Question

How do you find all solutions to `x^6-64i=0`?

Answer 1

Use de Moivre's theorem to find all `6` Complex roots.

Explanation:

Given:

`x^6-64i = 0`

Add `64i` to both sides to get:

`x^6 = 64i`

`color(white)(x^6) = 2^6(0 + 1i)`

`color(white)(x^6) = 2^6(cos (pi/2) + i sin (pi/2))`

From de Moivre's theorem, we have:

`(cos theta + i sin theta)^n = cos n theta + i sin n theta`

Hence principal root:

`x_1 = root(6)(64i) = 2(cos (pi/12) + i sin (pi/12))`

The other `5` roots can be found by multiplying by the primitive Complex `6`th root of `1`, i.e. `(cos (pi/3) + i sin (pi/3))` to find:

`x_2 = 2(cos ((5pi)/12) + i sin ((5pi)/12))`

`x_3 = 2(cos ((9pi)/12) + i sin ((9pi)/12))`

`x_4 = 2(cos ((13pi)/12) + i sin ((13pi)/12))`

`x_5 = 2(cos ((17pi)/12) + i sin ((17pi)/12))`

`x_6 = 2(cos ((21pi)/12) + i sin ((21pi)/12))`

These trigonometric values can be expressed in terms of square roots:

`cos (pi/12) = 1/4(sqrt(6)+sqrt(2)) " " sin (pi/12) = 1/4(sqrt(6)-sqrt(2))`

`cos ((5pi)/12) = 1/4(sqrt(6)-sqrt(2)) " " sin ((5pi)/12) = 1/4(sqrt(6)+sqrt(2))`

`cos ((9pi)/12) = -sqrt(2)/2 " " sin ((9pi)/12) = sqrt(2)/2`

`cos ((13pi)/12) = -1/4(sqrt(6)+sqrt(2)) " " sin ((13pi)/12) = -1/4(sqrt(6)-sqrt(2))`

`cos ((17pi)/12) = -1/4(sqrt(6)-sqrt(2)) " " sin ((17pi)/12) = -1/4(sqrt(6)+sqrt(2))`

`cos ((21pi)/12) = sqrt(2)/2 " " sin ((21pi)/12) = -sqrt(2)/2`

So here are the `6` roots in the Complex plane:

graph{((x-1/2(sqrt(6)+sqrt(2)))^2 + (y-1/2(sqrt(6)-sqrt(2)))^2 - 0.01)((x-1/2(sqrt(6)-sqrt(2)))^2 + (y-1/2(sqrt(6)+sqrt(2)))^2 - 0.01)((x+sqrt(2))^2+(y-sqrt(2))^2-0.01)((x+1/2(sqrt(6)+sqrt(2)))^2 + (y+1/2(sqrt(6)-sqrt(2)))^2 - 0.01)((x+1/2(sqrt(6)-sqrt(2)))^2 + (y+1/2(sqrt(6)+sqrt(2)))^2 - 0.01)((x-sqrt(2))^2+(y+sqrt(2))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}