Hier kommt die Aufgabe 2
Bei b kam ich nicht ganz weiter, ich habe glaube ich phi falsch berechnet:
Text erkannt:
\( \begin{array}{l} \text { Nr. 2a) } z=\frac{\sqrt{2}-2 j}{1-\sqrt{2} j}=\frac{(\sqrt{2}-2 j) \cdot(1+\sqrt{2} j)}{(1-\sqrt{2} j) \cdot(1+\sqrt{23})} \\ \quad=\frac{\sqrt{2}+2 j-2 j+2 \sqrt{2}}{1+\sqrt{2} j-\sqrt{2} j+2}=\frac{\sqrt{2}+2 \sqrt{2}}{3} \\ \operatorname{Re}(2)=\frac{\sqrt{2}+2 \sqrt{2}}{3} \\ \operatorname{Im}(2)=\text { existief } n \text { nicht } \end{array} \)
\( \begin{array}{l} \text { b.) } z^{3}=-\frac{4}{5}+\frac{4}{\sqrt{3}} j \\ z^{3}=\omega=-\frac{4}{3}+\frac{4}{\sqrt{5}}{ }^{j} \\ |\omega|=\sqrt{\left(-\frac{4}{3}\right)^{2}+\left(\frac{4}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{16}{9}+\frac{16}{3}} \\ =\sqrt{\frac{16}{9}+\frac{48}{9}-\frac{4}{3}}=\sqrt{\frac{54}{9}}=\sqrt{6} \\ * \varphi=\arccos \frac{-\frac{4}{\sqrt{6}}}{\sqrt{6}}=-\frac{4}{3 \sqrt{6}} \end{array} \)
Text erkannt:
\( \begin{array}{l} \text { C.) } p(x)=2 x^{3}+3 x^{2}-1 \\ \text { NST1 }=-1 \\ \left(2 x^{3}+3 x^{2}-1\right):(x+1)=2 x^{2}+x-1 \\ \frac{\left(2 x^{3}+2 x^{2}\right)}{-\frac{x^{2}}{2}+1} \\ \frac{-(-x)}{-x-1)} \\ 0 \\ 2 x^{2}+x-1=0 \mid: 2 \\ x^{2}+\frac{1}{2} x-\frac{1}{2}=0 \\ x_{1,2}=\frac{1}{4} \pm \sqrt{\frac{1}{16}+\frac{1}{2}} \\ x_{1,2}=-\frac{1}{4} \pm \sqrt{\frac{1}{16}+\frac{8}{16}} \\ x_{1,2}=-\frac{1}{4} \pm \sqrt{\frac{9}{16}} \\ x_{1,2}=-\frac{1}{4} \pm \frac{\sqrt{9}}{\sqrt{6}}=-\frac{1}{4} \pm \frac{3}{4} \\ x_{1}=-1 \quad x_{2}=\frac{2}{4}=\frac{1}{2} \end{array} \)
\begin{tabular}{c|cc}
UST & \( V F H \). \\
\hline-1 & 2 & \\
\hline\( \frac{1}{2} \) & 1 & \( 2 \cdot(x+1)^{2} \cdot\left(x-\frac{1}{2}\right) \) \\
& &
\end{tabular}
