Aus Online Mathematik Brückenkurs 2

Version vom 15:05, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.3:2d moved to Solution 3.3:2d: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 10:54, 24. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	If we use
-	<center> [[Image:3_3_2d-1(2).gif]] </center>	+	<math>w=z-\text{1}</math>
-	{{NAVCONTENT_STOP}}	+	as a new unknown and move the term
-	{{NAVCONTENT_START}}	+	<math>\text{4}</math>
-	<center> [[Image:3_3_2d-2(2).gif]] </center>	+	over to the right-hand side, we have a binomial equation,
-	{{NAVCONTENT_STOP}}	+
		+
		+	<math>w^{4}=-4</math>
		+
		+
		+	We can solve this equation in the usual way by using polar form and de Moivre's formula. We have
		+
		+
		+	<math>\begin{align}
		+	& w=r\left( \cos \alpha +i\sin \alpha \right) \\
		+	& -4=4\left( \cos \pi +i\sin \pi \right) \\
		+	\end{align}</math>
		+
		+
		+	and the equation becomes
		+
		+
		+	<math>r^{4}\left( \cos 4\alpha +i\sin 4\alpha \right)=4\left( \cos \pi +i\sin \pi \right)</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>,
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	r^{4}=4 \\
		+	4\alpha =\pi +2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
		+	\end{array} \right.</math>
		+
		+
		+	which gives us that
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	r=\sqrt[4]{2}=\sqrt{2} \\
		+	\alpha =\frac{\pi }{4}+\frac{n\pi }{2}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
		+	\end{array} \right.</math>
		+
		+
		+	for
		+	<math>n=0,\ 1,\ 2</math>
		+	and
		+	<math>3</math>, the argument
		+	<math>\alpha </math>
		+	assumes the four different values
		+
		+
		+	<math>\frac{\pi }{4},\ \frac{3\pi }{4},\ \frac{5\pi }{4}</math>
		+	and
		+	<math>\frac{7\pi }{4}</math>
		+
		+
		+	and for different 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,
		+
		+
		+	<math>w=\left\{ \begin{array}{*{35}l}
		+	\sqrt{2}\left( \cos {\pi }/{4}\;+i\sin {\pi }/{4}\; \right) \\
		+	\sqrt{2}\left( \cos {3\pi }/{4}\;+i\sin 3{\pi }/{4}\; \right) \\
		+	\sqrt{2}\left( \cos {5\pi }/{4}\;+i\sin 5{\pi }/{4}\; \right) \\
		+	\sqrt{2}\left( \cos 7{\pi }/{4}\;+i\sin 7{\pi }/{4}\; \right) \\
		+	\end{array} \right.=\left\{ \begin{array}{*{35}l}
		+	1+i \\
		+	-1+i \\
		+	-1-i \\
		+	1-i \\
		+	\end{array} \right.</math>
		+
		+
		+	and the original variable z is
		+
		+
		+	<math>z=\left\{ \begin{array}{*{35}l}
		+	2+i \\
		+	i \\
		+	-i \\
		+	2-i \\
		+	\end{array} \right.</math>

---

Version vom 10:54, 24. Okt. 2008

If we use \displaystyle w=z-\text{1} as a new unknown and move the term \displaystyle \text{4} over to the right-hand side, we have a binomial equation,

\displaystyle w^{4}=-4

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\left( \cos \alpha +i\sin \alpha \right) \\ & -4=4\left( \cos \pi +i\sin \pi \right) \\ \end{align}

and the equation becomes

\displaystyle r^{4}\left( \cos 4\alpha +i\sin 4\alpha \right)=4\left( \cos \pi +i\sin \pi \right)

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{array}{*{35}l} r^{4}=4 \\ 4\alpha =\pi +2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.

which gives us that

\displaystyle \left\{ \begin{array}{*{35}l} r=\sqrt[4]{2}=\sqrt{2} \\ \alpha =\frac{\pi }{4}+\frac{n\pi }{2}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.

for \displaystyle n=0,\ 1,\ 2 and \displaystyle 3, the argument \displaystyle \alpha assumes the four different values

\displaystyle \frac{\pi }{4},\ \frac{3\pi }{4},\ \frac{5\pi }{4} and \displaystyle \frac{7\pi }{4}

and for different 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{array}{*{35}l} \sqrt{2}\left( \cos {\pi }/{4}\;+i\sin {\pi }/{4}\; \right) \\ \sqrt{2}\left( \cos {3\pi }/{4}\;+i\sin 3{\pi }/{4}\; \right) \\ \sqrt{2}\left( \cos {5\pi }/{4}\;+i\sin 5{\pi }/{4}\; \right) \\ \sqrt{2}\left( \cos 7{\pi }/{4}\;+i\sin 7{\pi }/{4}\; \right) \\ \end{array} \right.=\left\{ \begin{array}{*{35}l} 1+i \\ -1+i \\ -1-i \\ 1-i \\ \end{array} \right.

and the original variable z is

\displaystyle z=\left\{ \begin{array}{*{35}l} 2+i \\ i \\ -i \\ 2-i \\ \end{array} \right.