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Lösung 3.1:1e

(Unterschied zwischen Versionen)

Version vom 10:29, 11. Mär. 2009

A suitable first step can be to work out the square term, (2−i)2, by expanding it,

(2−i)2=22−2

2i+i2=4−4i+i2=4−4i−1=3−4i.

After that, we calculate the remaining product,

(1+i)(3−4i)=1

3−1

4i+i

3−i

4i=3−4i+3i−4i2=3+(−4+3)i−4

(−1)=3−i+4=7−i.