Lösung 3.3:1e
Aus Online Mathematik Brückenkurs 2
K (Lösning 3.3:1e moved to Solution 3.3:1e: Robot: moved page) |
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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 |
| - | + | <math>1+i\sqrt{3}</math>, | |
| - | < | + | <math>\text{1}-i</math> |
| - | + | and | |
| + | <math>\sqrt{3}-i</math> | ||
| + | in polar form. | ||
[[Image:3_3_1_e.gif]] [[Image:3_3_1_e_text.gif]] | [[Image:3_3_1_e.gif]] [[Image:3_3_1_e_text.gif]] | ||
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| + | This shows that | ||
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| + | |||
| + | <math>\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}</math> | ||
| + | |||
| + | |||
| + | Now, with the help of de Moivre's formula, | ||
| + | |||
| + | |||
| + | <math>\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}</math> | ||
Version vom 08:21, 24. Okt. 2008
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.
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}
