Lösung 3.1:1f

Let's begin by calculating some powers of i:

i2i3i4=ii=−1=i2i=(−1)i=−i=i2i2=(−1)(−1)=1

Now, we observe that because i4=1, we can try to factorize i11 and i20 in terms of i4,

i2i11i20=ii=−1=i4+4+3=i4i4i3=11(−i)=−i=i4+4+4+4+4=i4i4i4i4i4=11111=1

The answer becomes

i20+i11=1−i