Lösung 3.1:1e

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A suitable first step can be to work out the square term, <math>(2-i)^2</math>, by expanding it,
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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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{{Displayed math||<math>\begin{align}
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After that, we calculate the remaining product:
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(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt]
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&= 4-4i+i^2\\[5pt]
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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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&= 4-4i-1\\[5pt]
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&=3-4i+3i-4i^2\\
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&= 3-4i\,\textrm{.}
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&=3+(-4+3)i-4\cdot (-1)\\
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\end{align}</math>}}
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&=3-i+4\\
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&=7-i.\end{align}</math>
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After that, we calculate the remaining product,
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt]
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&= 3-4i+3i-4i^2\\[5pt]
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&= 3+(-4+3)i-4\cdot (-1)\\[5pt]
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&= 3-i+4\\[5pt]
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&= 7-i\,\textrm{.}
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\end{align}</math>}}

Version vom 15:00, 29. Okt. 2008

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

\displaystyle \begin{align}

(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt] &= 4-4i+i^2\\[5pt] &= 4-4i-1\\[5pt] &= 3-4i\,\textrm{.} \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\\[5pt] &= 3-4i+3i-4i^2\\[5pt] &= 3+(-4+3)i-4\cdot (-1)\\[5pt] &= 3-i+4\\[5pt] &= 7-i\,\textrm{.} \end{align}

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