Lösung 3.3:2a

Aus Online Mathematik Brückenkurs 2

An equation of the type "zn=a complex number" is called a binomial equation and these are usually solved by going over to polar form and using de Moivre's formula.

We start by writing z and 1 in polar form

z1=r(cos+isin)=1(cos0+isin0).

The equation then becomes

r4(cos4+isin4)=1(cos0+isin0)

where we have used de Moivre's formula on the left-hand side. In order that both sides are equal, they must have the same magnitude and the same argument to within a multiple of 2, i.e.

r44=1=0+2n(n is an arbitrary integer).

This means that

r=1=2n(n is an arbitrary integer).

The solutions are thus (in polar form)

z=1cos2n+isin2nfor n=0 1 2...

but observe that the argument on the right-hand side essentially takes only four different values 0, 2, and 32, because other values of n give some of these values plus/minus a multiple of 2.

The equation's solutions are therefore

z=1(cos0+isin0)1(cos(2)+isin(2))1(cos+isin)1(cos(32)+isin(32))=1i−1−i.

Note: If we mark these solutions on the complex number plane, we see that they are corners in a regular quadrilateral.