sa se rezolve ecuatia:
va rog dacă ma puteti ajuta
![Sa Se Rezolve Ecuatiava Rog Dacă Ma Puteti Ajuta class=](https://ro-static.z-dn.net/files/dd7/aeed938d2c9c976d696bf53b28276f7a.jpg)
[tex]z^{2}=(x+yi)^2=x^{2}+2xyi+(y)^{2}(i)^{2}\\\\x^{2}+2xyi-y^{2}=5-12i\\\\Ce\ este\ fara\ i\ din\ membrul\ stang\ este\ egal\ cu\ ce\ este\ fara\ i\ din\ membrul\ drept,\\iar\ ce\ este\ cu\ i\ din\ membrul\ drept\ este\ egal\ cu\ ce\ este\ cu\ i\ din\ membrul\ stang,\ adica\\\\x^{2}-y^{2}=5\\\\2xy=-12\\\\Le\ ridicam\ pe\ amandoua\ la\ a\ doua,\ dupa\ le\ adunam.\\\\x^{4}-2x^{2}y{2}+y^{4}=25\\\\4x^{2}y^{2}=169\\\\x^{4}-2x^{2}y{2}-y^{4}+4x^{2}y^{2}=x^{4}+2x^{2}y{2}+y^{4}=(x^2+y^2)^{2}=169\\\\[/tex]
[tex]x^{2}+y^{2}=\sqrt{169}=13\\\\x^{2}-y^{2}=5\\\\Le\ adunam\ si\ pe\ acestea\\\\2x^{2}=18\\\\x^{2}=9\\\\x=-3\ sau\ x=3\\\\\\2xy=-12\ =>\ y=\frac{12}{2x}=\frac{6}{x}\\x=-3\ =>\ y=\frac{6}{-3}=-2\ =>\ z1=-3-2i\\\\x=3\ =>\ y=\frac{6}{3}=2\ =>\ z2=3+2i[/tex]