D

- Hướng dẫn giải

Đặt $z_1=x_1+i{y}_1,z_2=x_2+i{y}_2$.

Từ giả thiết ta suy ra

$$\begin{gathered} \left\{\begin{array}{l}{x_1}^2+{y_1}^2={x_2}^2+{y_2}^2=1\\ {\left(x_1+x_2\right)}^2+{\left(y_1+y_2\right)}^2=3\end{array}\right. \\ \Rightarrow 2\left(x_1{y}_1+x_2{y}_2\right)=1 \end{gathered}$$

Suy ra:

$$\begin{gathered} {\left|z_1-z_2\right|}^2 \\ ={\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2 \\ ={\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2 \\ -4\left(x_1{y}_1+x_2{y}_2\right) \\ =3-2=1 \end{gathered}$$

Vậy $\left|z_1-z_2\right|$ = 1

Đáp án D