Lösung 3.3:1e

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One part of the quotient has rather high exponents and this indicates that we ought to use polar form for the calculation.

First, we write \displaystyle 1+i\sqrt{3}, \displaystyle \text{1}-i and \displaystyle \sqrt{3}-i in polar form.


Image:3_3_1_e.gif Image:3_3_1_e_text.gif

This shows that


\displaystyle \begin{align} & 1+i\sqrt{3}=2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\ & \text{1}-i=\sqrt{2}\left( \cos \left( -\frac{\pi }{4} \right)+i\sin \left( -\frac{\pi }{4} \right) \right) \\ & \sqrt{3}-i=2\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right) \\ \end{align}


Now, with the help of de Moivre's formula,


\displaystyle \begin{align} & \frac{\left( 1+i\sqrt{3} \right)\left( \text{1}-i \right)^{8}}{\left( \sqrt{3}-i \right)^{9}}=\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\left( \sqrt{2}\left( \cos \left( -\frac{\pi }{4} \right)+i\sin \left( -\frac{\pi }{4} \right) \right) \right)^{8}}{\left( 2\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right) \right)^{9}} \\ & =\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\left( \sqrt{2} \right)^{8}\left( \cos \left( 8\centerdot \left( -\frac{\pi }{4} \right) \right)+i\sin \left( 8\centerdot \left( -\frac{\pi }{4} \right) \right) \right)}{2^{9}\left( \cos \left( 9\centerdot \left( -\frac{\pi }{6} \right) \right)+i\sin \left( 9\centerdot \left( -\frac{\pi }{6} \right) \right) \right)} \\ & =\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\centerdot 2^{\frac{1}{2}\centerdot 8}\left( \cos \left( -2\pi \right)+i\sin \left( -2\pi \right) \right)}{2^{9}\left( \cos \left( -\frac{3\pi }{2} \right)+i\sin \left( -\frac{3\pi }{2} \right) \right)} \\ & =\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\centerdot 2^{4}\left( 1+i\centerdot 0 \right)}{2^{9}\left( \cos \left( -\frac{3\pi }{2} \right)+i\sin \left( -\frac{3\pi }{2} \right) \right)} \\ & =\frac{2\centerdot 2^{4}}{2^{9}}\left( \cos \left( \frac{\pi }{3}-\left( -\frac{3\pi }{2} \right) \right)+i\sin \left( \frac{\pi }{3}-\left( -\frac{3\pi }{2} \right) \right) \right) \\ & =\frac{2^{5}}{2^{9}}\left( \cos \left( \frac{\pi }{3}+\frac{3\pi }{2} \right)+i\sin \left( \frac{\pi }{3}+\frac{3\pi }{2} \right) \right) \\ & =\frac{1}{2^{4}}\left( \cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6} \right) \\ & =\frac{1}{16}\left( \cos \frac{12\pi -\pi }{6}+i\sin \frac{12\pi -\pi }{6} \right) \\ & =\frac{1}{16}\left( \cos \left( 2\pi -\frac{\pi }{6} \right)+i\sin \left( 2\pi -\frac{\pi }{6} \right) \right) \\ & =\frac{1}{16}\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right) \\ & =\frac{1}{16}\left( \frac{\sqrt{3}}{2}-\frac{i}{2} \right)=\frac{1}{32}\left( \sqrt{3}-i \right) \\ \end{align}