If - z 2-1- z -2-1- then z lies onA- a circleB- a parabolaC- an ellipseD- a straight line

Let z = x + iy. Then |z^2 1|= |z|^2 + 1 ⇒ |(x^2 y^2 1) + 2ixy| = x^2 + y^2 + 1 ⇒ (x^2 y^2 1)^2 + 4x^2 y^2 = (x^2 + y^2 + 1)^2 ⇒ (x^2 y^2 1)^2 nbsp; ...

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Standard XII

Mathematics

Equality of 2 Complex Numbers

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

Question

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

a circle

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a parabola

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an ellipse

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a straight line

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Solution

The correct option is D a straight line
Let $z = x + i y .$ Then
$| z^{2} - 1 | = | z |^{2} + 1 \Rightarrow | (x^{2} - y^{2} - 1) + 2 i x y | = x^{2} + y^{2} + 1 \Rightarrow (x^{2} - y^{2} - 1)^{2} + 4 x^{2} y^{2} = (x^{2} + y^{2} + 1)^{2} \Rightarrow (x^{2} - y^{2} - 1)^{2} - (x^{2} + y^{2} + 1)^{2} + 4 x^{2} y^{2} = 0 \Rightarrow (2 x^{2}) (- 2 y^{2} - 2) + 4 x^{2} y^{2} = 0 \Rightarrow - 4 x^{2} = 0 ∴ x = 0$

Hence, $z$ lies on imaginary axis.

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Equality of 2 Complex Numbers

Standard XII Mathematics

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If |z2−1|=|z|2+1, then z lies on

The correct option is D a straight lineLet z=x+iy. Then |z2−1|=|z|2+1⇒|(x2−y2−1)+2ixy|=x2+y2+1⇒(x2−y2−1)2+4x2y2=(x2+y2+1)2⇒(x2−y2−1)2−(x2+y2+1)2+4x2y2=0⇒(2x2)(−2y2−2)+4x2y2=0⇒−4x2=0∴x=0 Hence, z lies on imaginary axis.

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