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

z= 1cos4+isin4 1cos43+isin43= 21+i −21+i