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

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Answer 1 · 2016-12-03T15:11:00

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

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

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(costheta+isintheta)^n=cosntheta+isinntheta$

Here,

$z = 2 + i$ $z=2+i$

$∥ ∥ z ∥ = \sqrt{4 + 1} = \sqrt{5}$ $∥z∥=sqrt(4+1)=sqrt5$

$z = \sqrt{5} (\frac{2}{\sqrt{5}} + \frac{i}{\sqrt{5}})$ $z=sqrt5(2/sqrt5+i/sqrt5)$

$z = \sqrt{5} (cos θ + i sin θ)$ $z=sqrt5(costheta+isintheta)$

Therefore,

$cos θ = \frac{2}{\sqrt{5}}$ $costheta=2/sqrt5$

$sin θ = \frac{1}{\sqrt{5}}$ $sintheta=1/sqrt5$

$θ = 0.46 r d$ $theta=0.46 rd$

$z = \sqrt{5} (cos 0.46 + i sin 0.46)$ $z=sqrt5(cos0.46+isin0.46)$

Therefore,

$z^{5} = {(\sqrt{5} (cos 0.46 + i sin 0.46))}^{5}$ $z^5=(sqrt5(cos0.46+isin0.46))^5$

$= {(\sqrt{5})}^{5} (cos 2.32 + i sin 2.32)$ $=(sqrt5)^5(cos2.32+ isin2.32)$

$= 55.9 (- 0.68 + 0.73 i)$ $=55.9(-0.68+0.73i)$

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