How do you use DeMoivre's Theorem to simplify $2 {(\sqrt{3} + i)}^{7}$ $2(sqrt3+i)^7$ ?

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Answer 1 · 2016-10-17T10:29:02

$2 {(\sqrt{3} + i)}^{7} = - 128 \sqrt{3} - 128 i$ $2(sqrt3+i)^7=-128sqrt3-128i$

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

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

Now $(\sqrt{3} + i)$ $(sqrt3+i)$

= $2 (\frac{\sqrt{3}}{2} + i \times \frac{1}{2}$ $2(sqrt3/2+ixx1/2$

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

and using DeMoivre's Theorem, ${(\sqrt{3} + i)}^{7}$ $(sqrt3+i)^7$

= $2^{7} (cos (\frac{7 π}{6}) + i sin (\frac{7 π}{6}))$ $2^7(cos((7pi)/6)+isin((7pi)/6))$

= $128 (cos (π + \frac{π}{6}) + i sin (π + \frac{π}{6}))$ $128(cos(pi+pi/6)+isin(pi+pi/6))$

And $2 {(\sqrt{3} + i)}^{7}$ $2(sqrt3+i)^7$

= $2 \times 128 (cos (π + \frac{π}{6}) + i sin (π + \frac{π}{6}))$ $2xx128(cos(pi+pi/6)+isin(pi+pi/6))$

= $256 (- cos (\frac{π}{6}) - i sin (\frac{π}{6}))$ $256(-cos(pi/6)-isin(pi/6))$

= $256 (- \frac{\sqrt{3}}{2} - \frac{i}{2})$ $256(-sqrt3/2-i/2)$

= $- 128 \sqrt{3} - 128 i$ $-128sqrt3-128i$

How do you use DeMoivre's Theorem to simplify 2(√3+i)72(sqrt3+i)^7?