Lösung 3.3:2c
Aus Online Mathematik Brückenkurs 2
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