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Partial fraction decomposition: One quick question

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Give the partial fraction decomposition of 1/z4+z2


    2. Relevant equations



    3. The attempt at a solution
    My question is about the final answer. The book gives the answer to be 1/z2+ 1/2i(z+i)- 1/2i(z-i). For my answer I keep getting a negative for both of the 1/2i coefficients, i.e my answer has a - where the book puts a +, and I can not for the life of me figure out what I am missing. My answer revolves around 1= A(z+i)(z-i)+ Bz2(z-i)+ Cz2(z+i) and then solving for the coefficients. How is B= 1/2i and not -(1/2i)?
     
  2. jcsd
  3. Apr 24, 2013 #2

    HallsofIvy

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    Staff Emeritus
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    First, write it correctly! What you wrote would normally be interpreted as [itex](1/z)^4+ z^2[/itex] but I am sure you mean [tex]1/(z^4+ z^2)= 1/[z^2(z+i)(z- i)][/tex]. Normally, I would look for partial fractions with real coefficitents, but with tthe "book answer" you give it must be of the form
    [tex]\frac{1}{z^2(z+ i)(z- i)}= \frac{A}{z}+ \frac{B}{z^2}+ \frac{C}{z+ i}+ \frac{D}{z- i}[/tex]

    Multiplying both sides by [tex]z^2(z+ i)(z- i)[/tex] gives [tex]1= Az(z- i)(z+ i)+ B(z+ i)(z- i)+ Cz^2(z- i)+ Dz^2(z+ i)[/tex]

    Taking z= 0, 1= B. Taking z= i, 1= D(-1)(2i) so that D= -1/2i= (1/2)i. Taking z= -i, 1= C(-1)(-2i)= 2iC so that C= 1/2i or -(1/2)i. I think that last is what you are asking about.
     
    Last edited by a moderator: Apr 25, 2013
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