Question

Exercițiul 9. Punctul 2
Multumesc!

Answer 1

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