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Multiplication of Laurent Series

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the expression of the Laurent series for [tex]f(z) = \frac{z}{{(2z - 1)(\frac{2}{z} - 1)}}[/tex] so that [tex]\sum\limits_{ - \infty }^\infty {{a_n}} \[/tex] converges.

    3. The attempt at a solution

    First, I find that z = 1/2 and z = 2 (and infinite) are poles of the function f. Then, I can rewrite [tex]f(z) = \frac{1}{{(2z - 1)(2 - z)}}[/tex] since it doesn't change anything in terms of poles (is this right?). After that, I calculate the first Laurent series for [tex]\frac{1}{{(2z - 1)}}[/tex] and for [tex]\frac{1}{{(2 - z)}}[/tex] (the one for |z| < 1/2) and I find that [tex]f(z) = - \frac{1}{2}\sum\limits_{n = 0}^\infty {{2^n}{z^n}} \sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}}} [/tex]. Using the Cauchy product rule, I find that an has an expression for even powers and another for odd powers. Therefore, it diverges (right?)

    I look for the second Laurent series (1/2 < |z| < 2). The expression of [tex]\frac{1}{{(2 - z)}}[/tex] is the same, but I have to change the other one. I find that [tex]f(z) = \frac{1}{{4z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} \sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}}} [/tex].

    Now, one of the series is all of negative powers and the other one of positive powers. Even more, they have the same powers in ||. My question is, can I just simplify the powers of z and the powers of 2? Or can I use the Cauchy product rule?

    Thanks
     
  2. jcsd
  3. Oct 18, 2009 #2
    Up...
     
  4. Oct 19, 2009 #3
    Does anybody have any ideas?
     
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