Lösung 3.2:5c

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Geometrically, the multiplication of two complex numbers means that there magnitudes are multiplied and their arguments are added. The product
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<center> [[Image:3_2_5c-1(2).gif]] </center>
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<math>\left( \sqrt{3}+i \right)\left( 1-i \right)</math>
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therefore has an argument which is the sum of the argument for the
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<math>\sqrt{3}+i</math>
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<center> [[Image:3_2_5c-2(2).gif]] </center>
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and
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<math>1-i</math>, i.e.
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<math>\arg \left( \left( \sqrt{3}+i \right)\left( 1-i \right) \right)=\arg \left( \sqrt{3}+i \right)+\arg \left( 1-i \right)</math>
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By drawing the factors in the complex plane, we can determine relatively easily the argument using simple trigonometry:
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[[Image:3_2_5_c.gif|center]]
[[Image:3_2_5_c.gif|center]]
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(Because
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<math>1-i</math>
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lies in the fourth quadrant, the argument equals
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<math>-\beta </math>
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and not
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<math>\beta </math>.)
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Hence,
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<math>\begin{align}
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& \arg \left( \left( \sqrt{3}+i \right)\left( 1-i \right) \right)=\arg \left( \sqrt{3}+i \right)+\arg \left( 1-i \right) \\
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& =\frac{\pi }{6}-\frac{\pi }{4}=-\frac{\pi }{12} \\
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\end{align}</math>
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NOTE: if you prefer to give the argument between
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<math>0</math>
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and
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<math>2\pi </math>, then the answer is
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<math>-\frac{\pi }{12}+2\pi =\frac{-\pi +24\pi }{12}=\frac{23\pi }{12}</math>

Version vom 08:55, 23. Okt. 2008

Geometrically, the multiplication of two complex numbers means that there magnitudes are multiplied and their arguments are added. The product \displaystyle \left( \sqrt{3}+i \right)\left( 1-i \right) therefore has an argument which is the sum of the argument for the \displaystyle \sqrt{3}+i and \displaystyle 1-i, i.e.


\displaystyle \arg \left( \left( \sqrt{3}+i \right)\left( 1-i \right) \right)=\arg \left( \sqrt{3}+i \right)+\arg \left( 1-i \right)


By drawing the factors in the complex plane, we can determine relatively easily the argument using simple trigonometry:



(Because \displaystyle 1-i lies in the fourth quadrant, the argument equals \displaystyle -\beta and not \displaystyle \beta .)

Hence,


\displaystyle \begin{align} & \arg \left( \left( \sqrt{3}+i \right)\left( 1-i \right) \right)=\arg \left( \sqrt{3}+i \right)+\arg \left( 1-i \right) \\ & =\frac{\pi }{6}-\frac{\pi }{4}=-\frac{\pi }{12} \\ \end{align}


NOTE: if you prefer to give the argument between \displaystyle 0 and \displaystyle 2\pi , then the answer is


\displaystyle -\frac{\pi }{12}+2\pi =\frac{-\pi +24\pi }{12}=\frac{23\pi }{12}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:5c“