Compute (1+i)^6

Question

1. 64(cos45° + i sin45°) 
2. 64(cos270° + i sin270°)

Answer 1

By Moivre's Theorem,

(a + ib)ⁿ = rⁿ [cos(nθ) + isin(nθ)] where r = √(a²+b²) and θ = tan^-1(b/a).

Therefore (1 + i)⁶ = √2⁶ [cos(6π/4) + isin(6π/4)] where r = √(1²+1²) = √(1+1) =√2 and θ = tan^-1(b/a) = tan^-1(1/1) = π/4.

(1 +i)⁶ = ((√2)²)³ [cos(3π/2) + isin(3π/2)]

= ((√2)²)³ [cos(270^o) + isin(270^o)]

= (2)³ [0 + i(-1)]

= 8 [- i]

= -8i

Therefore (1 +i)⁶ = -8i.

1) 64(cos45° + i sin45°)            [Since cos45° = cos45° =1/√2]

= 64(1/√2 + i √2)

= 64/√2 (1 + i)

= [64√2 / √2√2] (1 + i)

= [64√2 / 2] (1 + i)

= 32√2 (1 + i)

Therefore 64(cos45° + i sin45°) = 32√2 (1 + i).

2) 64(cos270° + i sin270°)

= 64(0 + i (-1))                          [Since cos270° = 0 and sin270° = -1]

= 64(- i )

= -64i

Therefore 64(cos270° + i sin270°) = -64i