Let $S=\left\{z \in \mathbb{C}: z^{2}+\bar{z}=0\right\}$. Then $\sum\limits_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z))$ is equal to ______________.
Answer (integer)
0
Solution
<p>$\because$ ${z^2} + \overline z = 0$</p>
<p>Let $z = x + iy$</p>
<p>$\therefore$ ${x^2} - {y^2} + 2ixy + x - iy = 0$</p>
<p>$({x^2} - {y^2} + x) + i(2xy - y) = 0$</p>
<p>$\therefore$ ${x^2} + {y^2} = 0$ and $(2x - 1)y = 0$</p>
<p>if $x = \, + \,{1 \over 2}$ then $y = \, \pm \,{{\sqrt 3 } \over 2}$</p>
<p>And if $y = 0$ then $x = 0, - 1$</p>
<p>$\therefore$ $$z = 0 + 0i, - 1 + 0i,{1 \over 2} + {{\sqrt 3 } \over 2}i,{1 \over 2} - {{\sqrt 3 } \over 2}i$$</p>
<p>$\therefore$ $\sum {\left( {{R_e}(z) + m(z)} \right) = 0}$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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