Solution 3.3:1b

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
Current revision (08:55, 30 October 2008) (edit) (undo)
m
 
(2 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
First, we write the number <math>\frac{1}{2}+i\frac{\sqrt{3}}{2}</math> in polar form.
-
<center> [[Image:3_3_1b.gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
-
[[Image:3_3_1_b.gif]] [[Image:3_3_1_b_text.gif]]
+
<center>[[Image:3_3_1_b.gif]] [[Image:3_3_1_b_text.gif]]</center>
 +
 
 +
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>}}
 +
 
 +
and de Moivre's formula gives
 +
 
 +
{{Displayed math||<math>\begin{align}
 +
\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\cdot (\cos 4\pi + i\sin 4\pi)\\[5pt]
 +
&= 1\cdot (1+i\cdot 0)\\[5pt]
 +
&= 1\,\textrm{.}
 +
\end{align}</math>}}

Current revision

First, we write the number \displaystyle \frac{1}{2}+i\frac{\sqrt{3}}{2} in polar form.

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

Thus,

\displaystyle \frac{1}{2}+i\frac{\sqrt{3}}{2} = 1\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)

and de Moivre's formula gives

\displaystyle \begin{align}

\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\cdot (\cos 4\pi + i\sin 4\pi)\\[5pt] &= 1\cdot (1+i\cdot 0)\\[5pt] &= 1\,\textrm{.} \end{align}

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_3.3:1b"