Solution 3.1:2d - 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 3.1:2d
			
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- {{NAVCONTENT_START}} + Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
- <center> [[Image:3_1_2d-1(2).gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\begin{align}
- {{NAVCONTENT_START}} + 5-\frac{1}{1+i}
- <center> [[Image:3_1_2d-2(2).gif]] </center> + &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
- {{NAVCONTENT_STOP}} + = \frac{5+5i-1}{1+i}
  + = \frac{4+5i}{1+i}\,,\\[5pt]
  + 3i+\frac{i}{2-3i}
  + &= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}
  + = \frac{6i-9i^2+i}{2-3i}
  + = \frac{9+7i}{2-3i}\,\textrm{.}
  + \end{align}</math>}}
  +  
  + Hence,
  +  
  + {{Displayed math||<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}
  +  
  + We multiply out the numerator and denominator
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}
  + &= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
  + &= \frac{8-12i+10i+15}{9+9i+7i-7}\\[5pt]
  + &= \frac{23-2i}{2+16i}\,\textrm{.}
  + \end{align}</math>}}
  +  
  + This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{23-2i}{2+16i}
  + &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
  + &= \frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\[5pt]
  + &= \frac{46-368i-4i-32}{4+256}\\[5pt]
  + &= \frac{14-372i}{260}\,\textrm{.}
  + \end{align}</math>}}
  +  
  + If we divide up the numbers into factors,
  +  
  + {{Displayed math||<math>\begin{align}
  + 14 &= 2\cdot 7\,,\\[5pt]
  + 372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
  + 260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\,,
  + \end{align}</math>}}
  +  
  + we can simplify the answers
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{14}{260}-\frac{372}{260}\,i
  + &= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]
  + &= \frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}\,i\\[5pt]
  + &= \frac{7}{130}-\frac{93}{65}\,i\,\textrm{.}
  + \end{align}</math>}}
Current revision
Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,





\displaystyle \begin{align}
5-\frac{1}{1+i}
&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
= \frac{5+5i-1}{1+i}
= \frac{4+5i}{1+i}\,,\\[5pt]
3i+\frac{i}{2-3i}
&= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}
= \frac{6i-9i^2+i}{2-3i}
= \frac{9+7i}{2-3i}\,\textrm{.}
\end{align}



Hence,





\displaystyle \frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}


We multiply out the numerator and denominator





\displaystyle \begin{align}
\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}
&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
&= \frac{8-12i+10i+15}{9+9i+7i-7}\\[5pt]
&= \frac{23-2i}{2+16i}\,\textrm{.}
\end{align}



This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,





\displaystyle \begin{align}
\frac{23-2i}{2+16i}
&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
&= \frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\[5pt]
&= \frac{46-368i-4i-32}{4+256}\\[5pt]
&= \frac{14-372i}{260}\,\textrm{.}
\end{align}



If we divide up the numbers into factors,





\displaystyle \begin{align}
14 &= 2\cdot 7\,,\\[5pt]
372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\,,
\end{align}



we can simplify the answers





\displaystyle \begin{align}
\frac{14}{260}-\frac{372}{260}\,i
&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]
&= \frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}\,i\\[5pt]
&= \frac{7}{130}-\frac{93}{65}\,i\,\textrm{.}
\end{align}





Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-ImperialCollege/index.php/Solution_3.1:2d"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
-<center> [[Image:3_1_2d-1(2).gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
-{{NAVCONTENT_START}}
+-\frac{1}{1+i}
-<center> [[Image:3_1_2d-2(2).gif]] </center>
+&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
-{{NAVCONTENT_STOP}}
+= \frac{5+5i-1}{1+i}
+= \frac{4+5i}{1+i}\,,\\[5pt]
+i+\frac{i}{2-3i}
+&= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}
+= \frac{6i-9i^2+i}{2-3i}
+= \frac{9+7i}{2-3i}\,\textrm{.}
+\end{align}</math>}}
+Hence,
+{{Displayed math||<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}
+We multiply out the numerator and denominator
+{{Displayed math||<math>\begin{align}
+\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}
+&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
+&= \frac{8-12i+10i+15}{9+9i+7i-7}\\[5pt]
+&= \frac{23-2i}{2+16i}\,\textrm{.}
+\end{align}</math>}}
+This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,
+{{Displayed math||<math>\begin{align}
+\frac{23-2i}{2+16i}
+&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
+&= \frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\[5pt]
+&= \frac{46-368i-4i-32}{4+256}\\[5pt]
+&= \frac{14-372i}{260}\,\textrm{.}
+\end{align}</math>}}
+If we divide up the numbers into factors,
+{{Displayed math||<math>\begin{align}
+&= 2\cdot 7\,,\\[5pt]
+&= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
+&= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\,,
+\end{align}</math>}}
+we can simplify the answers
+{{Displayed math||<math>\begin{align}
+\frac{14}{260}-\frac{372}{260}\,i
+&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]
+&= \frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}\,i\\[5pt]
+&= \frac{7}{130}-\frac{93}{65}\,i\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \begin{align}
5-\frac{1}{1+i}
&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
= \frac{5+5i-1}{1+i}
= \frac{4+5i}{1+i}\,,\\[5pt]
3i+\frac{i}{2-3i}
&= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}
= \frac{6i-9i^2+i}{2-3i}
= \frac{9+7i}{2-3i}\,\textrm{.}
\end{align}
 \displaystyle \frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}
 \displaystyle \begin{align}
\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}
&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
&= \frac{8-12i+10i+15}{9+9i+7i-7}\\[5pt]
&= \frac{23-2i}{2+16i}\,\textrm{.}
\end{align}
 \displaystyle \begin{align}
\frac{23-2i}{2+16i}
&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
&= \frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\[5pt]
&= \frac{46-368i-4i-32}{4+256}\\[5pt]
&= \frac{14-372i}{260}\,\textrm{.}
\end{align}
 \displaystyle \begin{align}
14 &= 2\cdot 7\,,\\[5pt]
372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\,,
\end{align}
 \displaystyle \begin{align}
\frac{14}{260}-\frac{372}{260}\,i
&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]
&= \frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}\,i\\[5pt]
&= \frac{7}{130}-\frac{93}{65}\,i\,\textrm{.}
\end{align}