Question

Hei am si eu o problema de mate la numere complexe.Asa suna:Sa se determine z a.î.: |z|=z+3+i

Answer 1

|z|∈R
deci z+3+i∈R adica a+bi+3+i∈R
a+3+(b+1)i∈R
b+1=0
b=-1
 deci daca z=a+bi
 b=-1
 atunci
√(a²+(-1)²)=a+3
ridicam la patrat
a²+1=a²+6a+9
0=6a+8
3a+4=0
a=-4/3

deci z=-4/3-i

verificare
√(16/9+1)=-4/3-i+3+i
√(25/9)=3-4/3
5/3=9/3-4/3
5/3=5/3 ADEVARAT
problema este bine rezolvata