Lösung 3.1:2d
Aus Online Mathematik Brückenkurs 2
Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator:
(1+i)−11+i=1+i5+5i−1=1+i4+5i
=2−3i3i(2−3i)+i2−3i=2−3i6i−9i2+i=2−3i9+7i
Hence,
We multiply out the numerator and denominator
1+9
i+7i
1+7i
i4
2−4
3i+5i
2−5i
3i=9+9i+7i−78−12i+10i+15=2+16i23−2i
This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator:
2−23
16i−2i
2+2i
16i=4+25646−368i−4i−32=26014−372i
If we divide up the numbers into factors,
7=2
186=2
2
93=2
2
3
31=10
26=2
5
13
2
we can simplify the answers:
72
2
5
13−2
2
5
132
2
3
31i=72
5
13−5
133
31i=7130−6593i