b) ( 3+i)(3-i)(1/5 + 1/10 i)= 2+1 ( 3+i)(3-i) = 9 - i^2 ==>9 +1 ==>10
10( 1 + i ) ==> 10(2+ i) ==> 2 +i
5 10 10
c) (raiz de 2 -i)-i(1-i raiz2)=-2i
(V2 - i ) - ( i -i^2V2)
V2 - i - i + i^2V2)
V2 - i - i -V2)
- 2i
d) (1-i) elevado a 4= -4
(1-i)^4 = (-2i)(-2i) ==> 4i^2 ==> 4(-1) = - 4
(1-i)^4 = (1-i)^2 . (1-i)^2
(1-i)^2 = 1 - 2i + i^2 = 1-2i-1 = - 2i
(1-i)^2 = 1 - 2i + i^2 = 1-2i-1 = - 2i