Let z-x-iy be a non-zero complex number such that z2-i - z -2- where i-1- Then z lies on the

Z=x+iy z^2=i | z |^2 ⇒ x^2 y^2+2i x y = i(x^2+y^2) Equating real terms, x^2 y^2 =0⇒ x^2=y^2 Equating imaginary terms, 2xy=x^2+y^2 ⇒ (x y)^2=0⇒ x=y ...

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Byju's Answer

Standard XI

Mathematics

Properties of Argument

Let z=x+iy be...

Question

Let $z = x + i y$ be a non-zero complex number such that $z^{2} = i {| z |}^{2}$ , where $i = \sqrt{- 1}$ . Then $z$ lies on the

line,

y = x

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real axis

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imaginary axis

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line,

y = - x

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Solution

The correct option is A line, $y = x$
$z = x + i y$
$z^{2} = i {| z |}^{2}$
$\Rightarrow x^{2} - y^{2} + 2 i x y = i (x^{2} + y^{2})$
Equating real terms,
$x^{2} - y^{2} = 0 \Rightarrow x^{2} = y^{2}$
Equating imaginary terms,
$2 x y = x^{2} + y^{2}$
$\Rightarrow (x - y)^{2} = 0 \Rightarrow x = y$

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Let z=x+iy be a non-zero complex number such that z2=i|z|2, where i=√−1. Then z lies on the

The correct option is A line, y=xz=x+iy z2=i|z|2 ⇒x2−y2+2ixy=i(x2+y2) Equating real terms, x2−y2=0⇒x2=y2 Equating imaginary terms, 2xy=x2+y2 ⇒(x−y)2=0⇒x=y

Let $z = x + i y$ be a non-zero complex number such that $z^{2} = i {| z |}^{2}$ , where $i = \sqrt{- 1}$ . Then $z$ lies on the

The correct option is A line, $y = x$
$z = x + i y$
$z^{2} = i {| z |}^{2}$
$\Rightarrow x^{2} - y^{2} + 2 i x y = i (x^{2} + y^{2})$
Equating real terms,
$x^{2} - y^{2} = 0 \Rightarrow x^{2} = y^{2}$
Equating imaginary terms,
$2 x y = x^{2} + y^{2}$
$\Rightarrow (x - y)^{2} = 0 \Rightarrow x = y$