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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Geometrically, the multiplication of two complex numbers means that there magnitudes are multiplied and their arguments are added. The product <math>(\sqrt{3}+i)(1-i)</math> therefore has an argument which is the sum of the argument for the <math>\sqrt{3}+i</math> and <math>1-i</math>, i.e.
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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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and
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<math>1-i</math>, i.e.
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{{Displayed math||<math>\arg \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i)\,\textrm{.}</math>}}
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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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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 <math>1-i</math> lies in the fourth quadrant, the argument equals
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(Because
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<math>-\beta</math> and not <math>\beta</math>.)
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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,
Hence,
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{{Displayed math||<math>\arg \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i) = \frac{\pi}{6} - \frac{\pi}{4} = -\frac{\pi}{12}\,\textrm{.}</math>}}
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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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Note: If you prefer to give the argument between <math>0</math> and <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>
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{{Displayed math||<math>-\frac{\pi}{12}+2\pi = \frac{-\pi+24\pi}{12} = \frac{23\pi}{12}\,\textrm{.}</math>}}

Version vom 12:39, 29. Okt. 2008

Geometrically, the multiplication of two complex numbers means that there magnitudes are multiplied and their arguments are added. The product \displaystyle (\sqrt{3}+i)(1-i) 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 \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i)\,\textrm{.}

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 \arg \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i) = \frac{\pi}{6} - \frac{\pi}{4} = -\frac{\pi}{12}\,\textrm{.}


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