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Lösung 3.3:2a

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 3: Zeile 3:
We start by writing <math>z</math> and <math>1</math> in polar form
We start by writing <math>z</math> and <math>1</math> in polar form
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
1 &= 1(\cos 0 + i\sin 0)\,\textrm{.}
1 &= 1(\cos 0 + i\sin 0)\,\textrm{.}
Zeile 10: Zeile 10:
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>}}
+
{{Abgesetzte Formel||<math>r^4(\cos 4\alpha + i\sin 4\alpha) = 1\,(\cos 0 + i\sin 0)\,,</math>}}
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.
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.
-
{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r^{4} &= 1\,,\\[5pt]
r^{4} &= 1\,,\\[5pt]
4\alpha &= 0+2n\pi\,,\quad (\text{n is an arbitrary integer})\,\textrm{.}
4\alpha &= 0+2n\pi\,,\quad (\text{n is an arbitrary integer})\,\textrm{.}
Zeile 21: Zeile 21:
This means that
This means that
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{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r &= 1\,,\\[5pt]
r &= 1\,,\\[5pt]
\alpha &= \frac{n\pi}{2}\,,\quad \text{(n is an arbitrary integer).}
\alpha &= \frac{n\pi}{2}\,,\quad \text{(n is an arbitrary integer).}
Zeile 28: Zeile 28:
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>}}
+
{{Abgesetzte Formel||<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>}}
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>.
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>.
Zeile 34: Zeile 34:
The equation's solutions are therefore
The equation's solutions are therefore
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{{Displayed math||<math>z=\left\{\begin{align}
+
{{Abgesetzte Formel||<math>z=\left\{\begin{align}
&1\cdot(\cos 0 + i\sin 0)\,,\\[5pt]
&1\cdot(\cos 0 + i\sin 0)\,,\\[5pt]
&1\cdot(\cos (\pi/2) + i\sin (\pi/2))\,,\\[5pt]
&1\cdot(\cos (\pi/2) + i\sin (\pi/2))\,,\\[5pt]

Version vom 13:11, 10. Mär. 2009

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“