Biết số phức z, w được biểu diễn bởi các điểm M, N và w = z

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{x}+\mathrm{yi} (\mathrm{x};\mathrm{y}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có}: \mathrm{w}=\frac{\mathrm{z}}{1-\mathrm{i}}=\frac{\mathrm{z}(1+\mathrm{i})}{1-{\mathrm{i}}^2}=\frac{\left(\mathrm{x}+\mathrm{yi}\right)\left(1+\mathrm{i}\right)}{2}=\frac{\mathrm{x}-\mathrm{y}+(\mathrm{x}+\mathrm{y})\mathrm{i}}{2} \\ \Rightarrow \mathrm{M}(\mathrm{x};\mathrm{y}); \mathrm{N}\left(\frac{\mathrm{x}-\mathrm{y}}{2};\frac{\mathrm{x}+\mathrm{y}}{2}\right) \\ \Rightarrow \overrightarrow{\mathrm{MN}}=\left(\frac{-\mathrm{x}-\mathrm{y}}{2};\frac{\mathrm{x}-\mathrm{y}}{2}\right) \\ \Rightarrow \mathrm{MN}=\sqrt{\frac{{\left(\mathrm{x}+\mathrm{y}\right)}^2}{4}+\frac{{\left(\mathrm{x}-\mathrm{y}\right)}^2}{4}}=\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\sqrt{2}}=\frac{\left|\mathrm{z}\right|}{\sqrt{2}} \\ \mathrm{Mặt} \mathrm{khác}: \mathrm{OM}=\left|\mathrm{z}\right| \mathrm{và} \mathrm{ON}=\sqrt{\frac{{\left(\mathrm{x}-\mathrm{y}\right)}^2}{4}+\frac{{\left(\mathrm{x}+\mathrm{y}\right)}^2}{4}}=\frac{\left|\mathrm{z}\right|}{\sqrt{2}} \\ \mathrm{Theo} \mathrm{đề} \mathrm{bài} \mathrm{chu} \mathrm{vi} ∆\mathrm{OMN}=2 \\ \iff \mathrm{OM}+\mathrm{ON}+\mathrm{MN}=2 \\ \iff \left|\mathrm{z}\right|+\frac{\left|\mathrm{z}\right|}{\sqrt{2}}+\frac{\left|\mathrm{z}\right|}{\sqrt{2}}=2 \\ \iff \left|\mathrm{z}\right|\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=2 \\ \iff \left|\mathrm{z}\right|=-2+2\sqrt{2} \end{gathered}$$