Let S=z C:z=i(z^2+Re(z)).

Question

Let $S={z\in C:\vec{z}=i({z}^{2}+Re(\vec{z}))}$. Then $\sum _{z\in S}|z{|}^{2}$ is equal to :

Accepted Answer

Let $z=x+iy$

$\begin{gathered} \vec{z}=i({z}^{2}+Re(\vec{z})) \ \Rightarrow x-iy=i({x}^{2}-{y}^{2}+2ixy+x) \ x-iy=i({x}^{2}-{y}^{2}+x)-2xy \ x=-2xy\Rightarrow x(2y+1)=0 \ \Rightarrow x=0,y=\frac{-1}{2}........(i) \ -y={x}^{2}-{y}^{2}+x...........(ii) \end{gathered}$

Case (I) $x=0$

$\Rightarrow -y=-{y}^{2}\Rightarrow {y}^{2}-y=0\Rightarrow y=0,1$

$	herefore $ $z=0,i$

Case (II) $y=\frac{-1}{2}$

$\Rightarrow \frac{1}{2}={x}^{2}-\frac{1}{4}+x\Rightarrow {x}^{2}+x-\frac{3}{4}=0$

$\Rightarrow$ $4x+4x-3=0\Rightarrow (2x-1)(2x+3)=0$

$\Rightarrow$ $x=\frac{1}{2},\frac{-3}{2}$

$\begin{gathered} z=\frac{1}{2}-\frac{1}{2}i,\frac{-3}{2}-\frac{1}{2}i \ \sum |z{|}^{2}=0+1+\frac{1}{2}+\frac{5}{2}=4 \end{gathered}$

Solutions · 1