Lösung 3.1:2d

Aus Online Mathematik Brückenkurs 2

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Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
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Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator:
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{{Displayed math||<math>\begin{align}
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5-\frac{1}{1+i}
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&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
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= \frac{5+5i-1}{1+i}
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= \frac{4+5i}{1+i}\,,\\[5pt]
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3i+\frac{i}{2-3i}
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&= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}
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= \frac{6i-9i^2+i}{2-3i}
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= \frac{9+7i}{2-3i}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}5-\frac{1}{1+i}&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i} = \frac{5+5i-1}{1+i} = \frac{4+5i}{1+i},\\
 
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3i+\frac{i}{2-3i} &= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}=\frac{6i-9i^2+i}{2-3i}=\frac{9+7i}{2-3i}.\end{align}</math>
 
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Hence,
Hence,
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{{Displayed math||<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}
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<math>\frac{5-\frac{1}{1+i}}{3i+\frac{i}{2-3i}}=\frac{\frac{4+5i}{1+i}}{\frac{9+7i}{2-3i}}=\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}.</math>
 
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We multiply out the numerator and denominator
We multiply out the numerator and denominator
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{{Displayed math||<math>\begin{align}
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\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}
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&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
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&= \frac{8-12i+10i+15}{9+9i+7i-7}\\[5pt]
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&= \frac{23-2i}{2+16i}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\
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This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,
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&=\frac{8-12i+10i+15}{9+9i+7i-7}\\
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&=\frac{23-2i}{2+16i}.\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\frac{23-2i}{2+16i}
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&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
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&= \frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\[5pt]
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&= \frac{46-368i-4i-32}{4+256}\\[5pt]
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&= \frac{14-372i}{260}\,\textrm{.}
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\end{align}</math>}}
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{{NAVCONTENT_STEP}}
 
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This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator:
 
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<math>\begin{align}\frac{23-2i}{2+16i}&=\frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\
 
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&=\frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\
 
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&=\frac{46-368i-4i-32}{4+256}\\
 
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&=\frac{14-372i}{260}\end{align}</math>
 
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{{NAVCONTENT_STEP}}
 
If we divide up the numbers into factors,
If we divide up the numbers into factors,
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{{Displayed math||<math>\begin{align}
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14 &= 2\cdot 7\,,\\[5pt]
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372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
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260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\,,
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\end{align}</math>}}
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<math>\begin{align}14 &= 2\cdot 7\\
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we can simplify the answers
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372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\\
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260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\end{align}</math>
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we can simplify the answers:
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<math>\begin{align}\frac{14}{260}-\frac{372}{260}i&=\frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}i\\
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&=\frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}i\\
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&=\frac{7}{130}-\frac{93}{65}i\end{align}</math>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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\frac{14}{260}-\frac{372}{260}\,i
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&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]
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&= \frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}\,i\\[5pt]
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&= \frac{7}{130}-\frac{93}{65}\,i\,\textrm{.}
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\end{align}</math>}}

Version vom 15:41, 29. Okt. 2008

Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,

\displaystyle \begin{align}

5-\frac{1}{1+i} &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i} = \frac{5+5i-1}{1+i} = \frac{4+5i}{1+i}\,,\\[5pt] 3i+\frac{i}{2-3i} &= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i} = \frac{6i-9i^2+i}{2-3i} = \frac{9+7i}{2-3i}\,\textrm{.} \end{align}

Hence,

\displaystyle \frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}

We multiply out the numerator and denominator

\displaystyle \begin{align}

\frac{(4+5i)(2-3i)}{(9+7i)(1+i)} &= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt] &= \frac{8-12i+10i+15}{9+9i+7i-7}\\[5pt] &= \frac{23-2i}{2+16i}\,\textrm{.} \end{align}

This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,

\displaystyle \begin{align}

\frac{23-2i}{2+16i} &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt] &= \frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\[5pt] &= \frac{46-368i-4i-32}{4+256}\\[5pt] &= \frac{14-372i}{260}\,\textrm{.} \end{align}

If we divide up the numbers into factors,

\displaystyle \begin{align}

14 &= 2\cdot 7\,,\\[5pt] 372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt] 260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\,, \end{align}

we can simplify the answers

\displaystyle \begin{align}

\frac{14}{260}-\frac{372}{260}\,i &= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt] &= \frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}\,i\\[5pt] &= \frac{7}{130}-\frac{93}{65}\,i\,\textrm{.} \end{align}

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