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

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

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Answer 1 · 2016-12-21T10:43:26

The answer is $= 64 (- \sqrt{3} - i)$ $=64(-sqrt3-i)$

Explanation:

De Moivre's theorem states

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

Let $z = \sqrt{3} + i$ $z=sqrt3+i$

$∥ ∥ z ∥ = \sqrt{3 + 1} = 2$ $∥z∥=sqrt(3+1)=2$

$z = 2 \cdot (\frac{\sqrt{3}}{2} + i \frac{1}{2})$ $z=2*(sqrt3/2+i1/2)$

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

$cos θ = \frac{\sqrt{3}}{2}$ $costheta=sqrt3/2$ , $\Rightarrow$ $=>$ , $θ = \frac{π}{6}$ $theta=pi/6$

$sin θ = \frac{1}{2}$ $sintheta=1/2$ , $\Rightarrow$ $=>$ , $θ = \frac{π}{6}$ $theta=pi/6$

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

Therefore,

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

$= 128 (cos (\frac{7}{6} π) + i sin (\frac{7}{6} π))$ $=128(cos(7/6pi)+isin(7/6pi))$

$= 128 (- \frac{\sqrt{3}}{2} - \frac{i}{2})$ $=128(-sqrt3/2-i/2)$

$= 64 (- \sqrt{3} - i)$ $=64(-sqrt3-i)$

Answer 2 · 2016-12-21T10:53:34

See explanation.

Explanation:

De Moivre's Theorem says that:

**If a complex number $z$ $z$ is given in trigonometric form: **

$z = r (cos φ + i sin φ)$ $z=r(cosvarphi+isinvarphi)$

Then $n - t h$ $n-th$ power of $z$ $z$ is given as:

$z^{n} = {| z |}^{n} \cdot (cos n φ + i sin n φ)$ $z^n=|z|^n*(cosnvarphi+isinnvarphi)$

So first thing to do is to change $z = \sqrt{3} + i$ $z=sqrt(3)+i$ into trigonometric form:

$| z | = \sqrt{{\sqrt{3}}^{2} + 1^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2$ $|z|=sqrt(sqrt(3)^2+1^2)=sqrt(3+1)=sqrt(4)=2$

$cos φ = \frac{r e (z)}{r} = \frac{\sqrt{3}}{2} \Rightarrow φ = 30^{o}$ $cosvarphi=(re(z))/r=sqrt(3)/2 => varphi=30^o$

So the trigonometric form of $z$ $z$ is:

$z = 2 (cos 30 + i sin 30)$ $z=2(cos30+isin30)$

Now we can calculate $z^{7}$ $z^7$ according to the de Moivre's theorem:

$z^{7} = 2^{7} \cdot (cos 7 \cdot 30 + i sin 7 \cdot 30)$ $z^7=2^7*(cos 7*30+isin7*30)$

$z^{7} = 128 \cdot (cos 210 + i sin 210)$ $z^7=128*(cos210+isin210)$

$z^{7} = 128 \cdot (cos (180 + 30) + i sin (180 + 30))$ $z^7=128*(cos(180+30)+isin(180+30))$

$z^{7} = 128 \cdot (- cos 30 - sin 30 i)$ $z^7=128*(-cos30-sin30i)$

$z^{7} = 128 \cdot (- \frac{\sqrt{3}}{2} - \frac{1}{2} i)$ $z^7=128*(-sqrt(3)/2-1/2i)$

Answer: $z^{7} = - 64 \sqrt{3} - 64 i$ $z^7=-64sqrt(3)-64i$

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