Solution 3.1:2a - 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 3.1:2a
			
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- {{NAVCONTENT_START}} + A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,
- <center> [[Image:3_1_2a-1(2).gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}
- {{NAVCONTENT_START}} +  
- <center> [[Image:3_1_2a-2(2).gif]] </center> + Then, the formula for the difference of two squares gives that the new denominator is a real number,
- {{NAVCONTENT_STOP}} +  
  + {{Displayed math||<math>\begin{align}
  + \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}
  + &= \frac{(3-2i)(1-i)}{(1+i)(1-i)}\\[5pt]
  + &= \frac{(3-2i)(1-i)}{1^2-i^2}\\[5pt]
  + &= \frac{(3-2i)(1-i)}{1+1}\\[5pt]
  + &= \frac{(3-2i)(1-i)}{2}\,\textrm{.}
  + \end{align}</math>}}
  +  
  + All that remains is to multiply together what is in the numerator,
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{(3-2i)(1-i)}{2}
  + &= \frac{3\cdot 1 -3\cdot i - 2i\cdot 1 - 2i\cdot(-i)}{2}\\[5pt]
  + &= \frac{3-3i-2i+2i^2}{2}\\[5pt]
  + &= \frac{3-(3+2)i+2\cdot (-1)}{2}\\[5pt]
  + &= \frac{1-5i}{2}\\[5pt]
  + &= \frac{1}{2}-\frac{5}{2}\,i\,\textrm{.}
  + \end{align}</math>}}
Current revision
A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,





\displaystyle \frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}


Then, the formula for the difference of two squares gives that the new denominator is a real number,





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



All that remains is to multiply together what is in the numerator,





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




Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_3.1:2a"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,
-<center> [[Image:3_1_2a-1(2).gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}
-{{NAVCONTENT_START}}
-<center> [[Image:3_1_2a-2(2).gif]] </center>
+Then, the formula for the difference of two squares gives that the new denominator is a real number,
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
+\frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}
+&= \frac{(3-2i)(1-i)}{(1+i)(1-i)}\\[5pt]
+&= \frac{(3-2i)(1-i)}{1^2-i^2}\\[5pt]
+&= \frac{(3-2i)(1-i)}{1+1}\\[5pt]
+&= \frac{(3-2i)(1-i)}{2}\,\textrm{.}
+\end{align}</math>}}
+All that remains is to multiply together what is in the numerator,
+{{Displayed math||<math>\begin{align}
+\frac{(3-2i)(1-i)}{2}
+&= \frac{3\cdot 1 -3\cdot i - 2i\cdot 1 - 2i\cdot(-i)}{2}\\[5pt]
+&= \frac{3-3i-2i+2i^2}{2}\\[5pt]
+&= \frac{3-(3+2)i+2\cdot (-1)}{2}\\[5pt]
+&= \frac{1-5i}{2}\\[5pt]
+&= \frac{1}{2}-\frac{5}{2}\,i\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}
 \displaystyle \begin{align}
\frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}
&= \frac{(3-2i)(1-i)}{(1+i)(1-i)}\\[5pt]
&= \frac{(3-2i)(1-i)}{1^2-i^2}\\[5pt]
&= \frac{(3-2i)(1-i)}{1+1}\\[5pt]
&= \frac{(3-2i)(1-i)}{2}\,\textrm{.}
\end{align}
 \displaystyle \begin{align}
\frac{(3-2i)(1-i)}{2}
&= \frac{3\cdot 1 -3\cdot i - 2i\cdot 1 - 2i\cdot(-i)}{2}\\[5pt]
&= \frac{3-3i-2i+2i^2}{2}\\[5pt]
&= \frac{3-(3+2)i+2\cdot (-1)}{2}\\[5pt]
&= \frac{1-5i}{2}\\[5pt]
&= \frac{1}{2}-\frac{5}{2}\,i\,\textrm{.}
\end{align}