Lösung 3.3:4a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.3:4a moved to Solution 3.3:4a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
This is a typical binomial equation which we solve in polar form.
-
<center> [[Image:3_3_4a.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
We write
 +
 
 +
 
 +
<math>\begin{align}
 +
& z=r\left( \cos \alpha +i\sin \alpha \right) \\
 +
& i=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right) \\
 +
\end{align}</math>
 +
 
 +
 
 +
and, on using de Moivre's formula, the equation becomes
 +
 
 +
 
 +
<math>r^{2}\left( \cos 2\alpha +i\sin 2\alpha \right)=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)</math>
 +
 
 +
 
 +
Both sides are equal when
 +
 
 +
 
 +
<math>\left\{ \begin{array}{*{35}l}
 +
r^{2}=1 \\
 +
2\alpha =\frac{\pi }{2}+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
 +
\end{array} \right.</math>
 +
 
 +
 
 +
which gives that
 +
 
 +
 
 +
<math>\left\{ \begin{array}{*{35}l}
 +
r=1 \\
 +
\alpha =\frac{\pi }{4}+n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
 +
\end{array} \right.</math>
 +
 
 +
 
 +
When
 +
<math>n=0</math>
 +
and
 +
<math>n=\text{1}</math>, we get two different arguments for
 +
<math>\alpha </math>, whilst different values of
 +
<math>n</math>
 +
only give these arguments plus/minus a multiple of
 +
<math>2\pi </math>.
 +
 
 +
The solutions to the equation are
 +
 
 +
 
 +
<math>z=\left\{ \begin{array}{*{35}l}
 +
\ 1\centerdot \left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\
 +
\ 1\centerdot \left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right) \\
 +
\end{array} \right.=\left\{ \begin{array}{*{35}l}
 +
\ \frac{1+i}{\sqrt{2}} \\
 +
\ -\frac{1+i}{\sqrt{2}} \\
 +
\end{array} \right.</math>

Version vom 09:37, 25. Okt. 2008

This is a typical binomial equation which we solve in polar form.

We write


\displaystyle \begin{align} & z=r\left( \cos \alpha +i\sin \alpha \right) \\ & i=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right) \\ \end{align}


and, on using de Moivre's formula, the equation becomes


\displaystyle r^{2}\left( \cos 2\alpha +i\sin 2\alpha \right)=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)


Both sides are equal when


\displaystyle \left\{ \begin{array}{*{35}l} r^{2}=1 \\ 2\alpha =\frac{\pi }{2}+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.


which gives that


\displaystyle \left\{ \begin{array}{*{35}l} r=1 \\ \alpha =\frac{\pi }{4}+n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\ \end{array} \right.


When \displaystyle n=0 and \displaystyle n=\text{1}, we get two different arguments for \displaystyle \alpha , whilst different values of \displaystyle n only give these arguments plus/minus a multiple of \displaystyle 2\pi .

The solutions to the equation are


\displaystyle z=\left\{ \begin{array}{*{35}l} \ 1\centerdot \left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\ \ 1\centerdot \left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right) \\ \end{array} \right.=\left\{ \begin{array}{*{35}l} \ \frac{1+i}{\sqrt{2}} \\ \ -\frac{1+i}{\sqrt{2}} \\ \end{array} \right.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:4a“