Lösung 3.3:2c

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We write \displaystyle z\text{ } and the right-hand side \displaystyle \text{-1-}i in polar form


\displaystyle \begin{align} & z=r\left( \cos \alpha +i\sin \alpha \right) \\ & \text{-1-}i=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right) \\ \end{align}


Using de Moivre's formula, the equation can now be written as


\displaystyle r^{5}\left( \cos 5\alpha +i\sin 5\alpha \right)=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right)


If we identify the magnitude and argument on both sides, we get


\displaystyle \left\{ \begin{array}{*{35}l} r^{5}=\sqrt{2} \\ 5\alpha =\frac{5\pi }{4}+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.


(The arguments \displaystyle 5\alpha and \displaystyle \frac{5\pi }{4} can differ by a multiple of \displaystyle 2\pi and still correspond to the same complex number.)

This gives that


\displaystyle \left\{ \begin{array}{*{35}l} r=\sqrt[5]{2}=\left( 2^{{1}/{2}\;} \right)^{{1}/{5}\;}=2^{{1}/{10}\;} \\ \alpha =\frac{1}{5}\left( \frac{5\pi }{4}+2n\pi \right)=\frac{\pi }{4}+\frac{2n\pi }{5}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.


If we investigate the argument \displaystyle \alpha more closely, we see that it assumes essentially only five different values,


\displaystyle \frac{\pi }{4},\ \frac{\pi }{4}+\frac{2\pi }{5},\ \frac{\pi }{4}+\frac{4\pi }{5},\ \frac{\pi }{4}+\frac{6\pi }{5} and \displaystyle \ \frac{\pi }{4}+\frac{8\pi }{5}


since these angle values then repeat to within a multiple of \displaystyle 2\pi .

In summary, the roots of the equation are


\displaystyle z=2^{{1}/{10}\;}\left( \cos \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right)+i\sin \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right) \right)


for \displaystyle n=0,\ 1,\ 2,\ 3 and \displaystyle 4