Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laurent Series Complex Analysis question

  1. Sep 27, 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) = Σ [(fj(0))÷(j!)] × zj

    for b) laurent series formula?

    3. The attempt at a solution

    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 j=0 to infinity)

    But don't understand how to calculate 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... or possibly just for the latter.

    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!
  2. jcsd
  3. Sep 27, 2011 #2


    User Avatar
    Homework Helper

    [tex] \frac{1}{1+z^2}= \frac{1}{(1+z)(1-z)}[/tex]

    so it has poles of order 1 at z=i & z=-i, thus the taylor series about z=0 will be convergent for |z|<1

    you can also find a laurent series, that will be convergent on the annulus |z|>1

    for a) finding the derivatives to find the taylor series is valid, however a quicker way would be to use partial fractions and consider geometric series

    similarly for b) use the partial fraction expansion to find the laurent series, taking a factor of 1/z outside the partial fraction, and using geometric.

    see example 38.2 here for a good summary on how it works
  4. Sep 27, 2011 #3
    Hey thanks for the advice!

    I was already pretty happy with part (a)

    the result looks quite similar for part b)

    are you able to tell me if i should end up with this? :

    (1/2i)*(j=0[itex]\sum[/itex][itex]\infty[/itex] [(i^j) - ((-i)^j)]/(z^j+1)

    I feel good about using the example you gave but would love to know for sure that what I've done is correct.

    also what am I doing wrong with the symbols on this site... I can't see them but maybe it's this pc... does that look like it should to you?
    Last edited: Sep 27, 2011
  5. Sep 28, 2011 #4


    User Avatar
    Homework Helper

    with the symbols try putting one set of tex tags around the whole equation
    - for fractions use \frac{}{}
    - for exponents longer than one character enclose with curly brackets eg z^{j+1}
    - right click on equations to see the source
  6. Sep 29, 2011 #5
    Just to echo lanedance--clever applications of the geometric series are usually much easier than actually calculating the derivatives. Basically, the inequality describing the region will tell you what expression in z is less than 1, making it a prime candidate for a geometric series expansion. So for part b), [itex] 1<|z|[/itex] means [itex] \left|\frac{1}{z}\right| <1 [/itex], so [itex]\left|\frac{1}{z^2}\right| < 1[/itex]. Does that suggest anything to you?

    Using geometric series also gives you neater answers. I'm pretty sure your answer for part a) is right because for odd j the two terms in the numerator cancel and for even j the numerator becomes [itex] 2 z^j [/itex] and then the 2 cancels with the 1/2 outside the sum. So can you see how to simplify your answer for part a)? You should get the same thing you'd get by using the geometric series.
  7. Sep 29, 2011 #6
    Hey thanks for the help guys, I see what you mean, much neater!

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook