4-1.Complex numbers
normal

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

A

$0$

B

$1$

C

$2$

D

$3$

Solution

$z^{2}+\bar{w}=z$     ……$(i)$

${w}^{2}+\overline{{z}}={w} \cdots({i} {i})$

Taking conjugate of $(ii)$

${z}=\overline{{w}}-\overline{{w}}^{2}$

from $(i)$

$z=\left(z-z^{2}\right)+\left(z-z^{2}\right)^{2} \Rightarrow z=0,1+i, 1-i$

from $( i )$

$({z}, {w}) \equiv(0,0),(1+{i}, 1+{i}),(1-{i}, 1-{i})$

Standard 11
Mathematics

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