Question

Find all the `8^(th)` roots of `3i-3`?

Find all the `8^(th)` roots of
`3i-3`

Also show any one of them in an Argand diagram.

Answer 1

Explanation:

Let `omega=-3+3i`, and let `z^8 = omega`

First, we will put the complex number into polar form:

`|omega| = sqrt((-3)^2+3^2) =3sqrt(2)`
`theta =arctan(3/(-3)) = arctan(-1) = -pi/4`
`=> arg \ omega = pi/2+pi/4 = (3pi)/4`

So then in polar form we have:

`omega = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4))`

We now want to solve the equation `z^8=omega` for `z` (to gain `8` solutions):

`z^8 = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4))`

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

`z^8 = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi)) \ \ \ n in ZZ`

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

`z = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8)`
`\ \ = (3sqrt(2))^(1/8)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8)`
`\ \ = 3^(1/8)2^(1/16) (cos(((3pi)/4+2npi)/8) + isin(((3pi)/4+2npi)/8))`
`\ \ = 3^(1/8)2^(1/16) (cos theta + isin theta )`

Where:

`theta =((3pi)/4+2npi)/8 = ((3+8n)pi)/32`

And we will get `8` unique solutions by choosing appropriate values of `n`. Working to 3dp, and using excel to assist, we get:

After which the pattern continues (due the above mentioned periodicity).

We can plot these solutions on the Argand Diagram: