#### Answer

$-16\sqrt 3+16i$

#### Work Step by Step

Since, $(\sqrt 3+i)=2(\cos 30^{\circ}+i \sin 30^{\circ})$
and $(\sqrt 3+i)^{5}=2(\cos 30^{\circ}+i \sin 30^{\circ})^{5}$
De Moivre’s Theorem states that when $ r (\cos\theta+i \sin\theta)$ is a complex number, and if $n$ is any real number, then the following relationship holds.
$[ r (\cos\theta+i \sin\theta)]^{n}=[ r^{n} (\cos n\theta+i \sin n\theta)]$
In compact form, this is written
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$2(\cos 30^{\circ}+i \sin 30^{\circ})^{5}=2^{5}(\cos 5\times30^{\circ}+i \sin 5\times30^{\circ})^{5}$
$=32(\cos 150^{\circ}+i \sin 150^{\circ})$
$=[3 2(\frac{-\sqrt 3}{2}+i.\frac{ 1}{2})]$
$=-16\sqrt 3+16i$