Answer
$1$
Work Step by Step
Since, $(\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i)=(\cos 315^{\circ}+i \sin 315^{\circ})$
and $(\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i)^{8}=(\cos 315^{\circ}+i \sin 315^{\circ})^{8}$
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
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$(\cos 315^{\circ}+i \sin 315^{\circ})^{8}=(\cos 8\times315^{\circ}+i \sin 8\times315^{\circ})$
$=(\cos 2520^{\circ}+i \sin 2520^{\circ})$
$=(\cos 0^{\circ}+i \sin 0^{\circ})$
$=[1+i.0)]$
$=1$
