Question #35611

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Answer 1 · 2017-07-18T04:54:58

$z^6=-64$

De Moivre's Theorem states that:

$z^n=(rcis(x))^n=r^ncis(nx)$

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

$r^ncis(nx)=r^n(cos(nx)+isin(nx))$

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.