Lösung 3.1:2d
Aus Online Mathematik Brückenkurs 2
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Version vom 10:31, 11. Mär. 2009
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} |
