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Complex analysis, taylor series, radius of convergence

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data

    For f(z) = 1/(1+z^2)

    a) find the taylor series centred at the origin and the radius of convergence.

    b)find the laurent series for the annulus centred at the origin with inner radius given by the r.o.c. from part a), and an arbitrarily large outer radius.

    2. Relevant equations

    for a) (sum from j = 0 to infinity)

    f(z) = [itex]\Sigma[/itex] [(f[itex]^{j}[/itex](0))[itex]\div[/itex](j!)] [itex]\times[/itex] z[itex]^{j}[/itex]

    for b) laurent series formula?

    3. The attempt at a solution

    From what I understand, the radius of convergence is from Zo (in this case, the origin) to the closest point where f(z) isn't analytic. f(z) isn't analytic at i or -i. This function is a circle, discontinuous at i and -i. So, by inspection(?), the r.o.c. should be 1.

    I don't get how to input the information I have into the formula for a). I think that in understanding this, finding the laurent series should be simplified.

  2. jcsd
  3. Sep 6, 2011 #2
    Indeed, the radius of convergence is 1.

    To find the Taylor/Laurent series; I suggest you factor the denominator first and split into partial fractions. Then you should use


    for |z|<1.
  4. Sep 6, 2011 #3


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    You CAN find the taylor series of f(z) by taking derivatives and putting z=0, as you said. But the easy way is the use the geometric sum formula 1/(1-u)=1+u+u^2+... What should u be?
  5. Sep 7, 2011 #4
    Hey guys,

    Thanks for your help, but I'm still confused. I have found f(z) = 1/(z+i)(z-i) and I think I've found the partial fractions representation:

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

    but don't know where to go from here... do I seperately apply taylor series expansion to each of these fractions...?
  6. Sep 8, 2011 #5
    Yes, that's the idea. Let me do an example

    [tex]\frac{1}{z-2} = -\frac{1}{2-z} = -\frac{1}{2}\frac{1}{1-\frac{z}{2}} = -\frac{1}{2}\sum_{n=0}^{+\infty} {\left(\frac{z}{2}\right)^2}[/tex]

    This is how to find the Laurent/Taylor series of [itex]\frac{1}{z-2}[/itex]. Can you do the same for your two dractions?
  7. Sep 21, 2011 #6
    I got part a)

    I'm fairly confident that the answer is f(z) = ([itex]\frac{1}{2}[/itex])0[itex]\sum[/itex][itex]\infty[/itex] ((zj) + (-z)j)/(ij)

    (sum from 0 to infinity)

    But don't understand how to do laurent series... I think I need to do it for the annulus centered at the origin with radius 1, and then again for the annulus centered at the origin but with arbitrarily large outer radius and inner radius of 1...

    Do I need to change the format of this sum? split it into two parts? take out the first few terms? any ideas would be great!
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