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Lösung 3.3:4a

Aus Online Mathematik Brückenkurs 2

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This is a typical binomial equation which we solve in polar form.
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<center> [[Image:3_3_4a.gif]] </center>
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We write
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<math>\begin{align}
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& z=r\left( \cos \alpha +i\sin \alpha \right) \\
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& i=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right) \\
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\end{align}</math>
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and, on using de Moivre's formula, the equation becomes
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<math>r^{2}\left( \cos 2\alpha +i\sin 2\alpha \right)=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)</math>
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Both sides are equal when
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<math>\left\{ \begin{array}{*{35}l}
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r^{2}=1 \\
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2\alpha =\frac{\pi }{2}+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 }{4}+n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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When
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<math>n=0</math>
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and
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<math>n=\text{1}</math>, we get two different arguments for
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<math>\alpha </math>, whilst different values of
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<math>n</math>
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only give these arguments plus/minus a multiple of
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<math>2\pi </math>.
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The solutions to the equation are
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<math>z=\left\{ \begin{array}{*{35}l}
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\ 1\centerdot \left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\
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\ 1\centerdot \left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right) \\
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\end{array} \right.=\left\{ \begin{array}{*{35}l}
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\ \frac{1+i}{\sqrt{2}} \\
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\ -\frac{1+i}{\sqrt{2}} \\
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\end{array} \right.</math>

Version vom 09:37, 25. Okt. 2008

This is a typical binomial equation which we solve in polar form.

We write


z=rcos+isini=1cos2+isin2 


and, on using de Moivre's formula, the equation becomes


r2cos2+isin2=1cos2+isin2 


Both sides are equal when


r2=12=2+2nn an arbitrary integer  


which gives that


r=1=4+nn an arbitrary integer  


When n=0 and n=1, we get two different arguments for , whilst different values of n only give these arguments plus/minus a multiple of 2.

The solutions to the equation are


z= 1cos4+isin4 1cos43+isin43= 21+i 21+i