Lösung 3.1:2a

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{{NAVCONTENT_START}}
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A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,
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A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator:
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<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\frac{1-i}{1-i}.</math>
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{{Displayed math||<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}
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Then, the conjugate rule gives that the new denominator is a real number
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<math>\begin{align}\frac{3-2i}{1+i}\frac{1-i}{1-i}&=\frac{(3-2i)(1-i)}{(1+i)(1-i)}\\
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Then, the formula for the difference of two squares gives that the new denominator is a real number,
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&=\frac{(3-2i)(1-i)}{1^2-i^2}\\
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&=\frac{(3-2i)(1-i)}{1+1}\\
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&=\frac{(3-2i)(1-i)}{2}\end{align}</math>
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{{NAVCONTENT_STEP}}
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All that remains is to multiply together what is in the numerator:
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{{Displayed math||<math>\begin{align}
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\frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}
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&= \frac{(3-2i)(1-i)}{(1+i)(1-i)}\\[5pt]
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&= \frac{(3-2i)(1-i)}{1^2-i^2}\\[5pt]
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&= \frac{(3-2i)(1-i)}{1+1}\\[5pt]
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&= \frac{(3-2i)(1-i)}{2}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}\frac{(3-2i)(1-i)}{2}&=\frac{3\cdot 1 -3\cdot i - 1\cdot 2i -2i\cdot(-i)}{2}\\
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All that remains is to multiply together what is in the numerator,
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&=\frac{3-3i-2i+2i^2}{2}\\
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&=\frac{3-(3+2)i+2(-1)}{2}\\
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&=\frac{1-5i}{2}\\
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&=\frac{1}{2}-\frac{5}{2}i.\end{align}</math>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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\frac{(3-2i)(1-i)}{2}
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&= \frac{3\cdot 1 -3\cdot i - 2i\cdot 1 - 2i\cdot(-i)}{2}\\[5pt]
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&= \frac{3-3i-2i+2i^2}{2}\\[5pt]
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&= \frac{3-(3+2)i+2\cdot (-1)}{2}\\[5pt]
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&= \frac{1-5i}{2}\\[5pt]
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&= \frac{1}{2}-\frac{5}{2}\,i\,\textrm{.}
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\end{align}</math>}}

Version vom 15:15, 29. Okt. 2008

A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,

\displaystyle \frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}

Then, the formula for the difference of two squares gives that the new denominator is a real number,

\displaystyle \begin{align}

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

All that remains is to multiply together what is in the numerator,

\displaystyle \begin{align}

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

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