Xét số phức:

C

- Hướng dẫn giải

Ta có

$$\begin{gathered} z=\frac{1-m}{\left(1-m^2\right)+2mi} \\ =\frac{\left(-m+i\right)\left(1-m^2-2mi\right)}{{\left(1-m^2\right)}^2+4m^2} \\ =\frac{-m\left(1-m^2\right)+2m+i}{{\left(1-m^2\right)}^2} \\ \frac{\left(1-m^2+2m^2\right)}{1} \\ =\frac{m\left(1+m^2\right)+i\left(1+m^2\right)}{{\left(1-m^2\right)}^2} \\ =\frac{m}{1+m^2}+\frac{1}{1+m^2}i \\ \Rightarrow \overline{z}=\frac{m}{1+m^2}-\frac{1}{1+m^2}i \end{gathered}$$

Do đó

$$\begin{gathered} z.\overline{z}=\frac{1}{2}\iff \frac{m^2+1}{{\left(m^2+1\right)}^2}=\frac{1}{2} \\ \iff \frac{1}{m^2+1}=\frac{1}{2} \\ \iff m^2+1=2 \\ \iff m=\pm 1 \end{gathered}$$

Đáp án C