Aus Online Mathematik Brückenkurs 2

The The equation \displaystyle z^{3}=-\text{1 } is a so-called binomial equation, which we solve by writing both sides in polar form. We have

\displaystyle \begin{align} & z=r\left( \cos \alpha +i\sin \alpha \right) \\ & -1=1\left( \cos \pi +i\sin \pi \right) \\ \end{align}

and, with the help of de Moivre's formula, the equation becomes

\displaystyle r^{3}\left( \cos 3\alpha +i\sin 3\alpha \right)=1\left( \cos \pi +i\sin \pi \right)

Both sides are equal when their magnitudes are equal and the arguments differ by a multiple of \displaystyle 2\pi ,

\displaystyle \left\{ \begin{array}{*{35}l} r^{3}=1 \\ 3\alpha =\pi +2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.

which gives that

\displaystyle \left\{ \begin{array}{*{35}l} r=1 \\ \alpha =\frac{\pi }{3}+\frac{2n\pi }{3}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.

For every third integer \displaystyle n, the solution formula gives in principal the same value for the argument (the difference is a multiple of \displaystyle 2\pi ), so the equation has in reality three solutions (for \displaystyle n=0,\ 1 and \displaystyle \text{2}):

\displaystyle z=\left\{ \begin{array}{*{35}l} 1\centerdot \left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\ 1\centerdot \left( \cos \pi +i\sin \pi \right) \\ 1\centerdot \left( \cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \right) \\ \end{array} \right.=\left\{ \begin{array}{*{35}l} \frac{1+i\sqrt{3}}{2} \\ -1 \\ \frac{1-i\sqrt{3}}{2} \\ \end{array} \right.

We obtain the typical behaviour that the solutions are corner points in a regular polygon (a triangle in this case because the degree of the equation is \displaystyle \text{3}).