vom calcularadacina de ordin 3, pt ca ese o ecuatie de grad 3
z³=-i
z³= cos3π/2+isin3π/2
|z³|=1⇒(|z|)^1/3=1
z1=cos ( (3π/2):3) +isin (3π/6)=cos π/2 +isinπ/2=0+i=i
z2= cos[ ( 3π/2+2π)/3] +isin (7π/6)=cos7π/6 +isin7π/6= -√3/2-i/2
z3=cos [(3π/2+4π)/3] +isin11π/6= cos11π/6 +isin 11π/6=√3/2-i/2
care verifica, auargumentele de 90°, 210°si, respectiv 330°, toate inmultite cu 3 duc la argumentul redus 270° (3π/2) al lui z³=-i
problema este bine rezolvata
