# Question 35611

Jul 18, 2017

${z}^{6} = - 64$

#### Explanation:

De Moivre's Theorem states that:

${z}^{n} = {\left(r c i s \left(x\right)\right)}^{n} = {r}^{n} c i s \left(n x\right)$

We will want to expand out the cis form to then convert to rectangular form:

${r}^{n} c i s \left(n x\right) = {r}^{n} \left(\cos \left(n x\right) + i \sin \left(n x\right)\right)$

Ok, now apply the theorem to the complex number:

z^6=2^6cis(6*30˚)=64cis(180˚)=64(cos(180˚)+isin(180˚))

From the unit circle:

cos(180˚)=-1

sin(180˚)=0

Therefore:

64(cos(180˚)+isin(180˚))=64(-1+0i)=-64#

Interestingly, the imaginary component is zero.