Răspuns:
1+i 1-i
l1+il=l1-il=√(1^2+1^2)=√2
2+i 2-i
l2+il=l2-il=√(4+1)=√5
3+4i 3-4i
l3+4il=l3-4il=√(9+16)=5
cosα+isinα cosα-isinα
lcosα+isinαl=lcosα-isinαl=√(cos²α+isin²α)=1
6-i 6+i
l6-il=l6+il=√(6²+1²)=√37
9-3i 9+3i
l9-3il=l9+3il=√(81+9)=√90=3√10
Explicație pas cu pas:
