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Lösung 3.3:1d

Aus Online Mathematik Brückenkurs 2

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K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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We obtain
We obtain
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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\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)
1+i &= \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)
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and
and
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1+i\sqrt{3}}{1+i}
\frac{1+i\sqrt{3}}{1+i}
&= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)}{\sqrt{2}\Bigl(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\Bigr)}\\[5pt]
&= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)}{\sqrt{2}\Bigl(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\Bigr)}\\[5pt]
Zeile 22: Zeile 22:
Finally, de Moivre's formula gives
Finally, de Moivre's formula gives
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(\frac{1+i\sqrt{3}}{1+i}\Bigr)^{12}
\Bigl(\frac{1+i\sqrt{3}}{1+i}\Bigr)^{12}
&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos\Bigl(12\cdot\frac{\pi}{12}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{12}\Bigr)\Bigr)\\[5pt]
&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos\Bigl(12\cdot\frac{\pi}{12}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{12}\Bigr)\Bigr)\\[5pt]

Version vom 13:11, 10. Mär. 2009

Because we are going to raise something to the power 12, the base in the expression should be written in polar form. In turn, the base consists of a quotient which is advantageous to calculate in polar form. Thus, it seems appropriate to write 1+i3  and 1+i in polar form right from the beginning and to carry out all calculations in polar form.

Image:3_3_1_d.gif Image:3_3_1_d_text.gif

We obtain

1+i31+i=2cos3+isin3=2cos4+isin4

and

1+i1+i3=2cos3+isin32cos4+isin4=22cos3−4+isin3−4=2cos12+isin12.

Finally, de Moivre's formula gives

1+i1+i312=212cos1212+isin1212=2(12)12(cos+isin)=26(−1+i0)=−64.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:1d“