#### Answer

$-128+128\sqrt 3i$

#### Work Step by Step

Since, $(2-2i\sqrt 3)=4(\cos 300^{\circ}+i \sin 300^{\circ})$
and $(2-2i\sqrt 3)^{4}=4(\cos 300^{\circ}+i \sin 300^{\circ})^{4}$
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
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$4(\cos 300^{\circ}+i \sin 300^{\circ})^{4}=4^{4}(\cos 4\times300^{\circ}+i \sin 4\times300^{\circ})$
$=256(\cos 1200^{\circ}+i \sin 1200^{\circ})$
$=[256(\frac{-1}{2}+i.\frac{ \sqrt 3}{2})]$
$=-128+128\sqrt 3i$