Question

How do you find all the cube roots of z, where `z = (sqrt3)/2 + 1/2i`?

Answer 1

`r_1 = cos(pi/18) + i sin(pi/18)`

`r_2 = cos((13pi)/18) + i sin((13pi)/18)`

`r_3 = cos((25pi)/18) + i sin((25pi)/18)`

Explanation:

This is an example of "Casus Irreducibilis" - a Complex cube root that is not expressible in the form `a+bi` where `a` and `b` are expressed in terms of Real `n`th roots.

You can express it in terms of trigonometric functions.

`z` itself lies on the unit circle:

`z = cos(pi/6) + i sin(pi/6)`

Use De Moivre's formula:

`(cos theta + i sin theta)^n = cos (n theta) + i sin (n theta)`

to deduce:

The cube roots also lie on the unit circle at intervals of `(2pi)/3`:

`r_1 = cos(pi/18) + i sin(pi/18)`

`r_2 = cos((13pi)/18) + i sin((13pi)/18)`

`r_3 = cos((25pi)/18) + i sin((25pi)/18)`