Solution 3.3:1c - Förberedande kurs i matematik 2

-                      Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Derivatives
                       
                        1.1 Introduction 
                        1.2 Rules of differentiation 
                        1.3 Max and min problems
                      
                    
                      2. Integrals
                       
                        2.1 Introduction 
                        2.2 Substitution 
                        2.3 Integration by parts
                      
                    
                      3. Complex numbers
                       
                        3.1 Calculations 
                        3.2 Polar form 
                        3.3 Powers and roots 
                        3.4 Complex polynomials
                      
                    
                    
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+			Solution 3.3:1c
			
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- {{NAVCONTENT_START}} + The calculation follows a fairly set pattern. We write the number <math>4\sqrt{3}-4i</math> in polar form and then use de Moivre's formula.
- <center> [[Image:3_3_1c.gif]] </center> +  
- {{NAVCONTENT_STOP}} + <center>[[Image:3_3_1_c.gif]] [[Image:3_3_1_c_text.gif]]</center>
  +  
  + 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>}}
  +  
  + and then we get, on using de Moivre's formula,
  +  
  + {{Displayed math||<math>\begin{align}
  + \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]
  + &= \bigl(2^3\bigr)^{22}\Bigl(\cos\Bigl(-\frac{11\pi}{3}\Bigr) + i\sin\Bigl(-\frac{11\pi}{3}\Bigr)\Bigr)\\[5pt]
  + &= 2^{3\cdot 22}\Bigl(\cos\Bigl(-\frac{12\pi -\pi }{3}\Bigr) + i\sin\Bigl(-\frac{12\pi -\pi}{3}\Bigr)\Bigr)\\[5pt]
  + &= 2^{66}\Bigl(\cos\Bigl(-4\pi+\frac{\pi}{3}\Bigr) + i\sin\Bigl(-4\pi+\frac{\pi}{3} \Bigr)\Bigr)\\[5pt]
  + &= 2^{66}\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt] 
  + &= 2^{66}\Bigl(\frac{1}{2} + i\frac{\sqrt{3}}{2}\Bigr)\\[5pt]
  + &= 2^{65}(1+i\sqrt{3}\,)\,\textrm{.}
  + \end{align}</math>}}
Current revision
The calculation follows a fairly set pattern. We write the number \displaystyle 4\sqrt{3}-4i in polar form and then use de Moivre's formula.

 
This gives





\displaystyle 4\sqrt{3}-4i = 8\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)


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





\displaystyle \begin{align}
\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]
&= \bigl(2^3\bigr)^{22}\Bigl(\cos\Bigl(-\frac{11\pi}{3}\Bigr) + i\sin\Bigl(-\frac{11\pi}{3}\Bigr)\Bigr)\\[5pt]
&= 2^{3\cdot 22}\Bigl(\cos\Bigl(-\frac{12\pi -\pi }{3}\Bigr) + i\sin\Bigl(-\frac{12\pi -\pi}{3}\Bigr)\Bigr)\\[5pt]
&= 2^{66}\Bigl(\cos\Bigl(-4\pi+\frac{\pi}{3}\Bigr) + i\sin\Bigl(-4\pi+\frac{\pi}{3} \Bigr)\Bigr)\\[5pt]
&= 2^{66}\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt] 
&= 2^{66}\Bigl(\frac{1}{2} + i\frac{\sqrt{3}}{2}\Bigr)\\[5pt]
&= 2^{65}(1+i\sqrt{3}\,)\,\textrm{.}
\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_3.3:1c"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+The calculation follows a fairly set pattern. We write the number <math>4\sqrt{3}-4i</math> in polar form and then use de Moivre's formula.
-<center> [[Image:3_3_1c.gif]] </center>
-{{NAVCONTENT_STOP}}
+<center>[[Image:3_3_1_c.gif]] [[Image:3_3_1_c_text.gif]]</center>
+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>}}
+and then we get, on using de Moivre's formula,
+{{Displayed math||<math>\begin{align}
+\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]
+&= \bigl(2^3\bigr)^{22}\Bigl(\cos\Bigl(-\frac{11\pi}{3}\Bigr) + i\sin\Bigl(-\frac{11\pi}{3}\Bigr)\Bigr)\\[5pt]
+&= 2^{3\cdot 22}\Bigl(\cos\Bigl(-\frac{12\pi -\pi }{3}\Bigr) + i\sin\Bigl(-\frac{12\pi -\pi}{3}\Bigr)\Bigr)\\[5pt]
+&= 2^{66}\Bigl(\cos\Bigl(-4\pi+\frac{\pi}{3}\Bigr) + i\sin\Bigl(-4\pi+\frac{\pi}{3} \Bigr)\Bigr)\\[5pt]
+&= 2^{66}\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt]
+&= 2^{66}\Bigl(\frac{1}{2} + i\frac{\sqrt{3}}{2}\Bigr)\\[5pt]
+&= 2^{65}(1+i\sqrt{3}\,)\,\textrm{.}
+\end{align}</math>}}
 \displaystyle 4\sqrt{3}-4i = 8\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)
 \displaystyle \begin{align}
\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]
&= \bigl(2^3\bigr)^{22}\Bigl(\cos\Bigl(-\frac{11\pi}{3}\Bigr) + i\sin\Bigl(-\frac{11\pi}{3}\Bigr)\Bigr)\\[5pt]
&= 2^{3\cdot 22}\Bigl(\cos\Bigl(-\frac{12\pi -\pi }{3}\Bigr) + i\sin\Bigl(-\frac{12\pi -\pi}{3}\Bigr)\Bigr)\\[5pt]
&= 2^{66}\Bigl(\cos\Bigl(-4\pi+\frac{\pi}{3}\Bigr) + i\sin\Bigl(-4\pi+\frac{\pi}{3} \Bigr)\Bigr)\\[5pt]
&= 2^{66}\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt] 
&= 2^{66}\Bigl(\frac{1}{2} + i\frac{\sqrt{3}}{2}\Bigr)\\[5pt]
&= 2^{65}(1+i\sqrt{3}\,)\,\textrm{.}
\end{align}