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Lösung 3.3:2a

Aus Online Mathematik Brückenkurs 2

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An equation of the type “
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An equation of the type "<math>z^{n} = \text{a complex number}</math>" is called a binomial equation and these are usually solved by going over to polar form and using de Moivre's formula.
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<math>z^{n}</math>
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= a complex number” is called a binomial equation and these are usually solved by going over to polar form and using de Moivre's formula.
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We start by writing
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We start by writing <math>z</math> and <math>1</math> in polar form
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<math>z\text{ }</math>
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and
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<math>\text{1}</math>
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in polar form
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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 0+i\sin 0 \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 0 + i\sin 0)\,\textrm{.}
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\end{align}</math>}}
The equation then becomes
The equation then becomes
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{{Displayed math||<math>r^4(\cos 4\alpha + i\sin 4\alpha) = 1\,(\cos 0 + i\sin 0)\,,</math>}}
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<math>r^{4}\left( \cos 4\alpha +i\sin 4\alpha \right)=1\left( \cos 0+i\sin 0 \right)</math>
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where we have used de Moivre's formula on the left-hand side. In order that both sides are equal, they must have the same magnitude and the same argument to within a multiple of <math>2\pi</math>, i.e.
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{{Displayed math||<math>\left\{\begin{align}
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where we have used de Moivre's formula on the left-hand side. In order that both sides are equal, they must have the same magnitude and the same argument to within a multiple of
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r^{4} &= 1\,,\\[5pt]
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<math>2\pi </math>, i.e.
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4\alpha &= 0+2n\pi\,,\quad (\text{n is an arbitrary integer})\,\textrm{.}
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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r^{4}=1 \\
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4\alpha =0+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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This means that
This means that
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{{Displayed math||<math>\left\{\begin{align}
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<math>\left\{ \begin{array}{*{35}l}
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r &= 1\,,\\[5pt]
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r=1 \\
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\alpha &= \frac{n\pi}{2}\,,\quad \text{(n is an arbitrary integer).}
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\alpha =\frac{n\pi }{2}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{align}\right.</math>}}
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\end{array} \right.</math>
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The solutions are thus (in polar form)
The solutions are thus (in polar form)
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{{Displayed math||<math>z = 1\cdot\Bigl(\cos\frac{n\pi}{2} + i\sin\frac{n\pi}{2}\Bigr)\,,\quad\text{for }n=0,\ \pm 1,\ \pm 2,\ldots</math>}}
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<math>z=1\centerdot \left( \cos \frac{n\pi }{2}+i\sin \frac{n\pi }{2} \right)</math>, for
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but observe that the argument on the right-hand side essentially takes only four different values <math>0</math>, <math>\pi/2</math>, <math>\pi</math> and <math>3\pi/2\,</math>, because other values of <math>n</math> give some of these values plus/minus a multiple of <math>2\pi\,</math>.
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<math>n=0,\ \pm 1,\ \pm 2,...</math>
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but observe that the argument on the right-hand side essentially takes only four different values
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<math>0,\ {\pi }/{2}\;,\ \pi </math>
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and
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<math>{3\pi }/{2}\;</math>, because other values of
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<math>n\text{ }</math>
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give some of these values plus/minus a multiple of
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<math>2\pi </math>.
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The equation's solutions are therefore
The equation's solutions are therefore
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{{Displayed math||<math>z=\left\{\begin{align}
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&1\cdot(\cos 0 + i\sin 0)\,,\\[5pt]
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&1\cdot(\cos (\pi/2) + i\sin (\pi/2))\,,\\[5pt]
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&1\cdot(\cos \pi + i\sin \pi)\,,\\[5pt]
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&1\cdot(\cos (3\pi/2) + i\sin (3\pi/2))\,,
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\end{align}\right.
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=
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\left\{ \begin{align}
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1\,,&\\[5pt]
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i\,,&\\[5pt]
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-1\,,&\\[5pt]
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-i\,\textrm{.}&
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\end{align}\right.</math>}}
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<math>z=\left\{ \begin{array}{*{35}l}
 
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1\centerdot \left( \cos 0+i\sin 0 \right) \\
 
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1\centerdot \left( \cos {\pi }/{2}\;+i\sin {\pi }/{2}\; \right) \\
 
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1\centerdot \left( \cos \pi +i\sin \pi \right) \\
 
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1\centerdot \left( \cos {3\pi }/{2}\;+i\sin {3\pi }/{2}\; \right) \\
 
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\end{array} \right.=\left\{ \begin{matrix}
 
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1 \\
 
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i \\
 
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-1 \\
 
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-i \\
 
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\end{matrix} \right.</math>
 
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NOTE: note that if we mark these solutions on the complex number plane, we see that they are corners in a regular quadrilateral.
 
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Note: If we mark these solutions on the complex number plane, we see that they are corners in a regular quadrilateral.
[[Image:3_3_2_a.gif|center]]
[[Image:3_3_2_a.gif|center]]

Version vom 11:58, 30. Okt. 2008

An equation of the type "zn=a complex number" is called a binomial equation and these are usually solved by going over to polar form and using de Moivre's formula.

We start by writing z and 1 in polar form

z1=r(cos+isin)=1(cos0+isin0).

The equation then becomes

r4(cos4+isin4)=1(cos0+isin0)

where we have used de Moivre's formula on the left-hand side. In order that both sides are equal, they must have the same magnitude and the same argument to within a multiple of 2, i.e.

r44=1=0+2n(n is an arbitrary integer). 

This means that

r=1=2n(n is an arbitrary integer).

The solutions are thus (in polar form)

z=1cos2n+isin2nfor n=0 1 2... 

but observe that the argument on the right-hand side essentially takes only four different values 0, 2, and 32, because other values of n give some of these values plus/minus a multiple of 2.

The equation's solutions are therefore

z=1(cos0+isin0)1(cos(2)+isin(2))1(cos+isin)1(cos(32)+isin(32))=1i−1−i.


Note: If we mark these solutions on the complex number plane, we see that they are corners in a regular quadrilateral.

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