Răspuns:
0
Explicație pas cu pas:
fie ecuatia
z^5=1=cos0+isin0
sau z^5-1=0 in forma al;gebrica
aceasta va avea radacibile
z1---5=cos (0+2kπ)/5+isin(0+2kπ/5) k=0;1;2;3;4
adica
x1=cos0+isin0=1
x2=cos2π/5+isin 2π/5
x3= cos4π/5+isin4π/5
x4=cos6π/5+isin 6π/5
x5=cos8π/5+isin 8π/5
x1+x2+x3+x4+x5= (Viete )= 0=0+0i ⇒
⇒1+cos (2pi/5)+cos (4pi/5)+ cos ( 6pi/5 )+cos( 8pi/5) =0
extra
si partea imaginara va fi tot 0, adica
sin (2pi/5)+sin (4pi/5)+ sin ( 6pi/5 )+sin( 8pi/5) +0
