Có bao nhiêu số phức z là nghiệm phương trình |z| + z^2 = 0

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có} \left|\mathrm{z}\right|+{\mathrm{z}}^2=0 \\ \iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}+{(\mathrm{a}+\mathrm{bi})}^2=0\iff \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}+{\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{abi}=0 \\ \iff \left\{\begin{array}{l}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}+{\mathrm{a}}^2-{\mathrm{b}}^2=0 (*)\\ 2\mathrm{ab}=0\end{array}\right. \\ \mathrm{TH}1: \mathrm{a}=0 \mathrm{thì} (*)\iff \sqrt{{\mathrm{b}}^2}-{\mathrm{b}}^2=0 \\ \iff \left|\mathrm{b}\right|={\mathrm{b}}^2 \\ \iff {\mathrm{b}}^2=\mathrm{b} \mathrm{hoặc} {\mathrm{b}}^2=-\mathrm{b} \\ \iff \left\{\begin{array}{l}\mathrm{b}=0\\ \mathrm{b}=\pm 1\end{array}\right. \\ \Rightarrow \mathrm{Có} 3 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \\ \mathrm{TH}2: \mathrm{b}=0 \mathrm{thì} (*)\iff \sqrt{{\mathrm{a}}^2}+{\mathrm{a}}^2=0\iff \mathrm{a}=0 \\ \mathrm{KL}: \mathrm{có} 3 \mathrm{số} \mathrm{phức} \mathrm{thỏa} \mathrm{mãn} \mathrm{yêu} \mathrm{cầu} \mathrm{bài} \mathrm{toán} \mathrm{là} \mathrm{z}=0; \mathrm{z}=\pm \mathrm{i} \end{gathered}$$