Lösung 3.3:2b

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The The equation
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The equation <math>z^3=-1</math> is a so-called binomial equation, which we solve by writing both sides in polar form. We have
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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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<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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{{Displayed math||<math>\begin{align}
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z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
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-1 &= 1\,(\cos\pi + i\sin\pi)\,,
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\end{align}</math>}}
and, with the help of de Moivre's formula, the equation becomes
and, with the help of de Moivre's formula, the equation becomes
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{{Displayed math||<math>r^3(\cos 3\alpha + i\sin 3\alpha) = 1\,(\cos\pi + i\sin\pi)\,\textrm{.}</math>}}
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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 <math>2\pi</math>,
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{{Displayed math||<math>\left\{\begin{align}
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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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r^3 &= 1\,,\\[5pt]
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<math>2\pi </math>,
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3\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),}
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\end{align}\right.</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
which gives that
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{{Displayed math||<math>\left\{\begin{align}
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r &= 1\,\\[5pt]
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\alpha &= \frac{\pi}{3}+\frac{2n\pi}{3}\quad\text{(n is an arbitrary integer).}
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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For every third integer <math>n</math>, the solution formula gives in principal the same value for the argument (the difference is a multiple of <math>2\pi</math>), so the equation has in reality three solutions (for <math>n=0</math>, <math>1</math> and <math>\text{2}</math>),
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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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{{Displayed math||<math>z=\left\{\begin{align}
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&1\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt]
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&1\cdot \Bigl(\cos\pi + i\sin\pi\Bigr)\\[5pt]
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&1\cdot \Bigl(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\Bigr)
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\end{align}\right.
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=
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\left\{\begin{align}
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&\frac{1+i\sqrt{3}}{2}\,,\\[5pt]
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&-1\vphantom{\bigl(}\,,\\[5pt]
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&\frac{1-i\sqrt{3}}{2}\,\textrm{.}
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\end{align} \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 3.
[[Image:3_3_2_b.gif|center]]
[[Image:3_3_2_b.gif|center]]

Version vom 12:13, 30. Okt. 2008

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

\displaystyle \begin{align}

z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] -1 &= 1\,(\cos\pi + i\sin\pi)\,, \end{align}

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

\displaystyle r^3(\cos 3\alpha + i\sin 3\alpha) = 1\,(\cos\pi + i\sin\pi)\,\textrm{.}

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

\displaystyle \left\{\begin{align}

r^3 &= 1\,,\\[5pt] 3\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),} \end{align}\right.

which gives that

\displaystyle \left\{\begin{align}

r &= 1\,\\[5pt] \alpha &= \frac{\pi}{3}+\frac{2n\pi}{3}\quad\text{(n is an arbitrary integer).} \end{align}\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, \displaystyle 1 and \displaystyle \text{2}),

\displaystyle z=\left\{\begin{align}

&1\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt] &1\cdot \Bigl(\cos\pi + i\sin\pi\Bigr)\\[5pt] &1\cdot \Bigl(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\Bigr) \end{align}\right. = \left\{\begin{align} &\frac{1+i\sqrt{3}}{2}\,,\\[5pt] &-1\vphantom{\bigl(}\,,\\[5pt] &\frac{1-i\sqrt{3}}{2}\,\textrm{.} \end{align} \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 3.

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