Question

Express the following in the form a + ib, a, b ∈ R, using De Moivre's theorem:

`(1 - sqrt(3)"i")^4`

Answer 1

Let z = `1 - sqrt(3)"i"`

This is of the form a + bi, where a = 1, b = `-sqrt(3)`

∴ r = `sqrt("a"^2 + "b"^2) = sqrt(1^2 + (-sqrt(3))^2`

= `sqrt(1 + 3)`

= 2

Also, cos θ = `"a"/"r" = 1/2`

and sin θ = `"b"/"r" = (-sqrt(3))/2`

∴ θ lies in the fourth quadrant.

∴ tan θ = `sintheta/costheta`

= `(((-sqrt3)/2))/((1/2)`

= `-sqrt(3)`

= `-tan  pi/3`

= `tan(2pi - pi/3)`

= `tan  (5pi)/3`

∴ θ = `(5pi)/3`

∴ z = `1 - sqrt(3)"i" ` = r(cos θ + i sin θ)

= `2(cos  (5pi)/3 + "i" sin  (5pi)/3)`

∴ `(1 - sqrt(3)"i")^4 = [2(cos  (5pi)/3 + "i" sin  (5pi)/3)]^4`

= `2^4[cos(2pi - pi/3) + "i" sin(2pi - pi/3)]^4`

= `16(cos  pi/3 - "i" sin  pi/3)^4`

= `16(cos  (4pi)/3 - "i" sin  (4pi)/3)`  ...[∵ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ]

= `16[cos(pi + pi/3) - "i" sin(pi + pi/3)]`

= `16(-cos  pi/3 + "i" sin  pi/3)`

= `16[-1/2 + "i"(sqrt(3)/2)]`

= `-8 + 8sqrt(3)"i"`.