Answer
$1$
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
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$[ (\cos 45^{\circ}+i \sin 45^{\circ})]^{8}=[ (\cos 8\times 45^{\circ}+i \sin 8\times 45^{\circ})]$
$=[ (\cos 360^{\circ}+i \sin 360^{\circ})]$
$=1+i.0$
$=1$
