Question

How do you find the cube root of `64 (cos (pi/5) + isin (pi/5))`?

Answer 1

`\ \ \ \ \ \ 4 (cos (pi/15) + isin (pi/15))`
`\ \ \ \ \ \ 4 (cos ((11pi)/15) + isin ((11pi)/15))`
`\ \ \ \ \ \ 4 (cos ((7pi)/5) + isin ((7pi)/5))`

Explanation:

Let `z^3=64 (cos (pi/5) + isin (pi/5))`

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^3=64 (cos (pi/5+2npi) + isin (pi/5+2npi)) \ \ \ n in NN`

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

`z = {64 (cos (pi/5+2npi) + isin (pi/5+2npi))}^(1/3)`
`\ \= 64^(1/3) (cos (pi/5+2npi) + isin (pi/5+2npi))}^(1/3`
`\ \= 64^(1/3) (cos ((pi/5+2npi)/3) + isin ((pi/5+2npi)/3))`
`\ \= 4 (cos ((pi/5+2npi)/3) + isin ((pi/5+2npi)/3))`

Put:

`n=0 => z = 4 (cos ((pi/5)/3) + isin ((pi/5)/3))`
`" " = 4 (cos (pi/15) + isin (pi/15))`

`n=1 => z = 4 (cos ((pi/5+2pi)/3) + isin ((pi/5+2pi)/3))`
`" " = 4 (cos ((11pi)/15) + isin ((11pi)/15))`

`n=2 => 4 (cos ((pi/5+4pi)/3) + isin ((pi/5+4pi)/3))`
`" " = 4 (cos ((7pi)/15) + isin ((7pi)/15))`

After which the pattern continues

So the three roots are:

`z = 4 (cos (pi/15) + isin (pi/15))`
`\ \ \ \ \ \ 4 (cos ((11pi)/15) + isin ((11pi)/15))`
`\ \ \ \ \ \ 4 (cos ((7pi)/5) + isin ((7pi)/5))`

We can view these roots on the Argand diagram: