Question

Записать число z=-2+2i√3 в тригонометрической форме

Answer 1

$z=-2+2i\sqrt3\\\\z=a+bi\; \; \to \; \; a=-2\; ,\; b=2\sqrt3\\\\|z|=\sqrt{a^2+b^2}=\sqrt{4+4\cdot 3}=\sqrt{16}=4\\\\a\ \textless \ 0\; ,b\ \textgreater \ 0\; \; \; \Rightarrow \; \; argz=arctg\frac{b}{a}+\pi =arctg(-\sqrt3)+\pi =\\\\=-arctg\sqrt3+\pi =-\frac{\pi}{3}+\pi =\frac{2\pi}{3}\\\\z=4\cdot (cos\frac{2\pi }{3}+i\, sin\frac{2\pi}{3})$