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Lösung 3.1:1f

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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,

i11i20=i4+4+3=i4i4i3=11(i)=i=i4+4+4+4+4=i4i4i4i4i4=11111=1.

The answer becomes

i20+i11=1i.