Возвести в степень n=100 комплексное число z=2i-2

Решите задачу:

$z=2i-2\\a=-2,\; \; b=2\\|z|= \sqrt{a^2+b^2}= \sqrt{(-2)^2+2^2}= \sqrt{8}=2 \sqrt{2} \\\\ \left \{ {{cos \alpha = \frac{a}{|z|} } \atop {sin \alpha = \frac{b}{|z|} }} \right. =\ \textgreater \ \left \{ {{cos \alpha =\frac{-2}{2 \sqrt{2}} } \atop {sin \alpha = \frac{2}{2 \sqrt{2} } }} \right. =\ \textgreater \ \left \{ {{cos \alpha =- \frac{ \sqrt{2} }{2} } \atop {sin \alpha = \frac{ \sqrt{2} }{2} }} \right.=\ \textgreater \ \alpha = \frac{3 \pi }{4}\\\\z=|z|(cos \alpha +isin \alpha )\\z=(2 \sqrt{2})(cos\frac{3 \pi }{4}+isin\frac{3 \pi }{4})$

$z^n=(|z|)^n(cos \alpha n+isin \alpha n)\\\\z^{100}=(2 \sqrt{2})^{100}(cos \frac{3 \pi *100}{4}+isin \frac{3 \pi *100}{4})=\\=2^{100}*2^{50}(cos75 \pi +isin75 \pi )=2^{150}(-1+i*0)=-2^{150}$