Question

Solve the equation `z^4+81i=0`?

Answer 1

There are four solutions;

`z = 3(-sqrt(2-sqrt(2))/2-sqrt(2+sqrt(2))/2i)`
`z = 3(sqrt(2+sqrt(2))/2-sqrt(2-sqrt(2))/2i)`
`z = 3(sqrt(2-sqrt(2))/2+sqrt(2+sqrt(2))/2i)`
`z = 3(-sqrt(2+sqrt(2))/2+sqrt(2-sqrt(2))/2i)`

Explanation:

Let `omega=-81i`, and let `z^4 = omega`

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

And we will put the complex number into polar form (visually):

`|omega| = 81`
`arg(omega) = -(pi)/2`

So then in polar form we have:

`omega = 81(cos(-(pi)/2) + isin(-(pi)/2))`

We now want to solve the equation for `z` (to gain `4` solutions):

`z^3 = 81(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 = 81cos(2npi-(pi)/2) + isin(2npi-(pi)/2)) \ \ \ n in ZZ`

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

`z = {cos(2npi-(pi)/2) + isin(2npi-(pi)/2)}^(1/4)`
`\ \ = {81}^(1/4){cos(2npi-(pi)/2) + isin(2npi-(pi)/2)}^(1/4)`
`\ \ = 3{cos((2npi-(pi)/2)/4) + isin((2npi-(pi)/2)/4)}`
`\ \ = 3(cos theta + isin theta)`

Where:

`theta = (2npi-(pi)/2)/4`
`\ \ = 1/8(4n-1)pi`

Put:

`n=-1 => theta = (-5pi)/8`
`" " :. z = 3(cos ((-5pi)/8)+ isin ((-5pi)/8))`
`" " :. z = 3(-sqrt(2-sqrt(2))/2-sqrt(2+sqrt(2))/2i)`

`n=0 => theta = (-pi)/8`
`" " :. z = 3(cos ((-pi)/8)+ isin ((-pi)/8))`
`" " :. z = 3(sqrt(2+sqrt(2))/2-sqrt(2-sqrt(2))/2i)`

`n=1 => theta = (3pi)/8`
`" " :. z = 3(cos ((3pi)/8)+ isin ((3pi)/8))`
`" " :. z = 3(sqrt(2-sqrt(2))/2+sqrt(2+sqrt(2))/2i)`

`n=2 => theta = (7pi)/8`
`" " :. z = 3(cos ((7pi)/8)+ isin ((7pi)/8))`
`" " :. z = 3(-sqrt(2+sqrt(2))/2+sqrt(2-sqrt(2))/2i)`

After which the pattern continues.

We can plot these solutions on the Argand Diagram