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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