Question

Zpatrat+2zconjugat+1=0 va rog !!!

Answer 1

(a+bi)²+2(a-bi)+1=0+0i
a²+2abi -b²+2a-2bi+1=0+0i
a²-b²+2a+1+2abi+bi=0+0i
egalam partile reale si imaginare
a²-b²+2a+1=0 (1)
2ab+2b=0       (2)
2b(a+1)=0

din (2)deducem  b1=0⇒a²+2a+1=0 ⇒a=-1  deci z1=-1
sau a=-1
care inlocuit in (1)ne da
1-b²-2+1=0
-b²=0
b=0
 a=-1 si b=0 , adica acelasi numar, z=-1
deci ecuatia are solutia dubla  z=-1∈R