Question

How do you find all solutions to `x^3+64i=0`?

Answer 1

`x = 4i,+-2sqrt(3)-2i`

Explanation:

Let `omega=-64i`, and then `x^3 = omega`

First let us plot the point `omega` on the Argand diagram:

And we will put the complex number into polar form (visually):
`|omega| = 64`
`arg(omega) = -(pi)/2`

So then in polar form we have:

`omega = 64(cos(-(pi)/2) + isin(-(pi)/2))`

We now want to solve the equation for `x` (to gain `3` solutions):

`x^3 = 64(cos(-(pi)/2) + isin(-(pi)/2))`

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential has a period of `2pi`, so we can equivalently write (incorporating the periodicity):

`x^3 = 64(cos(2npi-(pi)/2) + isin(2npi-(pi)/2)) \ \ \ n in ZZ`

By De Moivre's Theorem we can write this as:

`x = 64(cos(2npi-(pi)/2) + isin(2npi-(pi)/2))^(1/3)`
`\ \ = {64}^(1/3){cos(2npi-(pi)/2) + isin(2npi-(pi)/2)}^(1/3)`
`\ \ = 4{cos((2npi-(pi)/2)/3) + isin((2npi-(pi)/2)/3)}`
`\ \ = 4(cos theta + isin theta) \ \ \ \` where `theta=((4n-1)pi)/6`

Put:

`n=-1 => theta = (-5pi)/6`
`" " :. x = 4(cos ((-5pi)/6)+ isin ((-5pi)/6))`
`" " :. x = 4(-sqrt(3)/2-1/2i)`
`" " :. x = -2sqrt(3)-2i`

`n=0 => theta = (-pi)/6`
`" " :. x = 4(cos ((-pi)/6)+ isin ((-pi)/6))`
`" " :. x = 4(sqrt(3)/2-1/2i)`
`" " :. x = 2sqrt(3)-2i`

`n=1 => theta = (pi)/2`
`" " :. x = 4(cos ((pi)/2)+ isin ((pi)/2))`
`" " :. x = 4(0+i)`
`" " :. x = 4i`

After which the pattern continues.

We can plot these solutions on the Argand Diagram