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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