How do you use DeMoivre's Theorem to simplify ${(2 + 2 i)}^{6}$ $(2+2i)^6$ ?

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How do you use DeMoivre's Theorem to simplify ${(2 + 2 i)}^{6}$ $(2+2i)^6$ ?

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Answer 1 · 2016-10-07T12:58:12

${(2 + 2 i)}^{6} = - 512 i$ $(2+2i)^6=-512i$

Explanation:

De Moivre's Theorem states that if a complex number

$z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$ , then

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

Now, in $2 + 2 i$ $2+2i$ as its absolute value is $\sqrt{2^{2} + 2^{2}} = \sqrt{8} = 2 \sqrt{2}$ $sqrt(2^2+2^2)=sqrt8=2sqrt2$ , it can be written as

$2 + 2 i = 2 \sqrt{2} (\frac{1}{\sqrt{2}} + i \cdot \frac{1}{\sqrt{2}})$ $2+2i=2sqrt2(1/sqrt2+i*1/sqrt2)$

= $2 \sqrt{2} (cos (\frac{π}{4}) + i sin (\frac{π}{4}))$ $2sqrt2(cos(pi/4)+isin(pi/4))$

Hence ${(2 + 2 i)}^{6} = {(2 \sqrt{2})}^{6} (cos (\frac{π}{4} \times 6) + i sin (\frac{π}{4} \times 6))$ $(2+2i)^6=(2sqrt2)^6(cos(pi/4xx6)+isin(pi/4xx6))$

= $2^{6} \times 2^{3} (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $2^6xx2^3(cos((3pi)/2)+isin((3pi)/2))$

= $512 (0 - i) = - 512 i$ $512(0-i)=-512i$

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How do you use DeMoivre's Theorem to simplify ${(2 + 2 i)}^{6}$ $(2+2i)^6$ ?

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