Đặt \(z = x + yi\,\left( {x,y \in\mathbb{R} } \right)\) suy ra \(\bar z = x - yi.\)
Khi đó ta được: \(\left\{ \begin{array}{l} \left( {x + yi} \right)\left( {x - yi} \right) = 1\\ \left| {{{\left( {x + yi} \right)}^2} + 2\left( {x - yi} \right) - 1} \right| = \sqrt {\frac{8}{{27}}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {y^2} = 1 - {x^2}\\ 4{x^3} - {x^2} - 2x + \frac{{52}}{{27}} = 0 \end{array} \right.\)
\(\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{2}{3}\\ {y^2} = \frac{5}{9} \end{array} \right.\\ \left\{ \begin{array}{l} x = - \frac{{13}}{{12}}\\ {y^2} = - \frac{{25}}{{144}} \end{array} \right.\left( L \right) \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{2}{3}\\ y = \frac{{\sqrt 5 }}{3} \end{array} \right.\\ \left\{ \begin{array}{l} x = \frac{2}{3}\\ y = - \frac{{\sqrt 5 }}{3} \end{array} \right. \end{array} \right.\)
Suy ra \({z_1} = \frac{2}{3} + \frac{{\sqrt 5 }}{3}i,\,\,{z_2} = \frac{2}{3} - \frac{{\sqrt 5 }}{3}i\)
Vậy \(3{z_1} + 6{z_2} = 6 - \sqrt 5 i.\)