Lösung 3.1:1e

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<center> [[Image:3_1_1e.gif]] </center>
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A suitable first step can be to work out the square term, <math>(2-i)^2</math>, with the help of the square rule:
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<math>\begin{align}(2-i)^2&=2^2-2\cdot 2i + i^2=4-4i+i^2\\
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&=4-4i-1=3-4i\end{align}</math>
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After that, we calculate the remaining product:
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<math>\begin{align}(1+i)(3-4i)&=1\cdot3-1\cdot 4i +i \cdot 3 - i\cdot 4i\\
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&=3-4i+3i-4i^2\\
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&=3+(-4+3)i-4\cdot (-1)\\
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&=3-i+4\\
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&=7-i.\end{align}</math>
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Version vom 11:57, 18. Sep. 2008

A suitable first step can be to work out the square term, \displaystyle (2-i)^2, with the help of the square rule:

\displaystyle \begin{align}(2-i)^2&=2^2-2\cdot 2i + i^2=4-4i+i^2\\ &=4-4i-1=3-4i\end{align}

After that, we calculate the remaining product:

\displaystyle \begin{align}(1+i)(3-4i)&=1\cdot3-1\cdot 4i +i \cdot 3 - i\cdot 4i\\ &=3-4i+3i-4i^2\\ &=3+(-4+3)i-4\cdot (-1)\\ &=3-i+4\\ &=7-i.\end{align}

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