Răspuns:
Explicație pas cu pas:
a) [tex]t^{3}+t^{2}+t+1=0\\t^{2}(t+1)+(t+1)=0, (t+1)(t^{2}+1)=0, \left \{ {{t+1=0} \atop {t^{2}+1 =0}} \right, \left \{ {{t=-1} \atop {t^{2} =-1}} \right, \left \{ {{t=-1} \atop {t=+-i}} \right.\\ Raspuns: -1, -i, i\\[/tex]
atunci obtinem ecuatia t^{3}+t^{2}+t+1=0,[/tex]
care se rezolva analog ca la subpunctul a). Obtinem:
[tex]b) Fie t=[tex]\left \{ {{\frac{3z+1}{z-i}=-1} \atop {\frac{3z+1}{z-i}=+-i}} \right.[/tex]\\
Avem de rezolvat totalitatea a 3 ecuaţii:
1) 3z+1=-1*(z-i)⇔3z+1=-z+i⇔4z=-1+i⇔z=(-1+i)/4
2) 3z+1=-i*(z-i)⇔3z+1=-iz+i⇔3z+iz=-1+i⇔z(3+i)=-1+i⇔z=(-1+i)/(3+i)⇔ z=((-1+i)(3-i))/((3+i)(3-i))=(-3+i+3i+1)/10=(-2+4i)/10=(-1+2i)/5
3) 3z+1=i*(z-i)⇔3z+1=iz+1⇔3z-iz=0⇔z(3-i)=0⇔z=0.
Raspuns: { (-1+i)/4; (-1+2i)/5; 0}
