Processing Math: Done
Lösung 3.3:1e
Aus Online Mathematik Brückenkurs 2
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
3
3−i
This shows that
3=2
cos
3+isin
3
1−i=
2
cos
−
4
+isin
−
4
3−i=2
cos
−
6
+isin
−
6
Now, with the help of de Moivre's formula,
3−i
9
1+i
3
1−i
8=
2
cos
−
6
+isin
−
6
92
cos
3+isin
3
2
cos
−
4
+isin
−
4
8=29
cos
9
−
6
+isin
9
−
6
2
cos
3+isin
3
2
8
cos
8
−
4
+isin
8
−
4
=29
cos
−23
+isin
−23
2
cos
3+isin
3
221
8
cos
−2
+isin
−2
=29
cos
−23
+isin
−23
2
cos
3+isin
3
24
1+i
0
=292
24
cos
3−
−23
+isin
3−
−23
=2925
cos
3+23
+isin
3+23
=124
cos611
+isin611
=116
cos612
−
+isin612
−
=116
cos
2
−
6
+isin
2
−
6
=116
cos
−
6
+isin
−
6
=116
2
3−i2
=132
3−i