Aus Online Mathematik Brückenkurs 2

If we use \displaystyle w=z-\text{1} as a new unknown and move the term \displaystyle \text{4} over to the right-hand side, we have a binomial equation,

\displaystyle w^{4}=-4

We can solve this equation in the usual way by using polar form and de Moivre's formula. We have

\displaystyle \begin{align} & w=r\left( \cos \alpha +i\sin \alpha \right) \\ & -4=4\left( \cos \pi +i\sin \pi \right) \\ \end{align}

and the equation becomes

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

The only way that both sides can be equal is if the magnitudes agree and the arguments do not differ by anything other than a multiple of \displaystyle 2\pi ,

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

which gives us that

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

for \displaystyle n=0,\ 1,\ 2 and \displaystyle 3, the argument \displaystyle \alpha assumes the four different values

\displaystyle \frac{\pi }{4},\ \frac{3\pi }{4},\ \frac{5\pi }{4} and \displaystyle \frac{7\pi }{4}

and for different values of \displaystyle n we obtain values of \displaystyle \alpha which are equal to those above, apart from multiples of \displaystyle 2\pi . Thus, we have four solutions,

\displaystyle w=\left\{ \begin{array}{*{35}l} \sqrt{2}\left( \cos {\pi }/{4}\;+i\sin {\pi }/{4}\; \right) \\ \sqrt{2}\left( \cos {3\pi }/{4}\;+i\sin 3{\pi }/{4}\; \right) \\ \sqrt{2}\left( \cos {5\pi }/{4}\;+i\sin 5{\pi }/{4}\; \right) \\ \sqrt{2}\left( \cos 7{\pi }/{4}\;+i\sin 7{\pi }/{4}\; \right) \\ \end{array} \right.=\left\{ \begin{array}{*{35}l} 1+i \\ -1+i \\ -1-i \\ 1-i \\ \end{array} \right.

and the original variable z is

\displaystyle z=\left\{ \begin{array}{*{35}l} 2+i \\ i \\ -i \\ 2-i \\ \end{array} \right.