Answer
$-8-8i$
Work Step by Step
Since, $(-1+i)=\sqrt 2(\cos 135^{\circ}+i \sin 135^{\circ})$
and $(-1+i)^{7}=\sqrt 2(\cos 135^{\circ}+i \sin 135^{\circ})^{7}$
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
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$\sqrt 2(\cos 135^{\circ}+i \sin 135^{\circ})^{7}=(\sqrt 2)^{7}(\cos 7\times 135^{\circ}+i \sin 7\times 135^{\circ})$
$=8\sqrt 2(\cos 945^{\circ}+i \sin 945^{\circ})$
$=8\sqrt 2(\cos 225^{\circ}+i \sin 225^{\circ})$
$=8\sqrt 2(\frac{-1}{\sqrt 2}-i.\frac{ 1}{\sqrt 2})$
$=-8-8i$
