Lösung 3.3:2c

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{{NAVCONTENT_START}}
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We write
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<center> [[Image:3_3_2c-1(2).gif]] </center>
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<math>z\text{ }</math>
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{{NAVCONTENT_STOP}}
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and the right-hand side
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{{NAVCONTENT_START}}
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<math>\text{-1-}i</math>
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<center> [[Image:3_3_2c-2(2).gif]] </center>
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in polar form
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& z=r\left( \cos \alpha +i\sin \alpha \right) \\
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& \text{-1-}i=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right) \\
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\end{align}</math>
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Using de Moivre's formula, the equation can now be written as
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<math>r^{5}\left( \cos 5\alpha +i\sin 5\alpha \right)=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right)</math>
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If we identify the magnitude and argument on both sides, we get
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<math>\left\{ \begin{array}{*{35}l}
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r^{5}=\sqrt{2} \\
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5\alpha =\frac{5\pi }{4}+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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(The arguments
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<math>5\alpha </math>
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and
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<math>\frac{5\pi }{4}</math>
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can differ by a multiple of
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<math>2\pi </math>
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and still correspond to the same complex number.)
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This gives that
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<math>\left\{ \begin{array}{*{35}l}
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r=\sqrt[5]{2}=\left( 2^{{1}/{2}\;} \right)^{{1}/{5}\;}=2^{{1}/{10}\;} \\
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\alpha =\frac{1}{5}\left( \frac{5\pi }{4}+2n\pi \right)=\frac{\pi }{4}+\frac{2n\pi }{5}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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If we investigate the argument
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<math>\alpha </math>
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more closely, we see that it assumes essentially only five different values,
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<math>\frac{\pi }{4},\ \frac{\pi }{4}+\frac{2\pi }{5},\ \frac{\pi }{4}+\frac{4\pi }{5},\ \frac{\pi }{4}+\frac{6\pi }{5}</math>
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and
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<math>\ \frac{\pi }{4}+\frac{8\pi }{5}</math>
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since these angle values then repeat to within a multiple of
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<math>2\pi </math>.
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In summary, the roots of the equation are
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<math>z=2^{{1}/{10}\;}\left( \cos \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right)+i\sin \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right) \right)</math>
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for
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<math>n=0,\ 1,\ 2,\ 3</math>
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and
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<math>4</math>
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[[Image:3_3_2_c.gif|center]]
[[Image:3_3_2_c.gif|center]]

Version vom 10:27, 24. Okt. 2008

We write \displaystyle z\text{ } and the right-hand side \displaystyle \text{-1-}i in polar form


\displaystyle \begin{align} & z=r\left( \cos \alpha +i\sin \alpha \right) \\ & \text{-1-}i=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right) \\ \end{align}


Using de Moivre's formula, the equation can now be written as


\displaystyle r^{5}\left( \cos 5\alpha +i\sin 5\alpha \right)=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right)


If we identify the magnitude and argument on both sides, we get


\displaystyle \left\{ \begin{array}{*{35}l} r^{5}=\sqrt{2} \\ 5\alpha =\frac{5\pi }{4}+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.


(The arguments \displaystyle 5\alpha and \displaystyle \frac{5\pi }{4} can differ by a multiple of \displaystyle 2\pi and still correspond to the same complex number.)

This gives that


\displaystyle \left\{ \begin{array}{*{35}l} r=\sqrt[5]{2}=\left( 2^{{1}/{2}\;} \right)^{{1}/{5}\;}=2^{{1}/{10}\;} \\ \alpha =\frac{1}{5}\left( \frac{5\pi }{4}+2n\pi \right)=\frac{\pi }{4}+\frac{2n\pi }{5}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.


If we investigate the argument \displaystyle \alpha more closely, we see that it assumes essentially only five different values,


\displaystyle \frac{\pi }{4},\ \frac{\pi }{4}+\frac{2\pi }{5},\ \frac{\pi }{4}+\frac{4\pi }{5},\ \frac{\pi }{4}+\frac{6\pi }{5} and \displaystyle \ \frac{\pi }{4}+\frac{8\pi }{5}


since these angle values then repeat to within a multiple of \displaystyle 2\pi .

In summary, the roots of the equation are


\displaystyle z=2^{{1}/{10}\;}\left( \cos \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right)+i\sin \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right) \right)


for \displaystyle n=0,\ 1,\ 2,\ 3 and \displaystyle 4



Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:2c“