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Lösung 3.3:2d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 3.3:2d moved to Solution 3.3:2d: Robot: moved page)
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{{NAVCONTENT_START}}
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If we use
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<center> [[Image:3_3_2d-1(2).gif]] </center>
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<math>w=z-\text{1}</math>
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{{NAVCONTENT_STOP}}
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as a new unknown and move the term
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{{NAVCONTENT_START}}
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<math>\text{4}</math>
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<center> [[Image:3_3_2d-2(2).gif]] </center>
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over to the right-hand side, we have a binomial equation,
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{{NAVCONTENT_STOP}}
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<math>w^{4}=-4</math>
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We can solve this equation in the usual way by using polar form and de Moivre's formula. We have
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<math>\begin{align}
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& w=r\left( \cos \alpha +i\sin \alpha \right) \\
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& -4=4\left( \cos \pi +i\sin \pi \right) \\
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\end{align}</math>
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and the equation becomes
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<math>r^{4}\left( \cos 4\alpha +i\sin 4\alpha \right)=4\left( \cos \pi +i\sin \pi \right)</math>
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The only way that both sides can be equal is if the magnitudes agree and the arguments do not differ by anything other than 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^{4}=4 \\
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4\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 us that
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<math>\left\{ \begin{array}{*{35}l}
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r=\sqrt[4]{2}=\sqrt{2} \\
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\alpha =\frac{\pi }{4}+\frac{n\pi }{2}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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for
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<math>n=0,\ 1,\ 2</math>
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and
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<math>3</math>, the argument
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<math>\alpha </math>
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assumes the four different values
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<math>\frac{\pi }{4},\ \frac{3\pi }{4},\ \frac{5\pi }{4}</math>
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and
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<math>\frac{7\pi }{4}</math>
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and for different values of
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<math>n</math>
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we obtain values of
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<math>\alpha </math>
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which are equal to those above, apart from multiples of
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<math>2\pi </math>. Thus, we have four solutions,
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<math>w=\left\{ \begin{array}{*{35}l}
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\sqrt{2}\left( \cos {\pi }/{4}\;+i\sin {\pi }/{4}\; \right) \\
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\sqrt{2}\left( \cos {3\pi }/{4}\;+i\sin 3{\pi }/{4}\; \right) \\
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\sqrt{2}\left( \cos {5\pi }/{4}\;+i\sin 5{\pi }/{4}\; \right) \\
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\sqrt{2}\left( \cos 7{\pi }/{4}\;+i\sin 7{\pi }/{4}\; \right) \\
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\end{array} \right.=\left\{ \begin{array}{*{35}l}
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1+i \\
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-1+i \\
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-1-i \\
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1-i \\
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\end{array} \right.</math>
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and the original variable z is
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<math>z=\left\{ \begin{array}{*{35}l}
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2+i \\
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i \\
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-i \\
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2-i \\
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\end{array} \right.</math>

Version vom 10:54, 24. Okt. 2008

If we use w=z−1 as a new unknown and move the term 4 over to the right-hand side, we have a binomial equation,


w4=−4


We can solve this equation in the usual way by using polar form and de Moivre's formula. We have


w=rcos+isin−4=4cos+isin 


and the equation becomes


r4cos4+isin4=4cos+isin 


The only way that both sides can be equal is if the magnitudes agree and the arguments do not differ by anything other than a multiple of 2,


r4=44=+2nn an arbitrary integer  


which gives us that


r=42=2=4+2nn an arbitrary integer  


for n=0 1 2 and 3, the argument assumes the four different values


4 43 45 and 47


and for different values of n we obtain values of which are equal to those above, apart from multiples of 2. Thus, we have four solutions,


w=2cos4+isin42cos34+isin342cos54+isin542cos74+isin74=1+i−1+i−1−i1−i


and the original variable z is


z=2+ii−i2−i

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