(New page: == Introduction == * Complex numbers are typically represnted in the form <math>(a + bi)\!</math>, where <math>i=\sqrt{-1}\!</math> and <math>i^2=-1\!</math>. The variable <math>a\!</math...) |
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* Subtraction of two complex numbers is done by subtracting the real parts and the imaginary parts seperately: | * Subtraction of two complex numbers is done by subtracting the real parts and the imaginary parts seperately: | ||
<math>(a+bi)-(c+di)=(a-c)+(b-d)i\!</math> | <math>(a+bi)-(c+di)=(a-c)+(b-d)i\!</math> | ||
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| + | == Multiplying Complex Numbers == | ||
| + | * Multiplication of complex numbers follows the basic commutative and distributive laws. Keep in mind <math>i^2=-1\!</math>. | ||
| + | <math>(a+bi)(c+di)=a(c+di)+(bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i\!</math> | ||
| + | <br> | ||
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| + | == Dividing Complex Numbers == | ||
| + | * Division of complex numbers is usually done by multiplying the numerator and denominator by a complex number that will get rid of the <math>i\!</math> in the denominator: | ||
| + | <math>(a+bi)/(c+di)=((a+bi)(c-di))/((c+di)(c-di))=((ac+bd)+(bc-ad)i)/(c^2+d^2)\!</math> | ||
Latest revision as of 16:45, 3 September 2008
Contents
Introduction
- Complex numbers are typically represnted in the form $ (a + bi)\! $, where $ i=\sqrt{-1}\! $ and $ i^2=-1\! $. The variable $ a\! $ reprents the real part and the variable $ b\! $ represents the imaginary part.
Adding Complex Numbers
- Addition of two complex numbers is done by adding the real parts together and the imaginary parts together:
$ (a+bi)+(c+di)=(a+c)+(b+d)i\! $
Subtracting Complex Numbers
- Subtraction of two complex numbers is done by subtracting the real parts and the imaginary parts seperately:
$ (a+bi)-(c+di)=(a-c)+(b-d)i\! $
Multiplying Complex Numbers
- Multiplication of complex numbers follows the basic commutative and distributive laws. Keep in mind $ i^2=-1\! $.
$ (a+bi)(c+di)=a(c+di)+(bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i\! $
Dividing Complex Numbers
- Division of complex numbers is usually done by multiplying the numerator and denominator by a complex number that will get rid of the $ i\! $ in the denominator:
$ (a+bi)/(c+di)=((a+bi)(c-di))/((c+di)(c-di))=((ac+bd)+(bc-ad)i)/(c^2+d^2)\! $
