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4-1.Complex numbers
hard
ધારોકે $S=\left\{z \in C : z^{2}+\bar{z}=0\right\}$ છે. તો $\sum \limits_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z))$ is equal to$......$
A
$1$
B
$2$
C
$3$
D
$0$
(JEE MAIN-2022)
Solution
$S=\left\{z \in C: z^{2}+\bar{z}=0\right\}$
Let $z = x + iy$
$z ^{2}= x ^{2}- y ^{2}+2 ixy$
$\bar{z}=x-i y$
$z^{2}+\bar{z}=x^{2}-y^{2}+x+i(2 x y-y)=0$
$x^{2}+x-y^{2}=0 \text { \& } 2 x y-y=0$
$y=0$ or $x=\frac{1}{2}$
If $y =0 ; x =0,-1$
If $x=\frac{1}{2} ; y=\frac{\sqrt{3}}{2}, \frac{-\sqrt{3}}{2}$
$\sum_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z)=\left(0-1+\frac{1}{2}+\frac{1}{2}\right)$
$+0+0+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2})$
Standard 11
Mathematics
