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Lösung 3.3:1c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 5: Zeile 5:
This gives
This gives
-
{{Displayed math||<math>4\sqrt{3}-4i = 8\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)</math>}}
+
{{Abgesetzte Formel||<math>4\sqrt{3}-4i = 8\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)</math>}}
and then we get, on using de Moivre's formula,
and then we get, on using de Moivre's formula,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\bigl(4\sqrt{3}-4i\bigr)^{22}
\bigl(4\sqrt{3}-4i\bigr)^{22}
&= 8^{22}\Bigl(\cos\Bigl(22\cdot\Bigl(-\frac{\pi}{6}\Bigr)\Bigr) + i\sin\Bigl(22\cdot\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)\Bigr)\\[5pt]
&= 8^{22}\Bigl(\cos\Bigl(22\cdot\Bigl(-\frac{\pi}{6}\Bigr)\Bigr) + i\sin\Bigl(22\cdot\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)\Bigr)\\[5pt]

Version vom 13:11, 10. Mär. 2009

The calculation follows a fairly set pattern. We write the number 43−4i  in polar form and then use de Moivre's formula.

Image:3_3_1_c.gif Image:3_3_1_c_text.gif

This gives

43−4i=8cos−6+isin−6 

and then we get, on using de Moivre's formula,

43−4i22=822cos22−6+isin22−6=2322cos−311+isin−311=2322cos−312−+isin−312−=266cos−4+3+isin−4+3=266cos3+isin3=26621+i23=265(1+i3).
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:1c“