Lösung 3.3:4a

Aus Online Mathematik Brückenkurs 2

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

We write

z=rcos+isini=1cos2+isin2 

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

r2cos2+isin2=1cos2+isin2 

Both sides are equal when

r2=12=2+2nn an arbitrary integer  

which gives that

r=1=4+nn an arbitrary integer  

When n=0 and n=1, we get two different arguments for , whilst different values of n only give these arguments plus/minus a multiple of 2.

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.