#### Answer

$128+128i$

#### Work Step by Step

Since, $(-2-2i)=2\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})$
and $(-2-2i)^{5}=2\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})^{5}$
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
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$2\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})^{5}=(2\sqrt 2)^{5}(\cos 5\times 225^{\circ}+i \sin 5\times 225^{\circ})$
$=32\sqrt {32}(\cos 1125^{\circ}+i \sin 1125^{\circ})$
$=128\sqrt 2(\cos 45^{\circ}+i \sin 45^{\circ})$
$=128\sqrt 2(\frac{1}{\sqrt 2}+i.\frac{ 1}{\sqrt 2})$
$=128+128i$