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

Question

Let $S={z\in C:4{z}^{2}+\vec{z}=0}$. Then $\sum _{z\in S}|z{|}^{2}$ is equal to:

Accepted Answer

$4{z}^{2}=\vec{z}=0$

Let $z=x+iy$

$\begin{gathered} 4(x+iy{)}^{2}+x-iy=0 \ 4{x}^{2}-4{y}^{2}+8xyi+x-iy=0 \ 4{x}^{2}-4{y}^{2}+x=0y(8x-1)=0 \ \Rightarrow y=0x=\frac{1}{8} $

If $y=0,4{x}^{2}+x=0$

$ x=0,\frac{-1}{4} \ 	herefore {z}_{1}=0+0.i {|{z}_{1}|}^{2}=0 \ {z}_{2}=-\frac{1}{4}+0i {|{z}_{2}|}^{2}=\frac{1}{16} \end{gathered}$

If $x=\frac{1}{8}$

$\begin{gathered} 4	imes \frac{1}{64}-4{y}^{2}+\frac{1}{8}=0\Rightarrow 4{y}^{2}=\frac{3}{16}\Rightarrow y=\pm \frac{\sqrt{3}}{8} \ 	herefore {z}_{3}=\frac{1}{8}+\frac{\sqrt{3}}{8}i\, {|{z}_{3}|}^{2}=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \ {z}_{4}=\frac{1}{8}-\frac{\sqrt{3}}{8}i\, {|{z}_{4}|}^{2}=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \ 	herefore \sum _{i=1}^{n}{|{z}_{i}|}^{2}=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16} \end{gathered}$

Solutions · 1