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Homework Help: Working complex fractions without conjugate method

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Ive been given a question that requires an answer in polar form but the method I must use is the normal addition/subtraction of fractions. This throws me because i'm sure there is a simple method for the reciprocal of a complex number.

    Q. 1/(13- 5i) - 1/(2-3i)

    2. Relevant equations

    where z = a + bi
    i'm aware of how to change to polar for whole numbers and how to get where I want using the conjugate method but cant fathom the correct start point.


    3. The attempt at a solution

    I considered 1/13 - 1/5i using the rules of fractions but wasn't convinced.
    The alternative was to convert straight to polar using 1 / 0degrees over the (13-5i in polar form)
     
    Last edited: Aug 11, 2011
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  3. Aug 11, 2011 #2

    HallsofIvy

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    [tex]\frac{1}{a+ bi}= \frac{1}{a+ bi}\frac{a- bi}{a- bi}= \frac{a- bi}{(a+ bi)(a- bi)}[/tex]
    [tex]= \frac{a- bi}{a^2-b^2i^2}= \frac{a- bi}{a^2+ b^2}[/tex]

    (Even for real numbers,
    [tex]\frac{1}{a+ b}\ne \frac{1}{a}+ \frac{1}{b}[/tex])
     
  4. Aug 11, 2011 #3
    Doesn't that just proove the Complex Conjugate method? I can only solve the answer without the use of this method.

    The actual wording of the question is:
    Evaluate the following expression using normal addition/subtraction of fractions and/or polar form (i.e DO NOT use the complex conjugate method) giving the result in polar form.

    if I use (a-bi)/(a2 + b2) doesnt this just mean I've used the complex Conjugate Method? Or ar you saying I can do it this way:

    where a^2 + b^2 = 13^2 + 5^2 = 194

    13/194 + (5/194)i

    0.067 + 0.026i
    and the same for the second rectangular element of the question (1/(2-3i)
     
  5. Aug 11, 2011 #4

    Redbelly98

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    Welcome to PF.

    I think "normal addition/subtraction of fractions" means you find a common denominator for the two fractions, and then you can add the resulting numerators.
     
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