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Lösung 3.3:1e

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One part of the quotient has rather high exponents and this indicates that we ought to use polar form for the calculation.

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


Image:3_3_1_e.gif Image:3_3_1_e_text.gif

This shows that


1+i3=2cos3+isin31i=2cos4+isin43i=2cos6+isin6


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


3i91+i31i8=2cos6+isin692cos3+isin32cos4+isin48=29cos96+isin962cos3+isin328cos84+isin84=29cos23+isin232cos3+isin32218cos2+isin2=29cos23+isin232cos3+isin3241+i0=29224cos323+isin323=2925cos3+23+isin3+23=124cos611+isin611=116cos612+isin612=116cos26+isin26=116cos6+isin6=11623i2=1323i