How do you use DeMoivre's Theorem to simplify $(2+i)^5$ ?

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Answer 1 · 2016-12-03T15:11:00

The answer is $=55.9(-0.68+0.73i)$

To use DeMoivre's theorem, we need to change to the trigonometric form of the complex numbers.

$(costheta+isintheta)^n=cosntheta+isinntheta$

Here,

$z=2+i$

$∥z∥=sqrt(4+1)=sqrt5$

$z=sqrt5(2/sqrt5+i/sqrt5)$

$z=sqrt5(costheta+isintheta)$

Therefore,

$costheta=2/sqrt5$

$sintheta=1/sqrt5$

$theta=0.46 rd$

$z=sqrt5(cos0.46+isin0.46)$

Therefore,

$z^5=(sqrt5(cos0.46+isin0.46))^5$

$=(sqrt5)^5(cos2.32+ isin2.32)$

$=55.9(-0.68+0.73i)$

How do you use DeMoivre's Theorem to simplify (2+i)^5(2+i)^5?