Let z - - C satisfy z 2- z and -2- z - then number of ordered pairs of complex number z - - is equal to

Z^2+ω̅=z......(i) ω^2+z̅=ω.....(ii) Taking conjugate of (ii) nbsp; z=ω̅ ω̅^2 From (i) z=(z z^2)+(z z^2)^2 ⇒ z=0,(1+i),(1 i) From (i) (z,ω)=(0,0),(1+i,1+i),(1 ...

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Byju's Answer

Standard XII

Mathematics

Properties of Conjugate of a Complex Number

Let z , ω∈ C ...

Question

Let $z, ω \in C$ satisfy $z^{2} + ¯ ω = z$ and $ω^{2} + ¯ z = ω$ then number of ordered pairs of complex number $(z, ω)$ is equal to

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Solution

$z^{2} + ¯ ω = z . . . . . . (i)$
$ω^{2} + ¯ z = ω . . . . . (i i)$
Taking conjugate of $(i i)$
$z = ¯ ω - {¯ ω}^{2}$
From $(i)$
$z = (z - z^{2}) + (z - z^{2})^{2}$
$\Rightarrow z = 0, (1 + i), (1 - i)$
From $(i)$
$(z, ω) = (0, 0), (1 + i, 1 + i), (1 - i, 1 - i)$

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Let z,ω∈C satisfy z2+¯ω=z and ω2+¯z=ω then number of ordered pairs of complex number (z,ω) is equal to

z2+¯ω=z......(i) ω2+¯z=ω.....(ii) Taking conjugate of (ii) z=¯ω−¯ω2 From (i) z=(z−z2)+(z−z2)2 ⇒z=0,(1+i),(1−i) From (i) (z,ω)=(0,0),(1+i,1+i),(1−i,1−i)

Let $z, ω \in C$ satisfy $z^{2} + ¯ ω = z$ and $ω^{2} + ¯ z = ω$ then number of ordered pairs of complex number $(z, ω)$ is equal to

$z^{2} + ¯ ω = z . . . . . . (i)$
$ω^{2} + ¯ z = ω . . . . . (i i)$
Taking conjugate of $(i i)$
$z = ¯ ω - {¯ ω}^{2}$
From $(i)$
$z = (z - z^{2}) + (z - z^{2})^{2}$
$\Rightarrow z = 0, (1 + i), (1 - i)$
From $(i)$
$(z, ω) = (0, 0), (1 + i, 1 + i), (1 - i, 1 - i)$