Write the complex number in a + ib form using cube roots of unity: `(-(1)/(2) + sqrt(3)/(2)i)^(1000)`

Answer

`omega = - (1)/(2) + sqrt(3)/(2)i`" " (b) `(1)/(2^(8))((-1-isqrt(3))/(2))` " "(c) 0

Explanation

(a) Here, `-1//2 + (1//2) isqrt(3)` is one of the tow imaginary cube roots of unity. If the we denote it b `omega`, then
`omega^(100) = omega^(999) omega=(omega^(3))^(333) omega = omega = -(1)/(2) + (sqrt(3))/(2) i`
`z =(sqrt(3) + i)^(17)/((1-i)^(50))`
`= (1)/(i^(17))((isqrt(3) +i^(2))^(17))/([(1-i)^(2) ]^(25))`
`= (2^(17))/(i) (((isqrt(3)-1)/(2))^(7))/(-2i)^(25)`
`=(1)/(2^(8)) (omega)^(17)`
`= (1)/(2^(8)) (omega)^(2)`
`= (1)/(2^(8)) ((-1-isqrt(3))/(2))`
(c) `(i+sqrt(3))^(100) + (i-sqrt(3))^(100) + 2^(100)`
`=((i^(2) + isqrt(3))/(i))^(100) + ((i^(2)-isqrt(3))/(i))^(100) + 2^(100)`
`= (2^(100))/(i^(100)) ((-1+isqrt(3))/(2))^(100) + (2^(100))/(i^(100)) ((-1-isqrt(3))/(2)) +2^(100)`
`= 2^(100) (omega) ^(100) + 2^(100) (omega^(2))^(100) + 2^(100)`
`=2^(100) (omega^(100) + omega^(200)+1)`
` = 2^(100) (omega + omega^(2) + 1) =0`