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Lösung 3.1:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)

Wechseln zu: Navigation, Suche

Version vom 10:31, 11. Mär. 2009

We start by expanding the quadratic in the numerator,

1+i

3(2−i

3)2=1+i

322−2

3+(i

3)2=1+i

34−4

3i−3=1+i

31−4

3i.

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

1+i

31−4

3i=(1+i

3)(1−i

3)(1−4

3i)(1−i

3)=12−(i

3)21

1−1

3−4

1+4

3=1+(

3)21−i

3−4

3i+4(

3)2i2=1+31−(

3+4

3)i−4

3=41−12−(1+4)

3i=−411−45

3i.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:2c“

1. Differentialrechnung

2. Integralrechnung

3. Komplexe Zahlen

Suche

Lösung 3.1:2c

Aus Online Mathematik Brückenkurs 2

Version vom 10:31, 11. Mär. 2009