Let S=z C: z^2+ z=0.

Question

Let $S=\{z\in\mathbb C: z^2+\bar z=0\}$. Then $\displaystyle\sum_{z\in S}\big(\operatorname{Re}(z)+\operatorname{Im}(z)\big)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 0

Accepted Answer

With $z=x+iy$: $x^2-y^2+2ixy+x-iy=0$, giving
$$x^2-y^2+x=0 \quad\text{and}\quad (2x-1)y=0.$$

$y=0$: $x^2+x=0 \implies x=0,-1$.

$x=\dfrac12$: $\dfrac14-y^2+\dfrac12=0 \implies y^2=\dfrac34 \implies y=\pm\dfrac{\sqrt3}2$.

$$S=\left\{0,\ -1,\ \frac12+\frac{\sqrt3}2i,\ \frac12-\frac{\sqrt3}2i\right\}.$$

$$\sum(\operatorname{Re}+\operatorname{Im}) = 0+(-1)+\left(\frac12+\frac{\sqrt3}2\right)+\left(\frac12-\frac{\sqrt3}2\right) = 0-1+1=0.$$

Solutions · 1