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Partial Fractions but in Complex Analysis

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Use partial fractions to rewrite:

    (2z)/(z^2+3)



    2. Relevant equations
    none


    3. The attempt at a solution

    I did this:

    (2z)/(z^2+3) = (Az+B)/(z^2+3)

    2z = Az +B

    A = 2, B = 0......problem is that it just recreates the original

    Here is their example in the book:

    1/(z^2+1) = 1/(2i(z-i)) - 1/(2i(z+i))

    I don't understand how they came to their conclusion in their example.

    Any help understanding their example or my question is appreciated.
     
  2. jcsd
  3. Nov 27, 2011 #2
    factor 1/(z^2+1) as 1/(z+i)(z-i)
    then 1/(z^2+1)= A/(z+i)+ B/(z-i)
    now solve for A and B.... what do you get .
    and then you can do the same trick on your problem
     
  4. Nov 27, 2011 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Of course it does. You wrote it in exactly the same form. Why would you expect anything else?

    You titled this "Partical fractions but in Complex Analyis"- so use complex numbers:
    [itex]z^2+ 3= (z+ i\sqrt{3})(z- i\sqrt{3})[/itex].

    Your partial fractions should be
    [tex]\frac{2z}{z^2+ 3}= \frac{A}{z+ i\sqrt{3}}+ \frac{B}{z- i\sqrt{3}}[/tex]
     
  5. Nov 27, 2011 #4
    Thanks guys, oh my gosh

    I actually did their example and set it up all correctly

    I got to the end with A+B=0, A-B=0 and without solving it, it just looked like a contradiction and I didnt realize that B=-1.

    thanks for pointing that out for me, i was like 99% there and got stuck. i hate that
     
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