Question

Sa se calculeze(poza)

Answer 1

Răspuns:

Explicație pas cu pas:

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

continuarea e in imagine....

Answer 2

$z = (1-i\sqrt 3)^{2017}\\ \\\\r = \sqrt{1+3} = 2 \\ \\\\ 1-i\sqrt 3 = -(-1+i\sqrt 3) =-2(\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}) \\ \\ (1-i\sqrt 3)^{2017} = -2^{2017}(\cos \frac{2\cdot 2017\pi}{3}+i\sin \frac{2\cdot 2017\pi}{3}) \\ \\ z = -2^{2017}(\cos \frac{4034\pi}{3}+i\sin \frac{4034\pi}{3}) \\ \\\\4034 : 3 = 1344\text{ rest } 2\\ \\\\z = -2^{2017}\Big(\cos(1344\pi+\frac{2\pi}{3})+i\sin(1344\pi+\frac{2\pi}{3})\Big)\\ \\ z =-2^{2017}\cdot (-\frac{1}{2}+i\frac{\sqrt 3}{2}) \\ \\ z = 2^{2016}\cdot(1-i\sqrt 3)$