1+i√3 = 2•[cos(π/3)+isin(π/3)]
1-i = √2•[cos(7π/4)+isin(7π/4)]
Acum aplic formula lui De Moivre.
{2•[cos(π/3)+isin(π/3)]}¹⁰ =
= 2¹⁰•[cos(10π/3)+isin(10π/3)]
= 2¹⁰•[cos(3π+π/3)+isin(3π+π/3)]
= 2¹⁰•(-1/2+i√3/2)
= -2⁹+i•2⁹√3
{√2•[cos(7π/4)+isin(7π/4)]}¹⁰ =
= 2⁵•[cos(70π/4)+isin(70π/4)]
= 2⁵•[cos(17π+2π/4)+isin(17π+2π/4)]
= 2⁵•(0-i)
= -i•2⁵
Raportul este:
(-2⁹+i•2⁹√3)/(-i•2⁵) = -i•2⁴+2⁴√3 =
= 16√3 - 16i
