Let S=zC: 4z^2+ z=0.

Question

Let $S=\{z\in\mathbb{C}: 4z^2+\bar z=0\}$. Then $\displaystyle\sum_{z\in S}|z|^2$ is equal to:

Accepted Answer

Let $z=x+iy$. Then $4(x+iy)^2+(x-iy)=0$ gives
$4x^2-4y^2+8xyi+x-iy=0.$

Separating real and imaginary parts:
$$4x^2-4y^2+x=0,\qquad y(8x-1)=0.$$

Case $y=0$: $4x^2+x=0 \Rightarrow x=0,-\tfrac14$. Gives $z_1=0$ ($|z_1|^2=0$) and $z_2=-\tfrac14$ ($|z_2|^2=\tfrac1{16}$).

Case $x=\tfrac18$: $4\left(\tfrac1{64}\right)-4y^2+\tfrac18=0 \Rightarrow 4y^2=\tfrac3{16}\Rightarrow y=\pm\dfrac{\sqrt3}{8}$. Both give $|z|^2=\tfrac1{64}+\tfrac3{64}=\tfrac1{16}$.

$$\sum_{z\in S}|z|^2 = 0+\frac1{16}+\frac1{16}+\frac1{16}=\frac{3}{16}.$$

Solutions · 1