Question

How do you find the 6th root of `sqrt3-i`?

Answer 1

Let `omega=sqrt(3)-i`, and we we define `z` such that `z^6 = omega`

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

And we will put the complex number into polar form:

`|omega| = sqrt(3+1) = 2`
`arg(omega) = tan^-1(-1/sqrt(3)) = -(pi)/6`

So then in polar form we have:

`omega = 2(cos(-(pi)/6) + isin(-(pi)/6))`

We now want to solve the equation for `z` (to gain `6` solutions):

`z^6 = 2(cos(-(pi)/6) + isin(-(pi)/6))`

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):

`z^6 = 2(cos(2npi-(pi)/6) + isin(2npi-(pi)/6)) \ \ \ n in NN`

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

`z = {2(cos(2npi-(pi)/6) + isin(2npi-(pi)/6))}^(1/6)`
`\ \ = 2^(1/6)(cos((2npi-(pi)/6)/6) + isin((2npi-(pi)/6)/6))`
`\ \ = 2^(1/6)(cos theta + isin theta) \ \ \ \` where `theta=((12n-1)pi)/36`

Put:

`n=1 => theta = (11pi)/36 => z = 2^(1/6)(cos ((11pi)/36)+ isin ((11pi)/36))`

`n=2 => theta = (23pi)/36 => z = 2^(1/6)(cos ((23pi)/36)+ isin ((23pi)/36))`

`n=3 => theta = (35pi)/36=> z = 2^(1/6)(cos ((35pi)/36)+ isin ((35pi)/36))`

`n=4 => theta = (47pi)/36=> z = 2^(1/6)(cos ((47pi)/36)+ isin ((47pi)/36))`

`n=5 => theta = (59pi)/36=> z = 2^(1/6)(cos ((59pi)/36)+ isin ((59pi)/36))`

`n=6 => theta = (71pi)/36=> z = 2^(1/6)(cos ((71pi)/36)+ isin ((71pi)/36))`

After which the pattern continues.

We can plot these solutions on the Argand Diagram