If z is a complex number such that z 2- z 2 - thenA- None of theseB- z is purely realC- z is purely imaginaryD- Either z is purely real or purely imaginary

Let z = x + iy, then its conjugate z̅ = x iy Given that z^2 = (z̅)^2 ⇒ x^2 y^2 + 2ixy = x^2 y^2 2ixy ⇒ 4ixy = 0 If x ≠ then y = 0and if y ≠ 0 then x = ...

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Standard XII

Mathematics

Purely Real

If z is a com...

Question

If z is a complex number such that $z^{2} = (¯ z)^{2}),$ then

z is purely real

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z is purely imaginary

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Either z is purely real or purely imaginary

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None of these

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Solution

The correct option is C
Either z is purely real or purely imaginary

Let z = x + iy, then its conjugate \bar{z} = x - iy
Given that $z^{2} = (¯ z)^{2}$
$\Rightarrow x^{2} - y^{2} + 2 i x y = x^{2} - y^{2} - 2 i x y \Rightarrow 4 i x y = 0$
If x $\neq$ then y = 0and if y $\neq$ 0 then x = 0

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If z is a complex number such that z2=(¯z)2), then

The correct option is C Either z is purely real or purely imaginary Let z = x + iy, then its conjugate \bar{z} = x - iy Given that z2=(¯z)2 ⇒ x2−y2+2ixy=x2−y2−2ixy ⇒ 4ixy=0 If x ≠ then y = 0and if y ≠ 0 then x = 0

If z is a complex number such that $z^{2} = (¯ z)^{2}),$ then

The correct option is C
Either z is purely real or purely imaginary

Let z = x + iy, then its conjugate \bar{z} = x - iy
Given that $z^{2} = (¯ z)^{2}$
$\Rightarrow x^{2} - y^{2} + 2 i x y = x^{2} - y^{2} - 2 i x y \Rightarrow 4 i x y = 0$
If x $\neq$ then y = 0and if y $\neq$ 0 then x = 0