Lösung 3.3:1e

Aus Online Mathematik Brückenkurs 2

Version vom 10:43, 11. Mär. 2009

Some parts of the quotient have rather high exponents and this indicates that we ought to use polar form for the calculation.

First, we write 1+i3 , 1−i and 3−i  in polar form.

This shows that

1+i31−i3−i=2cos3+isin3=2cos−4+isin−4=2cos−6+isin−6.

Now, with the help of de Moivre's formula,

3−i91+i3(1−i)8=2cos−6+isin−692cos3+isin32cos−4+isin−48=29cos9−6+isin9−62cos3+isin328cos8−4+isin8−4=29cos−23+isin−232cos3+isin32(12)8cos(−2)+isin(−2)=29cos−23+isin−232cos3+isin324(1+i0)=29224cos3−−23+isin3−−23=2925cos3+23+isin3+23=124cos611+isin611=116cos612−+isin612−=116cos2−6+isin2−6=116cos−6+isin−6=11623−i2=1323−i.