ПОМОГИТЕ РЕШИТЬ КОМПЛЕКСНЫЙ ЧИСЛА

Решите задачу:

$(\sqrt{3}-i)^{20}=(2*(\frac{\sqrt 3}{2}-i*\frac{1}{2}))^{20}=\\\\=2^{20}*(\cos (-\frac{\pi}{6})+i*\sin(-\frac{\pi}{6}))^{20}=\\\\=2^{20}*(\cos(-\frac{10\pi}{3})+i*\sin(-\frac{10\pi}{3}))=\\\\=2^{20}*(\cos(\frac{2\pi}{3})+i*\sin(\frac{2\pi}{3}));$

$(1+i)^{15}=(\sqrt 2*(\frac{\sqrt 2}{2}+ i*\frac{\sqrt 2}{2}))^{15}=\\\\=(\sqrt 2)^{15}*(\cos(\frac{\pi}{4})+i*\sin(\frac{\pi}{4}))^{15}=\\\\=2^{\frac{15}{2}}*(\cos(\frac{15\pi}{4})+i*\sin(\frac{15\pi}{4}))=\\\\=2^{7.5}*(\cos(-\frac{\pi}{4})+i*\sin(-\frac{\pi}{4}));$

$(\sqrt 3 -i)^{20}(1+i)^{15}=\\\\=2^{20}*(\cos(\frac{2\pi}{3})+i*\sin(\frac{2\pi}{3}))*2^{7.5}*(\cos(-\frac{\pi}{4})+i*\sin(-\frac{\pi}{4}))=\\\\=2^{27.5}*(\cos(\frac{5\pi}{12})+i*\sin(\frac{5\pi}{12}))=\\\\=2^{26.5}*\sqrt{2-\sqrt 3}+i*2^{26.5}*\sqrt{2+\sqrt 3}$