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Lösung 3.3:2b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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
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
-
{{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\pi + i\sin\pi)\,,
-1 &= 1\,(\cos\pi + i\sin\pi)\,,
Zeile 8: Zeile 8:
and, with the help of de Moivre's formula, the equation becomes
and, with the help of de Moivre's formula, the equation becomes
-
{{Displayed math||<math>r^3(\cos 3\alpha + i\sin 3\alpha) = 1\,(\cos\pi + i\sin\pi)\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>r^3(\cos 3\alpha + i\sin 3\alpha) = 1\,(\cos\pi + i\sin\pi)\,\textrm{.}</math>}}
Both sides are equal when their magnitudes are equal and the arguments differ by a multiple of <math>2\pi</math>,
Both sides are equal when their magnitudes are equal and the arguments differ by a multiple of <math>2\pi</math>,
-
{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r^3 &= 1\,,\\[5pt]
r^3 &= 1\,,\\[5pt]
3\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),}
3\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),}
Zeile 19: Zeile 19:
which gives that
which gives 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{\pi}{3}+\frac{2n\pi}{3}\quad\text{(n is an arbitrary integer).}
\alpha &= \frac{\pi}{3}+\frac{2n\pi}{3}\quad\text{(n is an arbitrary integer).}
Zeile 26: Zeile 26:
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>),
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>),
-
{{Displayed math||<math>z=\left\{\begin{align}
+
{{Abgesetzte Formel||<math>z=\left\{\begin{align}
&1\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt]
&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\pi + i\sin\pi\Bigr)\\[5pt]

Version vom 13:11, 10. Mär. 2009

The equation z3=−1 is a so-called binomial equation, which we solve by writing both sides in polar form. We have

z−1=r(cos+isin)=1(cos+isin)

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

r3(cos3+isin3)=1(cos+isin).

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

r33=1=+2n(n is an arbitrary integer), 

which gives that

r=1=3+32n(n is an arbitrary integer).

For every third integer n, the solution formula gives in principal the same value for the argument (the difference is a multiple of 2), so the equation has in reality three solutions (for n=0, 1 and 2),

z=1cos3+isin31cos+isin1cos35+isin35=21+i3−121−i3.

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.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:2b“