Cho hai số phức z1z2 thỏa mãn z1 = z2 = z1 - z2 = 1 Tính z1 + z2

 Đáp án A

Ta có:  ${\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|}^2=\left({\mathrm{z}}_1-{\mathrm{z}}_2\right)\left(\overline{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)=\left({\mathrm{z}}_1-{\mathrm{z}}_2\right)\left(\overline{{\mathrm{z}}_1}-\overline{{\mathrm{z}}_2}\right)={\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2-\left({\mathrm{z}}_1.\overline{{\mathrm{z}}_2}+\overline{{\mathrm{z}}_1}.{\mathrm{z}}_2\right)$

$\Rightarrow {\mathrm{z}}_1.\overline{{\mathrm{z}}_2}+\overline{{\mathrm{z}}_1}.{\mathrm{z}}_2=1$ 

$\Rightarrow {\left|{\mathrm{z}}_1+{\mathrm{z}}_2\right|}^2=\left({\mathrm{z}}_1+{\mathrm{z}}_2\right)\left(\overline{{\mathrm{z}}_1+{\mathrm{z}}_2}\right)=\left({\mathrm{z}}_1+{\mathrm{z}}_2\right)\left(\overline{{\mathrm{z}}_1}+\overline{{\mathrm{z}}_2}\right)={\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2+\left({\mathrm{z}}_1.\overline{{\mathrm{z}}_2}+\overline{{\mathrm{z}}_1}.{\mathrm{z}}_2\right)=3.$ 

Từ đó suy ra $\left|{\mathrm{z}}_1+{\mathrm{z}}_2\right|=\sqrt{3}.$