Cho hai số phức z1, z2 thỏa mãn |z1| = |z2| = |z1 − z2| = 1

Trả lời

Ta có:  $\left|z_1-z_2\right|=1\Rightarrow \left(z_1-z_2\right)\left(\overline{z_1-z_2}\right)=1$

$\iff \left(z_1-z_2\right)\left(\overline{z_1}-\overline{z_2}\right)=1$

 $\iff z_1.\overline{z_1}+z_2.\overline{z_2}-z_1.\overline{z_2}-\overline{z_1}.z_2=1$

 $\iff {\left|z_1\right|}^2+{\left|z_2\right|}^2-z_1.\overline{z_2}-\overline{z_1}.z_2=1$

 $\iff 1^2+1^2-z_1.\overline{z_2}-\overline{z_1}.z_2=1$

 $\iff z_1.\overline{z_2}+\overline{z_1}.z_2=1$

Khi đó  $P={\left(\frac{z_1}{z_2}\right)}^2+{\left(\frac{z_2}{z_1}\right)}^2={\left(\frac{z_1}{z_2}+\frac{z_2}{z_1}\right)}^2-2\frac{z_1}{z_2}.\frac{z_2}{z_1}$

$={\left(\frac{z_1}{z_2}+\frac{z_2}{z_1}\right)}^2-2={\left(\frac{z_1.\overline{z_2}}{{\left|z_2\right|}^2}+\frac{\overline{z_1}.z_2}{{\left|z_1\right|}^2}\right)}^2-2$

 $={\left(z_1.\overline{z_2}+\overline{z_1}.z_2\right)}^2-2=1^2-2=-1$