Răspuns:
Explicație pas cu pas:
[tex]z=(1-i\sqrt{3})^{2017}\\fie~a=1-i\sqrt{3}.~Folosim~formula~lui~Moivre\\a^{n}=r^{n}(cosn\beta+isinn\beta),~unde~a=r(cos\beta +isin\beta )\\este~forma~trigonometrica~a~numarului~complex~a=x+iy=1-i\sqrt{3}\\r=|a|=\sqrt{x^{2}+y^{2}}=\sqrt{1^{2}+(-\sqrt{3})^{2}}=\sqrt{1+3}=2.\\cos\beta =\frac{x}{r}=\frac{1}{2},~sin\beta=\frac{y}{r} =\frac{-\sqrt{3}}{2} .\\cos\beta >0,~sin\beta <0~deci~a~are~imaginea~geometrica~in~cardanul~IV.\\atunci~\beta =2\pi -\frac{\pi}{3}=\frac{5\pi}{3}.\\[/tex]
continuarea e in imagine....
