |] \\ (z \bar{z} + z + \bar{z} + 1) - (z \bar{z} - z - \bar{z} + 1) = (z + \bar{z}) [| z - 1 | + | z + 1 |] \\ 2 (z + \bar{z}) = (z + \bar{z}) [| z + 1 | + | z - 1 |] \end{array}$

$(z + \bar{z}) [| z + 1 | + | z - 1 | - 2] = 0 \Rightarrow$ either
$z + \bar{z} = 0 \Rightarrow z$ is purely imaginary
$\Rightarrow z lies on y axis \Rightarrow x = 0 or | z + 1 | + | z - 1 | = 2 \Rightarrow z lie on the line segment joining (- 1, 0) and (1, 0) .$