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Equation of Conics in Complex Form

If |z2|=2Re...

Question

If $| z^{2} | = 2 R e (z)$ , then the locus of $z$ is:

the circle

x^{2} + y^{2} + 2 x = 0

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the circle

x^{2} + y^{2} - 2 x + 2 y = 0

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the circle

x^{2} + y^{2} - 2 x = 0

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the circle

x^{2} + y^{2} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is C the circle $x^{2} + y^{2} - 2 x = 0$
Taking $z = x + i y$
$| z^{2} | = | z |^{2} = 2 R e (z)$
$\Rightarrow x^{2} + y^{2} = 2 R e (x + i y) = 2 x$
Hence, the locus is $x^{2} + y^{2} - 2 x = 0$

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