How do you use DeMoivre's theorem to simplify (-1/2+sqrt3/2i)^3?

2 Answers
Narad T.
Nov 23, 2016

The answer is =1

Explanation:

Let z=-1/2+isqrt3/2

We have to write z in trigonometric form in order to apply DeMoivre's theorem

z=r(costheta+isintheta)

r=1

costheta=-1/2

sintheta=sqrt3/2

So, theta is located in the 2nd quadrant

theta=2pi/3

Therefore, z=cos((2pi)/3)+isin((2pi)/3)

We need z^3

We use DeMoivre 's theorem,
(costheta+isintheta)^n=cosntheta+isinntheta

z^3=(cos((2pi)/3)+isin((2pi)/3))^3

=cos(2pi/3*3)+isin(2pi/3*3)

=cos2pi+isin2pi

=1

Shwetank Mauria
Nov 23, 2016

(-1/2+sqrt3/2i)^3=1

Explanation:

According to DeMoivre's theorem, if a+ib=rcostheta+irsintheta

then (a+ib)^n=r^n(cosntheta+isinntheta)

Writing -1/2+sqrt3/2i in trigonometric form

-1/2+sqrt3/2i=cos((2pi)/3)+isin((2pi)/3)

Hence (-1/2+sqrt3/2i)^3=1^3(cos(3xx(2pi)/3)+isin(3xx(2pi)/3))

= cos(2pi)+isin(2pi)

= 1+i0

= 1

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