Question

Представить в тригонометрической форме:

Answer 1

$z_1=3+\sqrt{3}i\\\mod z=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}\\\arg z =\phi=\arctan\frac{\sqrt{3}}{3}=\frac{\pi}{6}\\z_1=2\sqrt{3}(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}))=2\sqrt{3}e^{i\frac{\pi}{6}}$

$z_2=-2-2i\\\mod z=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\arg z=\phi=\arctan{\frac{-2}{-2}}-\pi=\frac{\pi}{4}-\pi=-\frac{3\pi}{4}\\z_2=2\sqrt{2}(\cos(\frac{-3\pi}{4})+i\sin(\frac{-3\pi}{4}))=2\sqrt{2}e^{-\frac{3\pi}{4}i}$

$\frac{z_1}{z_2}=\frac{3+\sqrt{3}i}{-2-2i}=\frac{(3+\sqrt{3})(-2+2i)}{(-2-2i)(-2+2i)}=\frac{-6+6i-2\sqrt{3}i-2\sqrt{3}}{4+4}=\frac{-6-2\sqrt{3}}{8}+\frac{6-2\sqrt{3}}{8}=\frac{-3-\sqrt{3}}{4}+\frac{3-\sqrt{3}}{4}i$