If - z 2-1- z -2-1- then z lies onA- real axisB- line x - yC- line x-2 yD- imaginary axis

The correct option is D imaginary axisLet z=x+iy |z^2 1|=|z|^2+1 ⇒|x^2 y^2+2xyi 1|=x^2+y^2+1 ⇒(x^2 y^2 1)^2+4x^2y^2=(x^2+y^2+1)^2 ⇒(x^2+y^2+1)^2 (x^2 y^2 1)^2= ...

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Byju's Answer

Standard XII

Mathematics

Equation of Conics in Complex Form

If | z 2-1|=|...

Question

If $| z^{2} - 1 | = | z |^{2} + 1$ , then $z$ lies on

imaginary axis

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real axis

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line

x = y

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line

x = 2 y

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Solution

The correct option is A imaginary axis
Let $z = x + i y$
$| z^{2} - 1 | = | z |^{2} + 1$
$\Rightarrow | x^{2} - y^{2} + 2 x y i - 1 | = x^{2} + y^{2} + 1$
$\Rightarrow (x^{2} - y^{2} - 1)^{2} + 4 x^{2} y^{2} = (x^{2} + y^{2} + 1)^{2}$
$\Rightarrow (x^{2} + y^{2} + 1)^{2} - (x^{2} - y^{2} - 1)^{2} = 4 x^{2} y^{2} \Rightarrow 2 x^{2} (2 y^{2} + 2) = 4 x^{2} y^{2} \Rightarrow x = 0$
$z$ lies on imaginary axis

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If |z2−1|=|z|2+1, then z lies on

The correct option is A imaginary axisLet z=x+iy |z2−1|=|z|2+1 ⇒|x2−y2+2xyi−1|=x2+y2+1 ⇒(x2−y2−1)2+4x2y2=(x2+y2+1)2 ⇒(x2+y2+1)2−(x2−y2−1)2=4x2y2⇒2x2(2y2+2)=4x2y2⇒x=0 z lies on imaginary axis

If $| z^{2} - 1 | = | z |^{2} + 1$ , then $z$ lies on