Biết số phức z thỏa mãn (z+1)/(z-1) là số ảo. Khi đó ta phải

* Hướng dẫn giải

Đáp án C

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}= \mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \Rightarrow \frac{\mathrm{z}+1}{\mathrm{z}-1}=1+\frac{2}{\mathrm{z}-1}=1+\frac{2}{\mathrm{a}-1+\mathrm{bi}} \\ =1+\frac{2(\mathrm{a}-1-\mathrm{bi})}{{(\mathrm{a}-1)}^2+{\mathrm{b}}^2}=1+\frac{2(\mathrm{a}-1)}{{(\mathrm{a}-1)}^2+{\mathrm{b}}^2}-\frac{2\mathrm{b}}{{(\mathrm{a}-1)}^2+{\mathrm{b}}^2}\mathrm{i} \\ \mathrm{Để} \frac{\mathrm{z}+1}{\mathrm{z}-1} \mathrm{là} \mathrm{số} \mathrm{thuần} \mathrm{ảo} \mathrm{thì} \left\{\begin{array}{l}1+\frac{2(\mathrm{a}-1)}{{(\mathrm{a}-1)}^2+{\mathrm{b}}^2}=0\\ \frac{-2\mathrm{b}}{{(\mathrm{a}-1)}^2+{\mathrm{b}}^2}\ne 0\end{array}\right. \\ \iff \left\{\begin{array}{l}\mathrm{b}\ne 0\\ {(\mathrm{a}-1)}^2+{\mathrm{b}}^2+2(\mathrm{a}-1)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}\mathrm{b}\ne 0\\ {\mathrm{a}}^2+{\mathrm{b}}^2=1\end{array}\right. \\ \Rightarrow \left|\mathrm{z}\right|=1 \end{gathered}$$