Processing Math: Done
Lösung 3.3:4a
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 3.3:4a moved to Solution 3.3:4a: Robot: moved page) |
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- | {{ | + | This is a typical binomial equation which we solve in polar form. |
- | < | + | |
- | {{ | + | 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
cos
+isin
i=1
cos
2+isin
2
and, on using de Moivre's formula, the equation becomes
cos2
+isin2
=1
cos
2+isin
2
Both sides are equal when
r2=12
=
2+2n
n an arbitrary integer
which gives that
r=1
=
4+n
n an arbitrary integer
When
The solutions to the equation are
1
cos
4+isin
4
1
cos43
+isin43
=
21+i −
21+i