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

1 Answer
Shwetank Mauria
Oct 2, 2016

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))

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