wykazać tożsamość

[paola-91]

Witam! Czy ktoś umie wykazać podaną niżej tożsamość:

\(\displaystyle{ \left| z_1(1+\left| z_2\right|^2) -z_2(1+\left| z_1\right|^2)\right|^2=\left| z_1-z_2\right|^2\left| 1-z_1\bar{z_2}\right|^2-(z_1\bar{z_2}-\bar{z_1}z_2)^2}\)

Wiem, że

\(\displaystyle{ \left| a-b\right|^2=(a-b)(\bar{a}-\bar{b})}\)

Próbowałam lewą i prawą stronę powyliczać, jednak nie wychodzi mi to samo...

[octahedron]

\(\displaystyle{ L=\left|z_1(1+|z_2|^2)-z_2(1+|z_1|^2)\right|=\\\\
=\left(z_1(1+|z_2|^2)-z_2(1+|z_1|^2)\right)\left(\overline{z}_1(1+|z_2|^2)-\overline{z}_2(1+|z_1|^2)\right)=\\\\
=|z_1|^2(1+|z_2|^2)^2+|z_2|^2(1+|z_1|^2)^2-(z_1\overline{z}_2+\overline{z}_1z_2)(1+|z_1|^2)(1+|z_2|^2)=\\\\
=|z_1|^2+|z_2|^2+4|z_1|^2|z_2|^2+|z_1|^2|z_2|^4+|z_1|^4|z_2|^2-\\\\-(z_1\overline{z}_2+\overline{z}_1z_2)(1+|z_1|^2)(1+|z_2|^2)=\\\\
=(|z_1|^2+|z_2|^2)(1+|z_1|^2|z_2|^2)+4|z_1|^2|z_2|^2-(z_1\overline{z}_2+\overline{z}_1z_2)(1+|z_1|^2)(1+|z_2|^2)\\\\
P=|z_1-z_2|^2|1-z_1\overline{z}_2|^2-(z_1\overline{z}_2-\overline{z}_1z_2)^2=\\\\
=(z_1-z_2)(\overline{z}_1-\overline{z}_2)(1-z_1\overline{z}_2)(1-\overline{z}_1z_2)-(z_1\overline{z}_2-\overline{z}_1z_2)^2=\\\\
=(|z_1|^2+|z_2|^2-z_1\overline{z}_2-\overline{z}_1z_2)(1+|z_1|^2|z_2|^2-z_1\overline{z}_2-\overline{z}_1z_2)-(z_1\overline{z}_2-\overline{z}_1z_2)^2=\\\\
=(|z_1|^2+|z_2|^2)(1+|z_1|^2|z_2|^2)-(z_1\overline{z}_2+\overline{z}_1z_2)(1+|z_1|^2+|z_2|^2+|z_1|^2|z_2|^2)+\\\\+(z_1\overline{z}_2+\overline{z}_1z_2)^2-(z_1\overline{z}_2-\overline{z}_1z_2)^2=\\\\
=(|z_1|^2+|z_2|^2)(1+|z_1|^2|z_2|^2)-(z_1\overline{z}_2+\overline{z}_1z_2)(1+|z_1|^2)(1+|z_2|^2)+\\\\+(z_1\overline{z}_2+\overline{z}_1z_2+z_1\overline{z}_2-\overline{z}_1z_2)(z_1\overline{z}_2+\overline{z}_1z_2-z_1\overline{z}_2+\overline{z}_1z_2)=\\\\
=(|z_1|^2+|z_2|^2)(1+|z_1|^2|z_2|^2)-(z_1\overline{z}_2+\overline{z}_1z_2)(1+|z_1|^2)(1+|z_2|^2)+4|z_1|^2|z_2|^2\\\\
L=P}\)

Możliwe, że się da prościej, ale jakoś nie widzę żadnego sprytnego przekształcenia.