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Lösung 3.3:1e

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 7: Zeile 7:
This shows that
This shows that
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\,,\\[5pt]
1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\,,\\[5pt]
1-i &= \sqrt{2}\Bigl(\cos\Bigl(-\frac{\pi}{4}\Bigr) + i\sin\Bigl(-\frac{\pi}{4}\Bigr)\Bigr)\,,\\[5pt]
1-i &= \sqrt{2}\Bigl(\cos\Bigl(-\frac{\pi}{4}\Bigr) + i\sin\Bigl(-\frac{\pi}{4}\Bigr)\Bigr)\,,\\[5pt]
Zeile 15: Zeile 15:
Now, with the help of de Moivre's formula,
Now, with the help of de Moivre's formula,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{\bigl(1+i\sqrt{3}\,\bigr)(1-i)^8}{\bigl(\sqrt{3}-i\bigr)^9}
\frac{\bigl(1+i\sqrt{3}\,\bigr)(1-i)^8}{\bigl(\sqrt{3}-i\bigr)^9}
&= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\Bigl(\sqrt{2}\Bigl(\cos\Bigl(-\dfrac{\pi}{4}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr)\Bigr)^8}{\Bigl( 2\Bigl(\cos\Bigl(-\dfrac{\pi}{6}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)\Bigr)^9}\\[5pt]
&= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\Bigl(\sqrt{2}\Bigl(\cos\Bigl(-\dfrac{\pi}{4}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr)\Bigr)^8}{\Bigl( 2\Bigl(\cos\Bigl(-\dfrac{\pi}{6}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)\Bigr)^9}\\[5pt]

Version vom 13:11, 10. Mär. 2009

Some parts of the quotient have rather high exponents and this indicates that we ought to use polar form for the calculation.

First, we write 1+i3 , 1−i and 3−i  in polar form.

Image:3_3_1_e.gif Image:3_3_1_e_text.gif

This shows that

1+i31−i3−i=2cos3+isin3=2cos−4+isin−4=2cos−6+isin−6.

Now, with the help of de Moivre's formula,

3−i91+i3(1−i)8=2cos−6+isin−692cos3+isin32cos−4+isin−48=29cos9−6+isin9−62cos3+isin328cos8−4+isin8−4=29cos−23+isin−232cos3+isin32(12)8cos(−2)+isin(−2)=29cos−23+isin−232cos3+isin324(1+i0)=29224cos3−−23+isin3−−23=2925cos3+23+isin3+23=124cos611+isin611=116cos612−+isin612−=116cos2−6+isin2−6=116cos−6+isin−6=11623−i2=1323−i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:1e“