Question

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

Answer 1

`(sqrt(3)/2 + "i"/2)^5 + (sqrt(3)/2 - "i"/2)^5`

Fnding the polar form of `sqrt(3)/2 + "i"/2`

Let `sqrt(3)/2 + "i"/2` = r(cosθ + i sinθ)

r cos θ = `sqrt(3)/2` (+ ve)

r cos θ = `1/2` (+ ve)

r²cos²θ + r²sin²θ = `3/4 + 1/4` = 1

r² = 1

r = 1

Since, cos θ and sin θ are + ve, θ lies in I quadrant

Now, cos θ =  `sqrt(3)/2`

sin θ = `1/2`

Argument θ = `pi/6`

∴ Polar form `sqrt(3)/2 + "i"/2 = (cos  pi/6 + "i" sin  pi/6)`

And `sqrt(3)/2 - "i"/2 = cos  pi/6 - "i" sin  pi/6`

Now `(sqrt(3)/2 + "i"/2)^5 + (sqrt(3)/2 - "i"/2)^5`

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

By Using de Moivre's theorem

= `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 (sqrt(3)/2)`

= `- sqrt(3)`