Aus Online Mathematik Brückenkurs 2

This is a typical binomial equation which we solve in polar form.

We write

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

and, on using de Moivre's formula, the equation becomes

\displaystyle r^{2}\left( \cos 2\alpha +i\sin 2\alpha \right)=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)

Both sides are equal when

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

When \displaystyle n=0 and \displaystyle n=\text{1}, we get two different arguments for \displaystyle \alpha , whilst different values of \displaystyle n only give these arguments plus/minus a multiple of \displaystyle 2\pi .

The solutions to the equation are

\displaystyle z=\left\{ \begin{array}{*{35}l} \ 1\centerdot \left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\ \ 1\centerdot \left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right) \\ \end{array} \right.=\left\{ \begin{array}{*{35}l} \ \frac{1+i}{\sqrt{2}} \\ \ -\frac{1+i}{\sqrt{2}} \\ \end{array} \right.