Lösung 3.3:4a

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Zeile 3: Zeile 3:
We write
We write
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
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& z=r\left( \cos \alpha +i\sin \alpha \right) \\
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i &= 1\,\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,,
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& i=1\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right) \\
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\end{align}</math>}}
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\end{align}</math>
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and, on using de Moivre's formula, the equation becomes
and, on using de Moivre's formula, the equation becomes
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{{Displayed math||<math>r^2(\cos 2\alpha + i\sin 2\alpha) = 1\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,\textrm{.}</math>}}
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<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>
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Both sides are equal when
Both sides are equal when
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{{Displayed math||<math>\left\{\begin{align}
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<math>\left\{ \begin{array}{*{35}l}
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r^2 &= 1\,,\\[5pt]
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r^{2}=1 \\
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2\alpha &= \frac{\pi}{2}+2n\pi\,,\quad\text{(n is an arbitrary integer),}
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2\alpha =\frac{\pi }{2}+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{align}\right.</math>}}
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\end{array} \right.</math>
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which gives that
which gives that
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{{Displayed math||<math>\left\{\begin{align}
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r &= 1\,,\\[5pt]
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\alpha &= \frac{\pi}{4} + n\pi\,,\quad\text{(n is an arbitrary integer).}
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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When <math>n=0</math> and <math>n=1</math>, we get two different arguments for
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r=1 \\
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<math>\alpha</math>, whilst different values of <math>n</math> only give these arguments plus/minus a multiple of <math>2\pi</math>.
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\alpha =\frac{\pi }{4}+n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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When
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<math>n=0</math>
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and
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<math>n=\text{1}</math>, we get two different arguments for
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<math>\alpha </math>, whilst different values of
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<math>n</math>
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only give these arguments plus/minus a multiple of
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<math>2\pi </math>.
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The solutions to the equation are
The solutions to the equation are
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{{Displayed math||<math>z=\left\{\begin{align}
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<math>z=\left\{ \begin{array}{*{35}l}
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&1\cdot\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt]
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\ 1\centerdot \left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\
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&1\cdot\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr)
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\ 1\centerdot \left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right) \\
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\end{align}\right.
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\end{array} \right.=\left\{ \begin{array}{*{35}l}
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=
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\ \frac{1+i}{\sqrt{2}} \\
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\left\{\begin{align}
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\ -\frac{1+i}{\sqrt{2}} \\
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&\frac{1+i}{\sqrt{2}}\,,\\[5pt]
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\end{array} \right.</math>
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&-\frac{1+i}{\sqrt{2}}\,\textrm{.}
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\end{align}\right.</math>}}

Version vom 14:18, 30. Okt. 2008

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

We write

\displaystyle \begin{align}

z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] i &= 1\,\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,, \end{align}

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

\displaystyle r^2(\cos 2\alpha + i\sin 2\alpha) = 1\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,\textrm{.}

Both sides are equal when

\displaystyle \left\{\begin{align}

r^2 &= 1\,,\\[5pt] 2\alpha &= \frac{\pi}{2}+2n\pi\,,\quad\text{(n is an arbitrary integer),} \end{align}\right.

which gives that

\displaystyle \left\{\begin{align}

r &= 1\,,\\[5pt] \alpha &= \frac{\pi}{4} + n\pi\,,\quad\text{(n is an arbitrary integer).} \end{align}\right.

When \displaystyle n=0 and \displaystyle n=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{align}

&1\cdot\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt] &1\cdot\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr) \end{align}\right. = \left\{\begin{align} &\frac{1+i}{\sqrt{2}}\,,\\[5pt] &-\frac{1+i}{\sqrt{2}}\,\textrm{.} \end{align}\right.

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