Lösung 3.3:1d

Aus Online Mathematik Brückenkurs 2

Version vom 10:43, 11. Mär. 2009

Because we are going to raise something to the power 12, the base in the expression should be written in polar form. In turn, the base consists of a quotient which is advantageous to calculate in polar form. Thus, it seems appropriate to write 1+i3  and 1+i in polar form right from the beginning and to carry out all calculations in polar form.

We obtain

1+i31+i=2cos3+isin3=2cos4+isin4

and

1+i1+i3=2cos3+isin32cos4+isin4=22cos3−4+isin3−4=2cos12+isin12.

Finally, de Moivre's formula gives

1+i1+i312=212cos1212+isin1212=2(12)12(cos+isin)=26(−1+i0)=−64.