Question

с р о ч н о комплексные числа​

Answer 1

$z_1=1+2i; ; ,; ; z_2=-3i\\a); ; frac{z_2}{z_1}=frac{-3i}{1+2i}=frac{-3i(1-2i)}{1-4i^2}=frac{-2i+6i^2}{1+4}=frac{-6-3i}{5}=-frac{6}{5}-frac{3}{5}cdot i$

$b); ; Big (frac{overline {z_1}-i, z_2}{2z_2}Big )^6=Big (frac{1-2i+3i^2}{-6i}Big )^6=Big (frac{-2-2i}{-6i}Big )^6=Big (frac{(-2-2i)(-6i)}{-6icdot 6i}Big )^6=\\=Big (frac{12i+12i^2}{-36i^2}Big )^6=Big (frac{-12+12i}{36}Big )^6=(-frac{1}{3}+frac{1}{3}, i)^6\\z=-frac{1}{3}+frac{1}{3}, i; ,; ; |z|=sqrt{frac{1}{9}+frac{1}{9}}=frac{sqrt2}{3}; ,\\varphi =pi +arctg(-frac{3}{3})=pi -frac{pi}{4}=frac{3pi }{4}\\z=frac{sqrt2}{3}cdot (cosfrac{3pi }{4}+i, sinfrac{3pi }{4})$

$z^6=(frac{sqrt2}{3})^6cdot Big (cosfrac{6cdot 3pi }{4}+i, sinfrac{6cdot 3pi }{4}Big )=frac{8}{729}cdot Big (cosfrac{9pi }{2}+i, sinfrac{9pi }{2}Big )=\\=frac{8}{729}cdot Big (cosfrac{pi}{2}+i, sinfrac{pi }{2}Big )=frac{8}{729}, i$