Lösung 3.3:4a

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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We write
We write
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
i &= 1\,\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,,
i &= 1\,\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,,
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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>}}
+
{{Abgesetzte Formel||<math>r^2(\cos 2\alpha + i\sin 2\alpha) = 1\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,\textrm{.}</math>}}
Both sides are equal when
Both sides are equal when
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{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r^2 &= 1\,,\\[5pt]
r^2 &= 1\,,\\[5pt]
2\alpha &= \frac{\pi}{2}+2n\pi\,,\quad\text{(n is an arbitrary integer),}
2\alpha &= \frac{\pi}{2}+2n\pi\,,\quad\text{(n is an arbitrary integer),}
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which gives that
which gives that
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{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r &= 1\,,\\[5pt]
r &= 1\,,\\[5pt]
\alpha &= \frac{\pi}{4} + n\pi\,,\quad\text{(n is an arbitrary integer).}
\alpha &= \frac{\pi}{4} + n\pi\,,\quad\text{(n is an arbitrary integer).}
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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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{{Abgesetzte Formel||<math>z=\left\{\begin{align}
&1\cdot\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt]
&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)
&1\cdot\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr)

Version vom 13:13, 10. Mär. 2009

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“