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Laurent series

  1. Apr 17, 2017 #1
    1. The problem statement, all variables and given/known data
    expand [itex] f(z)=\frac{1}{z(z-1)} [/itex] in a laurent series valid for the given annular domain.
    |z|> 3

    2. Relevant equations


    3. The attempt at a solution
    first I do partial fractions to get
    [itex] \frac{-1}{3z} +\frac{1}{3(z-3)} [/itex]
    then in the second fraction I factor out a z in the denominator to give me a geometric series
    where 3/z is the common ration and converges for 3/z<1 and then 3<z.

    so I get for my series [itex] \frac{-1}{3z}+\frac{1}{3z}[1+\frac{3}{z}+\frac{3^2}{z^2}+..... ] [/itex]
     
  2. jcsd
  3. Apr 17, 2017 #2
    Hi cragar:

    When I combine -1/3z + 1/3(z-3) I get 1/z(z-3).

    Regards,
    Buzz
     
  4. Apr 17, 2017 #3
    oh right i mistyped the problem it has a z-3 factor in the original factor
     
  5. Apr 17, 2017 #4
    Hi cragar:

    The series simplifies to 3/z + 32/z2 + 33/z3 ...
    The problem statement requests a series that converges for |z| > 3. Be careful to keep the absolute value notation.The region of convergence consists of all complex numbers outside the circle of radius 3 with its center at the origin z=0.

    Regards,
    Buzz
     
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