If a complex number is written in trigonometric form:
[tex]z=r(\cos \theta+i\cdot\sin \theta)[/tex]
Then, the De Moivre's theorem says that:
[tex]z^n=r^n(\cos (n\theta+i\cdot\sin (n\theta))[/tex]
For this particular case, we have that:
[tex]\begin{gathered} r=\sqrt[]{3} \\ \theta=\frac{5\pi}{6} \end{gathered}[/tex]
Then, using the De Moivre's formula:
[tex]\begin{gathered} \lbrack\sqrt[]{3}(\cos \frac{5\pi}{6}+i\cdot\sin \frac{5\pi}{6})\rbrack^{10}=\sqrt[]{3}^{10}(\cos 10\cdot\frac{5\pi}{6}+i\cdot\sin 10\cdot\frac{5\pi}{6}) \\ =3^{10/2}(\cos \frac{50\pi}{6}+i\cdot\sin \frac{50\pi}{6}) \\ =3^5(\cos \frac{25\pi}{3}+i\cdot\sin \frac{25\pi}{3}) \\ =243(\cos (8+\frac{1}{3})\pi+i\cdot\sin (8+\frac{1}{3})\pi) \\ =243(\cos \frac{\pi}{3}+i\cdot\sin \frac{\pi}{3}) \\ =243(\frac{1}{2}+i\cdot\frac{\sqrt[]{3}}{2}) \\ =\frac{243}{2}+i\cdot\frac{243\sqrt[]{3}}{2} \end{gathered}[/tex]
Therefore:
[tex]\lbrack\sqrt[]{3}(\cos \frac{5\pi}{6}+i\cdot\sin \frac{5\pi}{6})\rbrack^{10}=\frac{243}{2}+\frac{243\cdot\sqrt[]{3}}{2}i[/tex]
