«11C» 10. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 10. Hausaufgabe

0.0.1.1 ↑ Selbstgestellte Aufgabe

z_1 = -1 + \mathrm{i}\sqrt{3} = 2E(\frac{2}{3}\pi);$z_1=-1+\text{i}\sqrt{3}=2E\left(\frac{2}{3}\pi \right);$

z_2 = 4E(\frac{11}{6}\pi);$z_2=4E\left(\frac{11}{6}\pi \right);$

⇒ z_1z_2 = 8E(\frac{2}{3}\pi + \frac{11}{6}\pi) = 8E(\frac{5}{2}\pi) = 8E(\frac{\pi}{4}) = 8\mathrm{i};$z_1{z}_2=8E\left(\frac{2}{3}\pi +\frac{11}{6}\pi \right)=8E\left(\frac{5}{2}\pi \right)=8E\left(\frac{\pi }{4}\right)=8\text{i};$

⇒ \frac{z_1}{z_2} = \frac{1}{2}E(\frac{2}{3}\pi - \frac{11}{6}\pi) = \frac{1}{2}E(-\frac{7}{6}\pi) = \frac{1}{2}E(\frac{5}{6}\pi) = -\frac{1}{4}\sqrt{3} + \frac{1}{4}\mathrm{i};$\frac{z_1}{z_2}=\frac{1}{2}E\left(\frac{2}{3}\pi -\frac{11}{6}\pi \right)=\frac{1}{2}E\left(-\frac{7}{6}\pi \right)=\frac{1}{2}E\left(\frac{5}{6}\pi \right)=-\frac{1}{4}\sqrt{3}+\frac{1}{4}\text{i};$

⇒ z_1^{-1} = \frac{E(0)}{z_1} = \frac{1}{2}E(-\frac{2}{3}\pi) = \frac{1}{2}E(\frac{4}{3}\pi) = -\frac{1}{4} - \frac{1}{4}\sqrt{3}\mathrm{i};$z_1^{-1}=\frac{E\left(0\right)}{z_1}=\frac{1}{2}E\left(-\frac{2}{3}\pi \right)=\frac{1}{2}E\left(\frac{4}{3}\pi \right)=-\frac{1}{4}-\frac{1}{4}\sqrt{3}\text{i};$

⇒ z_2^{-1} = \frac{E(0)}{z_2} = \frac{1}{4}E(-\frac{11}{6}\pi) = \frac{1}{4}E(\frac{\pi}{6}) = \frac{1}{8}\sqrt{3} + \frac{1}{8}\mathrm{i};$z_2^{-1}=\frac{E\left(0\right)}{z_2}=\frac{1}{4}E\left(-\frac{11}{6}\pi \right)=\frac{1}{4}E\left(\frac{\pi }{6}\right)=\frac{1}{8}\sqrt{3}+\frac{1}{8}\text{i};$