Question

Demonstrati ca:

$|1-z_1\cdot \overline{z_2}|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$

Unde z1 si z2 sunt numere complexe.


Va rog, cu explicatii sau rezolvare detaliata. Am incercat sa-l fac dar ceva nu imi iasa.

Answer 1

Răspuns:

Explicație pas cu pas:

Tot jmenul consta in a folosi formula :

$|z|^2=z\cdot \overline{z}$

Si eventual si faptul ca :

$\overline{\overline{z}}=z$

Voi porni din L.H.S si voi ajunge in R.H.S :

$|1-z_1\cdot \overline{z_2}|^2-|z_1-z_2|^2=(1-z_1\cdot\overline{z_2})(\overline{1-z_1\overline{z_2}})-(z_1-z_2)(\overline{z_1-z_2})=\\=(1-z_1\cdot\overline{z_2})(1-\overline{z_1}\cdot z_2)-(z_1-z_2)(\overline{z_1}-\overline{z_2})=\\=1-z_2\cdot\overline{z_1}-\overline{z_1}\cdot z_2+|z_1\cdot z_2|^2-|z_1|^2+z_2\cdot\overline{z_1}+z_1\cdot\overline{z_2}-|z_2|^2=\\=1+|z_1\cdot z_2|^2-|z_1|^2-|z_2|^2=\\ =(1-|z_1|^2)-|z_2|^2(1-|z_1|^2)=(1-|z_1|^2)(1-|z_2|^2), Q.E.D.$