If $z = x + i y$ is a complex number such that $| z | = R e (i z) + 1$ , then the locus of $z$ is

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

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x^{2} = 2 y - 1

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y^{2} = 2 x - 1

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y^{2} = 1 - 2 x

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x^{2} = 1 - 2 y

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Solution

The correct option is E $x^{2} = 1 - 2 y$
It is given that, $z = x + i y$
Now, $| z | = R e (i z) + 1$
$\Rightarrow \sqrt{x^{2} + y^{2}} = R e (x i + i^{2} y) + 1$
$\Rightarrow \sqrt{x^{2} + y^{2}} = R e (- y + x i) + 1$
$\Rightarrow \sqrt{x^{2} + y^{2}} = - y + 1$
On squaring both sides, we get
$x^{2} + y^{2} = {(1 - y)}^{2}$
$\Rightarrow x^{2} + y^{2} = 1 + y^{2} - 2 y$
$\Rightarrow x^{2} + 2 y - 1 = 0$

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