Cho số phức z thỏa mãn

A

- Hướng dẫn giải

Đáp án A

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

$\begin{array}{l}\Rightarrow \overline{z}=\frac{m}{m^2+1}-\frac{1}{m^2+1}i\\ \Rightarrow z.\overline{z}=\frac{1}{5}\end{array}$

$\begin{array}{l}\iff \frac{m^2}{{\left(m^2+1\right)}^2}+\frac{1}{{\left(m^2+1\right)}^2}\\ =\frac{1}{m^2+1}=\frac{1}{5}\\ \Rightarrow m^2+1=5\\ \iff m=\pm 2\end{array}$