How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{5}$ $(3-2i)^5$ ?

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How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{5}$ $(3-2i)^5$ ?

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Answer 1 · 2017-02-04T06:45:18

${(3 - 2 i)}^{5} = 69 \sqrt{13} (cos 5 α + i sin 5 α)$ $(3-2i)^5=69sqrt13(cos5alpha+isin5alpha)$ ,

where $α = {tan}^{- 1} (- \frac{2}{3})$ $alpha=tan^(-1)(-2/3)$

Explanation:

DeMoivre's Theorem states that if a complex number $z$ $z$ in polar form is given by $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

then $z^{n} = r^{n} (cos n θ + i sin n θ)$ $z^n=r^n(cosntheta+isinntheta)$

Here $| 3 - 2 i | = \sqrt{3^{2} + 2^{2}} = \sqrt{13}$ $|3-2i|=sqrt(3^2+2^2)=sqrt13$ , hence we can write
$3 - 2 i$ $3-2i$ as $\sqrt{13} (cos α + i sin α)$ $sqrt13(cosalpha+isinalpha)$ , where $α = {tan}^{- 1} (- \frac{2}{3})$ $alpha=tan^(-1)(-2/3)$

Hence ${(3 - 2 i)}^{5}$ $(3-2i)^5$

= ${(\sqrt{13})}^{5} (cos 5 α + i sin 5 α)$ $(sqrt13)^5(cos5alpha+isin5alpha)$

= $169 \sqrt{13} (cos 5 α + i sin 5 α)$ $169sqrt13(cos5alpha+isin5alpha)$

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How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{5}$ $(3-2i)^5$ ?

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How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{5}$ $(3-2i)^5$ ?