Convert [tex]-1+i[/tex] to polar form.
[tex]z = -1 + i \implies \begin{cases}|z| = \sqrt{(-1)^2 + 1^2} = \sqrt2 \\\\ \arg(z) = \pi + \tan^{-1}\left(\dfrac1{-1}\right) = \dfrac{3\pi}4 \end{cases}[/tex]
By de Moivre's theorem,
[tex]\left(-1+i\right)^7 = \left(\sqrt2 \, e^{i\,\frac{3\pi}4}\right)^7 \\\\ ~~~~~~~~ = \left(\sqrt2\right)^7 e^{i\,\frac{21\pi}4} \\\\ ~~~~~~~~ = 8\sqrt2 \, e^{-i\,\frac{3\pi}4} \\\\ ~~~~~~~~ = 8\sqrt2 \left(\cos\left(\dfrac{3\pi}4\right) - i \sin\left(\dfrac{3\pi}4\right)\right) \\\\ ~~~~~~~~ = 8\sqrt2 \left(-\dfrac1{\sqrt2} - \dfrac1{\sqrt2}\,i\right) \\\\ ~~~~~~~~ = -8 (1 + i)[/tex]
QED
