Số phức z nào dưới đây không phải là nghiệm phương trình z^2

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} \left(\mathrm{a},\mathrm{b}\in \mathrm{ℝ}\right) \\ {\mathrm{z}}^2+\overline{\mathrm{z}}=0 \\ \iff {\left(\mathrm{a}+\mathrm{bi}\right)}^2+\mathrm{a}-\mathrm{bi}=0 \\ \iff {\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}+\left(2\mathrm{ab}-\mathrm{b}\right)\mathrm{i}=0 \\ \iff \left\{\begin{array}{l}{\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}=0\\ \mathrm{b}(2\mathrm{a}-1)=0\end{array}\right. \\ \mathrm{TH}1: \mathrm{b}=0 \Rightarrow {\mathrm{a}}^2+\mathrm{a}=0\iff \left\{\begin{array}{l}\mathrm{a}=0\\ \mathrm{a}=-1\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{z}=0\\ \mathrm{z}=-1\end{array}\right.\right. \\ \mathrm{TH}2: \mathrm{a}=\frac{1}{2}\Rightarrow \frac{3}{4}-{\mathrm{b}}^2=0\iff \mathrm{b}=\pm \frac{\sqrt{3}}{2} \\ \Rightarrow \left\{\begin{array}{l}\mathrm{z}=\frac{1}{2}+\mathrm{i}.\frac{\sqrt{3}}{2}\\ \mathrm{z}=\frac{1}{2}-\mathrm{i}.\frac{\sqrt{3}}{2}\end{array}\right. \end{gathered}$$