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Lösung 3.3:1b

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:
Thus,
Thus,
-
{{Displayed math||<math>\frac{1}{2}+i\frac{\sqrt{3}}{2} = 1\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)</math>}}
+
{{Abgesetzte Formel||<math>\frac{1}{2}+i\frac{\sqrt{3}}{2} = 1\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)</math>}}
and de Moivre's formula gives
and de Moivre's formula gives
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Bigr)^{12}
\Bigl(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Bigr)^{12}
&= 1^{12}\cdot\Bigl(\cos\Bigl(12\cdot\frac{\pi}{3}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{3}\Bigr)\Bigr)\\[5pt]
&= 1^{12}\cdot\Bigl(\cos\Bigl(12\cdot\frac{\pi}{3}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{3}\Bigr)\Bigr)\\[5pt]

Version vom 13:10, 10. Mär. 2009

First, we write the number 21+i23  in polar form.

Image:3_3_1_b.gif Image:3_3_1_b_text.gif

Thus,

21+i23=1cos3+isin3 

and de Moivre's formula gives

21+i2312=112cos123+isin123=1(cos4+isin4)=1(1+i0)=1.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:1b“