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Lösung 3.1:2d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 3.1:2d moved to Solution 3.1:2d: Robot: moved page)
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<center> [[Image:3_1_2d-1(2).gif]] </center>
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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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<center> [[Image:3_1_2d-2(2).gif]] </center>
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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,
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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
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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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&=\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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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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If we divide up the numbers into factors,
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<math>\begin{align}14 &= 2\cdot 7\\
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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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Version vom 10:24, 23. Sep. 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:


5−11+i3i+i2−3i=1+i5(1+i)−11+i=1+i5+5i−1=1+i4+5i=2−3i3i(2−3i)+i2−3i=2−3i6i−9i2+i=2−3i9+7i


Hence,


5−11+i3i+i2−3i=2−3i9+7i1+i4+5i=(9+7i)(1+i)(4+5i)(2−3i)


We multiply out the numerator and denominator


(9+7i)(1+i)(4+5i)(2−3i)=91+9i+7i1+7ii42−43i+5i2−5i3i=9+9i+7i−78−12i+10i+15=2+16i23−2i


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


2+16i23−2i=(2+16i)(2−16i)(23−2i)(2−16i)=22−(16i)2232−2316i−2i2+2i16i=4+25646−368i−4i−32=26014−372i


If we divide up the numbers into factors,


14372260=27=2186=2293=22331=1026=25132


we can simplify the answers:


14260−260372i=2722513−2251322331i=72513−513331i=7130−6593i

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