Đặt \(z=x+yi\Rightarrow\left\{{}\begin{matrix}\overline{z}=x-yi\\z^2=x^2-y^2+2xy.i\end{matrix}\right.\)
\(\overline{z}=\sqrt{3}z^2\)
\(\Leftrightarrow x-yi=\sqrt{3}\left(x^2-y^2\right)+2\sqrt{3}xy.i\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3}\left(x^2-y^2\right)\\-y=2\sqrt{3}xy\end{matrix}\right.\)
TH1: \(y=0\Rightarrow x=0\Rightarrow z=0\)
TH2: \(2\sqrt{3}x=-1\Rightarrow x=-\frac{1}{2\sqrt{3}}\)
Tổng phần thực là \(-\frac{1}{2\sqrt{3}}\)