Question

Ex A2 subpunctul a!!

Answer 1

$z_{1}=1+i\\\\z_{1}=\sqrt{2}*(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=\ \sqrt{2}*(cos45+i*sin45)\\\\z_{2}=1=i\\\\z_{2}=\sqrt{2}*(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)=\ \sqrt{2}*(cos45-i*sin45)\\\\z_{2}=\sqrt{2}*(cos(360-45)+i*sin(360-45))\\\\z_{2}=\sqrt{2}*(cos315+i*sin315)\\\\z=\frac{\sqrt{2}*(cos45+i*sin45)}{\sqrt{2}*(cos315+i*sin315)}\\\\z=cos(45-315)+i*sin(45-315)=cos(-270)+i*sin(-270)\\\\z=cos270-i*sin270=cos(360*270)+i*sin(360-270)\\\\z=cos90+i*sin90=cos\frac{\pi}{2}+i*sin(\frac{\pi}{2})$

$z_k=cos(\frac{\frac{\pi}{2}+2k\pi}{4})+i*sin(\frac{\frac{\pi}{2}+2k\pi}{4}),k\ ia\ volori\ de\ la\ 0-3\\\\z_0=cos(\frac{\frac{\pi}{2}}{4})+i*sin(\frac{\frac{\pi}{2}}{4})\\\\z_0=cos(\frac{\pi}{8})+i*sin(\frac{\pi}8})\\\\z_1=cos(\frac{\frac{\pi}{2}+2\pi}{4})+i*sin(\frac{\frac{\pi}{2}+2\pi}{4})\\\\z_1=cos(\frac{5\pi}{8})+i*sin(\frac{5\pi}{8})\\\\z_2=cos(\frac{\frac{\pi}{2}+4\pi}{4})+i*sin(\frac{\frac{\pi}{2}+4\pi}{4})\\\\z_2=cos(\frac{9\pi}{8})+i*sin(\frac{9\pi}{8})$

$z_{3}=cos(\frac{\frac{\pi}{2}+6\pi}{4})+i*sin(\frac{\frac{\pi}{2}+6\pi}{4})\\\\z_{3}=cos(\frac{13\pi}{8})+i*sin(\frac{13\pi}{8})$