How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{8}$ $(3-2i)^8$ ?

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How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{8}$ $(3-2i)^8$ ?

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Answer 1 · 2017-01-02T06:57:46

${(3 - 2 i)}^{8} = 13^{4} (cos 8 θ + i sin 8 θ)$ $(3-2i)^8=13^4(cos8theta+isin8theta)$ , where $θ = {tan}^{- 1} (- \frac{2}{3}))$ $theta=tan^(-1)(-2/3))$

Explanation:

According to DeMoivre's theorem, if $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

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

Hence to simplify ${(3 - 2 i)}^{8}$ $(3-2i)^8$ , we should first put $(3 - 2 i)$ $(3-2i)$ in polar form.

As $r cos θ = 3$ $rcostheta=3$ and $r sin θ = - 2$ $rsintheta=-2$ , $r = \sqrt{3^{2} + 2^{2}} = \sqrt{13}$ $r=sqrt(3^2+2^2)=sqrt13$

and $θ = {tan}^{- 1} (- \frac{2}{3})$ $theta=tan^(-1)(-2/3)$

Hence ${(3 - 2 i)}^{8} = {(\sqrt{13})}^{8} (cos 8 θ + i sin 8 θ)$ $(3-2i)^8=(sqrt13)^8(cos8theta+isin8theta)$

= $13^{4} (cos 8 θ + i sin 8 θ)$ $13^4(cos8theta+isin8theta)$ , where $θ = {tan}^{- 1} (- \frac{2}{3}))$ $theta=tan^(-1)(-2/3))$

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How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{8}$ $(3-2i)^8$ ?

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How do you use DeMoivre's Theorem to simplify ${(3 - 2 i)}^{8}$ $(3-2i)^8$ ?