Có bao nhiêu số phức z thỏa mãn z^2 + z ngang = 0 Có 1 số

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{PT}: {\mathrm{z}}^2+\overline{\mathrm{z}}=0\iff {\left(\mathrm{a}+\mathrm{bi}\right)}^2+\mathrm{a}-\mathrm{bi}=0 \\ \Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}+(2\mathrm{ab}-\mathrm{b})\mathrm{i}=0 \\ \iff \left\{\begin{array}{l}{\mathrm{a}}^2-{\mathrm{b}}^2+\mathrm{a}=0 \left(1\right)\\ 2\mathrm{ab}-\mathrm{b}=0 \left(2\right)\end{array}\right. \\ \mathrm{Từ} \left(2\right)\Rightarrow \mathrm{b}=0 \mathrm{hoặc} \mathrm{a}=\frac{1}{2} \\ \mathrm{TH}1: \mathrm{b}=0 \mathrm{thì} \left(1\right) \mathrm{trở} \mathrm{thành} {\mathrm{a}}^2+\mathrm{a}=0\iff \mathrm{a}=0 \mathrm{hoặc} \mathrm{a}=-1 \\ \Rightarrow \mathrm{Có} 2 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{là} \mathrm{z}=0 \mathrm{và} \mathrm{z}=-1 \\ \mathrm{TH}2: \mathrm{a}=\frac{1}{2} \mathrm{thì} \left(1\right) \mathrm{trở} \mathrm{thành} \frac{3}{4}-{\mathrm{b}}^2=0\iff \mathrm{b}=\pm \frac{\sqrt{3}}{2} \\ \Rightarrow \mathrm{Có} 2 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{là} \mathrm{z}=\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2}\mathrm{và} \mathrm{z}=\frac{1}{2}-\mathrm{i}\frac{\sqrt{3}}{2} \\ \mathrm{KL}: \mathrm{có} 4 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{bài} \mathrm{toán} \end{gathered}$$