#### Answer

$-16$

#### Work Step by Step

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
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$[ 2 (\cos 135^{\circ}+i \sin135^{\circ})]^{4}=[ 2^{4} (\cos 4\times 135^{\circ}+i \sin 4\times 135^{\circ})]$
$=[ 16 (\cos 540^{\circ}+i \sin 540^{\circ})]$
$=[ 16 (-1+i.0]$
$=-16$