Question

Show that `(sqrt(3)/2 + i/2)^5 + (sqrt(3)/2 - i/2)^5 = - sqrt(3)`

Answer 1

LHS = `(sqrt3/2 + i/2)^5 + (sqrt3/2 - i/2)^5`

Polar form of `sqrt3/2 + i/2 = r (cos theta + i sin theta)` 

`r = sqrt(x^2 + y^2) = sqrt(3/4 + 1/4) = 1`

`alpha = tan^-1 |y/x| = tan^-1 |1/sqrt3| = pi/6`

`theta = alpha = pi/6`

`sqrt3/2 + i/2 = 1 (cos pi/6 + i sin pi/6)`

Similarly `sqrt3/2 - i/2 = (cos pi/6 - i sin pi/6)`

LHS = `(cos  pi/6 + i sin  pi/6)^5 + (cos  pi/6 - i sin  pi/6)^5`

`= cos  (5pi)/6 + i sin  (5pi)/6 + cos  (5pi)/6 - i sin  (5pi)/6`

`= 2 cos  (5pi)/6 = 2 cos (pi - pi/6) = -2 cos  pi/6`

`= -2 xx sqrt3/2 = -sqrt3` = RHS