Lösung 3.1:2c

We start by expanding the quadratic in the numerator with the square rule:

1+i3(2−i3)2=1+i322−22i3+(i3)2=1+i34−43i−3=1+i31−43i

Then, we multiply top and bottom by the complex conjugate of the numerator:

1+i31−43i=(1+i3)(1−i3)(1−43i)(1−i3)=12−(i3)211−1i3−143i+43ii3=1+(3)2)1−i3−43i+4(3)2i2=1+31−(3+43)i−43=41−12−(1+4)3i=−411−453i