Lösung 3.1:2d - Online Mathematik Brückenkurs 2

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                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
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			Lösung 3.1:2d
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
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				Version vom 15:41, 29. Okt. 2008 (bearbeiten)
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← Zum vorherigen Versionsunterschied
				Version vom 13:06, 10. Mär. 2009 (bearbeiten) (rückgängig)
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Zum nächsten Versionsunterschied →
			
		Zeile 1:
Zeile 1:
 Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,  Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 5-\frac{1}{1+i}  5-\frac{1}{1+i}
 &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}  &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
Zeile 14:
Zeile 14:
 Hence,  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>}} + {{Abgesetzte Formel||<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  We multiply out the numerator and denominator
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}  \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{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
Zeile 27:
Zeile 27:
 This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,  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} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{23-2i}{2+16i}  \frac{23-2i}{2+16i}
 &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]  &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
Zeile 37:
Zeile 37:
 If we divide up the numbers into factors,  If we divide up the numbers into factors,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 14 &= 2\cdot 7\,,\\[5pt]  14 &= 2\cdot 7\,,\\[5pt]
 372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]  372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
Zeile 45:
Zeile 45:
 we can simplify the answers  we can simplify the answers
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{14}{260}-\frac{372}{260}\,i  \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{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]
Version vom 13:06, 10. Mär. 2009
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}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:2d“
                       Willkommen zum Kurs
                       
                        Information über den Kurs 
                        Information über Prüfungen
                      
                    
                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
                      Kurs als PDF
                                            
                    
                    
		       Suche
		       
		         

                           
                            
			 
		       
		    
		    
		  
		  
		  
		    
		      
		        
			
			Lösung 3.1:2d
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 15:41, 29. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 13:06, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 1:
Zeile 1:
 Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,  Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 5-\frac{1}{1+i}  5-\frac{1}{1+i}
 &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}  &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
Zeile 14:
Zeile 14:
 Hence,  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>}} + {{Abgesetzte Formel||<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  We multiply out the numerator and denominator
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}  \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{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
Zeile 27:
Zeile 27:
 This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,  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} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{23-2i}{2+16i}  \frac{23-2i}{2+16i}
 &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]  &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
Zeile 37:
Zeile 37:
 If we divide up the numbers into factors,  If we divide up the numbers into factors,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 14 &= 2\cdot 7\,,\\[5pt]  14 &= 2\cdot 7\,,\\[5pt]
 372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]  372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
Zeile 45:
Zeile 45:
 we can simplify the answers  we can simplify the answers
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{14}{260}-\frac{372}{260}\,i  \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{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]
Version vom 13:06, 10. Mär. 2009
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}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:2d“
-                      Willkommen zum Kurs
                       
                        Information über den Kurs 
                        Information über Prüfungen
                      
                    
                      1. Differentialrechnung
                       
                        1.1 Einführung 
                        1.2 Ableitungsregeln 
                        1.3 Max/Min probleme
                      
                    
                      2. Integralrechnung
                       
                        2.1 Einführung 
                        2.2 Substitution 
                        2.3 Partielle Integration
                      
                    
                      3. Komplexe Zahlen
                       
                        3.1 Rechnungen 
                        3.2 Polarform 
                        3.3 Potenzen und Wurzeln 
                        3.4 Komplexe Polynome
                      
                    
                      Kurs als PDF
                                            
                    
                    
		       Suche
+			Lösung 3.1:2d
			
			  Aus Online Mathematik Brückenkurs 2
			  (Unterschied zwischen Versionen)
			  			                            Wechseln zu: Navigation, Suche
                          
		          
			
			
			
			
			
			
				Version vom 15:41, 29. Okt. 2008 (bearbeiten)
Tek  (Diskussion | Beiträge)
K 
← Zum vorherigen Versionsunterschied
				Version vom 13:06, 10. Mär. 2009 (bearbeiten) (rückgängig)
Tekbot  (Diskussion | Beiträge) 
K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel))
Zum nächsten Versionsunterschied →
			
		Zeile 1:
Zeile 1:
 Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,  Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 5-\frac{1}{1+i}  5-\frac{1}{1+i}
 &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}  &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
Zeile 14:
Zeile 14:
 Hence,  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>}} + {{Abgesetzte Formel||<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  We multiply out the numerator and denominator
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}  \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{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
Zeile 27:
Zeile 27:
 This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,  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} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{23-2i}{2+16i}  \frac{23-2i}{2+16i}
 &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]  &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
Zeile 37:
Zeile 37:
 If we divide up the numbers into factors,  If we divide up the numbers into factors,
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 14 &= 2\cdot 7\,,\\[5pt]  14 &= 2\cdot 7\,,\\[5pt]
 372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]  372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
Zeile 45:
Zeile 45:
 we can simplify the answers  we can simplify the answers
  
- {{Displayed math||<math>\begin{align} + {{Abgesetzte Formel||<math>\begin{align}
 \frac{14}{260}-\frac{372}{260}\,i  \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{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]
Version vom 13:06, 10. Mär. 2009
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}





Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:2d“
@@ Zeile 1: / Zeile 1: @@
 Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 -\frac{1}{1+i}
 &=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
@@ Zeile 14: / Zeile 14: @@
 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>}}
+{{Abgesetzte Formel||<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}
+{{Abgesetzte Formel||<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]
@@ Zeile 27: / Zeile 27: @@
 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}
+{{Abgesetzte Formel||<math>\begin{align}
 \frac{23-2i}{2+16i}
 &= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
@@ Zeile 37: / Zeile 37: @@
 If we divide up the numbers into factors,
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<math>\begin{align}
 &= 2\cdot 7\,,\\[5pt]
 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
@@ Zeile 45: / Zeile 45: @@
 we can simplify the answers
-{{Displayed math||<math>\begin{align}
+{{Abgesetzte Formel||<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]
 \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}