How do you use DeMoivre's theorem to simplify ${(\sqrt{e^{10 i}})}^{6}$ $(sqrt(e^(10i)))^6$ ?

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Answer 1 · 2017-04-04T12:42:21

${(\sqrt{e^{10 i}})}^{6} = cos 30 + i sin 30 = \frac{\sqrt{3}}{2} + i \frac{1}{2}$ $(sqrt(e^(10i)))^6=cos30+isin30=sqrt3/2+i1/2$

According to DeMoivre's Theorem if

$z = r e^{i θ} = r (cos θ + i sin θ)$ $z=re^(itheta)=r(costheta+isintheta)$ , then for all $n \in Q$ $ninQ$

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

Hence, ${(\sqrt{e^{10 i}})}^{6}$ $(sqrt(e^(10i)))^6$

= ${({(e^{10 i})}^{\frac{1}{2}})}^{6}$ $((e^(10i))^(1/2))^6$

= ${(e^{10 i})}^{\frac{1}{2} \times 6}$ $(e^(10i))^(1/2xx6)$

= ${(e^{10 i})}^{3}$ $(e^(10i))^3$

= $e^{10 i \times 3}$ $e^(10ixx3)$

= $e^{30 i}$ $e^(30i)$

= $cos 30 + i sin 30$ $cos30+isin30$

= $\frac{\sqrt{3}}{2} + i \frac{1}{2}$ $sqrt3/2+i1/2$

How do you use DeMoivre's theorem to simplify (√e10i)6(sqrt(e^(10i)))^6?