Lösung 3.3:2b

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The The equation
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<center> [[Image:3_3_2b-1(2).gif]] </center>
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<math>z^{3}=-\text{1 }</math>
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is a so-called binomial equation, which we solve by writing both sides in polar form. We have
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<center> [[Image:3_3_2b-2(2).gif]] </center>
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<math>\begin{align}
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& z=r\left( \cos \alpha +i\sin \alpha \right) \\
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& -1=1\left( \cos \pi +i\sin \pi \right) \\
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\end{align}</math>
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and, with the help of de Moivre's formula, the equation becomes
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<math>r^{3}\left( \cos 3\alpha +i\sin 3\alpha \right)=1\left( \cos \pi +i\sin \pi \right)</math>
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Both sides are equal when their magnitudes are equal and the arguments differ by a multiple of
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<math>2\pi </math>,
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<math>\left\{ \begin{array}{*{35}l}
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r^{3}=1 \\
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3\alpha =\pi +2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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which gives that
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<math>\left\{ \begin{array}{*{35}l}
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r=1 \\
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\alpha =\frac{\pi }{3}+\frac{2n\pi }{3}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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For every third integer
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<math>n</math>, the solution formula gives in principal the same value for the argument (the difference is a multiple of
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<math>2\pi </math>), so the equation has in reality three solutions (for
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<math>n=0,\ 1</math>
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and
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<math>\text{2}</math>):
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<math>z=\left\{ \begin{array}{*{35}l}
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1\centerdot \left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\
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1\centerdot \left( \cos \pi +i\sin \pi \right) \\
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1\centerdot \left( \cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \right) \\
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\end{array} \right.=\left\{ \begin{array}{*{35}l}
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\frac{1+i\sqrt{3}}{2} \\
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-1 \\
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\frac{1-i\sqrt{3}}{2} \\
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\end{array} \right.</math>
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We obtain the typical behaviour that the solutions are corner points in a regular polygon (a triangle in this case because the degree of the equation is
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<math>\text{3}</math>).
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[[Image:3_3_2_b.gif|center]]
[[Image:3_3_2_b.gif|center]]

Version vom 10:01, 24. Okt. 2008

The The equation \displaystyle z^{3}=-\text{1 } is a so-called binomial equation, which we solve by writing both sides in polar form. We have


\displaystyle \begin{align} & z=r\left( \cos \alpha +i\sin \alpha \right) \\ & -1=1\left( \cos \pi +i\sin \pi \right) \\ \end{align}


and, with the help of de Moivre's formula, the equation becomes


\displaystyle r^{3}\left( \cos 3\alpha +i\sin 3\alpha \right)=1\left( \cos \pi +i\sin \pi \right)


Both sides are equal when their magnitudes are equal and the arguments differ by a multiple of \displaystyle 2\pi ,


\displaystyle \left\{ \begin{array}{*{35}l} r^{3}=1 \\ 3\alpha =\pi +2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.

which gives that


\displaystyle \left\{ \begin{array}{*{35}l} r=1 \\ \alpha =\frac{\pi }{3}+\frac{2n\pi }{3}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.


For every third integer \displaystyle n, the solution formula gives in principal the same value for the argument (the difference is a multiple of \displaystyle 2\pi ), so the equation has in reality three solutions (for \displaystyle n=0,\ 1 and \displaystyle \text{2}):


\displaystyle z=\left\{ \begin{array}{*{35}l} 1\centerdot \left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\ 1\centerdot \left( \cos \pi +i\sin \pi \right) \\ 1\centerdot \left( \cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \right) \\ \end{array} \right.=\left\{ \begin{array}{*{35}l} \frac{1+i\sqrt{3}}{2} \\ -1 \\ \frac{1-i\sqrt{3}}{2} \\ \end{array} \right.

We obtain the typical behaviour that the solutions are corner points in a regular polygon (a triangle in this case because the degree of the equation is \displaystyle \text{3}).


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