Question

How do you find the cube roots of `8(cos((2pi)/3)+isin((2pi)/3))`?

Answer 1

Cube roots of `[8(cos((2pi)/3)+isin((2pi)/3))]` are `2(cos((2pi)/9)+isin((2pi)/9))`
`2(cos((8pi)/9)+isin((8pi)/9))` and
`2(cos((14pi)/9)+isin((14pi)/9))`

Explanation:

Accoding to De Moivre's theorem

If `z=r(costheta+isintheta)`

`z^n=r^n(cosntheta+isinntheta)`

Here `[8(cos((2pi)/3)+isin((2pi)/3))]^(1/3)`

= `8^(1/3)(cos(((2pi)/3)/3)+isin(((2pi)/3)/3)]`

= `2(cos((2pi)/9)+isin((2pi)/9))`

But this is only one cube root of `[8(cos((2pi)/3)+isin((2pi)/3))]^(1/3)`

for others we have

`2[cos((2pi+(2pi)/3)/3)+isin((2pi+(2pi)/3)/3)]` i.e.`2(cos((8pi)/9)+isin((8pi)/9))` and

`2[cos((4pi+(2pi)/3)/3)+isin((4pi+(2pi)/3)/3)]` i.e.`2(cos((14pi)/9)+isin((14pi)/9))`