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 \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}