Lösung 3.3:2d
Aus Online Mathematik Brückenkurs 2
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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, | ||
| - | {{ | + | {{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 | ||
| - | {{ | + | {{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 | ||
| - | {{ | + | {{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>, | ||
| - | {{ | + | {{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 | ||
| - | {{ | + | {{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 | ||
| - | {{ | + | {{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, | ||
| - | {{ | + | {{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] | ||
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and the original variable z is | and the original variable z is | ||
| - | {{ | + | {{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 \displaystyle w=z-1 as a new unknown and move the term \displaystyle 4 over to the right-hand side, we have a binomial equation,
| \displaystyle w^4=-4\,\textrm{.} |
We can solve this equation in the usual way by using polar form and de Moivre's formula. We have
| \displaystyle \begin{align}
w &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] -4 &= 4(\cos\pi + i\sin\pi)\,, \end{align} |
and the equation becomes
| \displaystyle r^4(\cos 4\alpha + i\sin 4\alpha) = 4(\cos\pi + i\sin\pi)\,\textrm{.} |
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 \displaystyle 2\pi,
| \displaystyle \left\{\begin{align}
r^4 &= 4\,,\\[5pt] 4\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),} \end{align} \right. |
which gives us that
| \displaystyle \left\{\begin{align}
r &= \sqrt[4]{4} = \sqrt{2}\,,\\[5pt] \alpha &= \frac{\pi}{4}+\frac{n\pi}{2}\,,\quad\text{(n is an arbitrary integer).} \end{align}\right. |
For \displaystyle n=0, \displaystyle 2 and \displaystyle 3, the argument \displaystyle \alpha assumes the four different values
| \displaystyle \frac{\pi}{4}, \displaystyle \quad\frac{3\pi}{4}, \displaystyle \quad\frac{5\pi}{4}\quadand\displaystyle \quad\frac{7\pi}{4}\,, |
and for other values of \displaystyle n we obtain values of \displaystyle \alpha which are equal to those above, apart from multiples of \displaystyle 2\pi. Thus, we have four solutions,
| \displaystyle w=\left\{\begin{align}
&\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{5\pi}{4}+i\sin\frac{5\pi}{4}\Bigr)\\[5pt] &\sqrt{2}\Bigl(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\Bigr) \end{align}\right. = \left\{\begin{align} 1+i\,,&\\[5pt] -1+i\,,&\\[5pt] -1-i\,,&\\[5pt] 1-i\,\textrm{,} \end{align}\right. |
and the original variable z is
| \displaystyle z=\left\{\begin{align}
&2+i\,,\\[5pt] &i\,,\\[5pt] &-i\,,\\[5pt] &2-i\,\textrm{.} \end{align}\right. |
