Question

с р о ч н о комплексные числа​

Answer 1

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

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

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