Solutions of equation in z , | z |^2 -(z + bar{z}) + i(z - bar{z}) + 2 = 0 are (i = √{-1})

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4-1.Complex numbers

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The solutions of equation in $z$, $| z |^2 -(z + \bar{z}) + i(z - \bar{z})$ + $2$ = $0$ are $(i = \sqrt{-1})$

A

$2 + i$, $1 -i$

B

$1 + i$, $1 -i$

C

$1 + 2i$, $-1 -i$

D

$1 + i$, $1 + i$

Solution

$z \bar{z}-z(1-i)-\bar{z}(1+i)+(1+i)(1-i)=0$

$(z-(1+i))(\bar{z}-(1-i))=0$

$\Rightarrow z=1+i, \bar{z}=1-i$

Standard 11

Mathematics

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