Cho hai số phức \(z_1, z_2\) thỏa mãn \({z_1},{z_2} \ne 0; {z_1} + {z_2} \ne 0\) và \(\frac{1}{{{z_1} + {z_2}}} = \frac{1}{{{z_1}}}

* Hướng dẫn giải

Đặt \(x = \frac{{{z_1}}}{{{z_2}}} \Rightarrow {z_1} = x.{z_2}\) và \(\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = \left| x \right|\)

Từ giả thiết \(\frac{1}{{{z_1} + {z_2}}} = \frac{1}{{{z_1}}} + \frac{2}{{{z_2}}} \Leftrightarrow \frac{1}{{x.{z_2} + {z_2}}} = \frac{1}{{x.{z_2}}} + \frac{2}{{{z_2}}}\)

\(\begin{array}{l}
 \Leftrightarrow \frac{1}{{{z_2}\left( {x + 1} \right)}} = \frac{1}{{{z_2}}}\left( {\frac{1}{x} + 2} \right)\\
 \Leftrightarrow \frac{1}{{x + 1}} = \frac{1}{x} + 2\\
 \Leftrightarrow 2{x^2} + 2x + 1 = 0 \Leftrightarrow x =  - \frac{1}{2} \pm \frac{1}{2}i \Rightarrow \left| x \right| = \frac{{\sqrt 2 }}{2}
\end{array}\)