Lösung 3.1:2a

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<center> [[Image:3_1_2a-1(2).gif]] </center>
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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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<center> [[Image:3_1_2a-2(2).gif]] </center>
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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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&=\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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All that remains is to multiply together what is in the numerator:
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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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&=\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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Version vom 09:39, 23. Sep. 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}\frac{1-i}{1-i}.

Then, the conjugate rule gives that the new denominator is a real number

\displaystyle \begin{align}\frac{3-2i}{1+i}\frac{1-i}{1-i}&=\frac{(3-2i)(1-i)}{(1+i)(1-i)}\\ &=\frac{(3-2i)(1-i)}{1^2-i^2}\\ &=\frac{(3-2i)(1-i)}{1+1}\\ &=\frac{(3-2i)(1-i)}{2}\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 - 1\cdot 2i -2i\cdot(-i)}{2}\\ &=\frac{3-3i-2i+2i^2}{2}\\ &=\frac{3-(3+2)i+2(-1)}{2}\\ &=\frac{1-5i}{2}\\ &=\frac{1}{2}-\frac{5}{2}i.\end{align}

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