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Lösung 3.3:2d

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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If we use <math>w=z-1</math> as a new unknown and move the term <math>4</math> over to the right-hand side, we have a binomial equation,
If we use <math>w=z-1</math> as a new unknown and move the term <math>4</math> over to the right-hand side, we have a binomial equation,
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{{Displayed math||<math>w^4=-4\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>w^4=-4\,\textrm{.}</math>}}
We can solve this equation in the usual way by using polar form and de Moivre's formula. We have
We can solve this equation in the usual way by using polar form and de Moivre's formula. We have
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
w &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
w &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
-4 &= 4(\cos\pi + i\sin\pi)\,,
-4 &= 4(\cos\pi + i\sin\pi)\,,
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and the equation becomes
and the equation becomes
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{{Displayed math||<math>r^4(\cos 4\alpha + i\sin 4\alpha) = 4(\cos\pi + i\sin\pi)\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>r^4(\cos 4\alpha + i\sin 4\alpha) = 4(\cos\pi + i\sin\pi)\,\textrm{.}</math>}}
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 <math>2\pi</math>,
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 <math>2\pi</math>,
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{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r^4 &= 4\,,\\[5pt]
r^4 &= 4\,,\\[5pt]
4\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),}
4\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),}
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which gives us that
which gives us that
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{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r &= \sqrt[4]{4} = \sqrt{2}\,,\\[5pt]
r &= \sqrt[4]{4} = \sqrt{2}\,,\\[5pt]
\alpha &= \frac{\pi}{4}+\frac{n\pi}{2}\,,\quad\text{(n is an arbitrary integer).}
\alpha &= \frac{\pi}{4}+\frac{n\pi}{2}\,,\quad\text{(n is an arbitrary integer).}
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For <math>n=0<math>, <math>1</math>, <math>2</math> and <math>3</math>, the argument <math>\alpha</math> assumes the four different values
For <math>n=0<math>, <math>1</math>, <math>2</math> and <math>3</math>, the argument <math>\alpha</math> assumes the four different values
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{{Displayed math||<math>\frac{\pi}{4}</math>, <math>\quad\frac{3\pi}{4}</math>, <math>\quad\frac{5\pi}{4}\quad</math>and<math>\quad\frac{7\pi}{4}\,,</math>}}
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{{Abgesetzte Formel||<math>\frac{\pi}{4}</math>, <math>\quad\frac{3\pi}{4}</math>, <math>\quad\frac{5\pi}{4}\quad</math>and<math>\quad\frac{7\pi}{4}\,,</math>}}
and for other values of <math>n</math> we obtain values of <math>\alpha</math> which are equal to those above, apart from multiples of <math>2\pi</math>. Thus, we have four solutions,
and for other values of <math>n</math> we obtain values of <math>\alpha</math> which are equal to those above, apart from multiples of <math>2\pi</math>. Thus, we have four solutions,
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{{Displayed math||<math>w=\left\{\begin{align}
+
{{Abgesetzte Formel||<math>w=\left\{\begin{align}
&\sqrt{2}\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt]
&\sqrt{2}\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt]
&\sqrt{2}\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr)\\[5pt]
&\sqrt{2}\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr)\\[5pt]
Zeile 50: Zeile 50:
and the original variable z is
and the original variable z is
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{{Displayed math||<math>z=\left\{\begin{align}
+
{{Abgesetzte Formel||<math>z=\left\{\begin{align}
&2+i\,,\\[5pt]
&2+i\,,\\[5pt]
&i\,,\\[5pt]
&i\,,\\[5pt]

Version vom 13:12, 10. Mär. 2009

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−4=r(cos+isin)=4(cos+isin)

and the equation becomes

r4(cos4+isin4)=4(cos+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,

r44=4=+2n(n is an arbitrary integer), 

which gives us that

r=44=2=4+2n(n is an arbitrary integer).

For n=01, 2 and 3, the argument assumes the four different values

4, 43, 45and47

and for other 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+isin42cos43+isin432cos45+isin452cos47+isin47=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“