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A Simple Pole

  1. Jul 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Why is the pole at [itex]z=+i[/itex] simple for the function [itex]1/(1+z^{2})[/itex]?

    2. Relevant equations

    The function can be written as:
    [itex]h(z)=1/(1+z^{2})=1/[(z-i)(z+i)][/itex]

    It can be expanded as a Laurent series:
    [itex]h(z)=\sum^{\infty}_{-\infty}a_{n}(z-z_{0})^{n}[/itex]

    The function is defined as "simple" if terms for [itex]n<-1[/itex] do not contribute to the Laurent series. That is:
    [itex]h(z)=\sum^{\infty}_{-1}a_{n}(z-z_{0})^{n}[/itex]

    3. The attempt at a solution

    Been struggling with this one for a few days now. Tried to do a Taylor series expansion of the function but ended up with something like:

    [itex]h(z)=1-2z/(1+z^{2})^{2}+....[/itex]

    So it looks like z has exponents less than -1, which to my simplistic mind suggests like the Laurent series has terms contributing for n<-1. This suggests to me that there is a problem in my understanding somewhere along the line.

    My intuition is telling me to take a different tack, I think the solution will probably have something to do with the fact that we are showing that the pole at +i is simple. So perhaps I can then sub [itex]z_{0}=i[/itex]?

    In which case:
    [itex]h(z)=\sum^{\infty}_{-\infty}a_{n}(z-i)^{n}=\frac{1}{(z-i)(z+i)}[/itex]

    Hmmm.. (I literally just thought of this as typing) .. is this moving in the right direction?
     
  2. jcsd
  3. Jul 11, 2011 #2

    micromass

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    You will need to develop a Laurent series around a certain point. So here you must find the Laurent series around i. This means you must be able to write your function in the form

    [tex]f(z)=...+\frac{a_{-2}}{(z-i)^2}+\frac{a_{-1}}{z-i}+a_0+a_1(z-i)+a_2(z-i)^2+...[/tex]

    So, you must write your function [itex]f(z)=\frac{1}{1+z^2}[/itex] in this form. To do this, you must start with splitting your function into partial fractions...
     
  4. Jul 13, 2011 #3
    How do you go about expanding that function into partial fractions?
     
  5. Jul 13, 2011 #4

    micromass

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    You'll need to find constants A and B such that

    [tex]\frac{1}{z^2+1}=\frac{A}{z+i}+\frac{B}{z-i}[/tex]

    Look at http://en.wikipedia.org/wiki/Partial_fractions for some nice examples!
     
  6. Jul 13, 2011 #5
    Thank you kind sir.

    I believe the partial fraction expansion looks like this:

    [itex]\frac{1}{2i(z-i)}-\frac{1}{2i(z+i)}[/itex]


    Ah yes and if we observe that [itex]z+i=(z-i)+2i[/itex] then.... urrrr ... how do we clean up that [itex]z+i[/itex] term? (I think my brain is not working today!)



    Do we just ignore the [itex]z+i[/itex] term and note that the exponent of the [itex]z-i[/itex] is [itex]-1[/itex]?

    Or can we say
     
  7. Jul 13, 2011 #6

    micromass

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    Seems good, and you probably notices that

    [tex]\frac{1}{2i(z-i)}[/tex]

    is in the correct form, but the other term is not. You will have to write the term

    [tex]\frac{1}{2i(z+i)}[/tex]

    in a series that involves only terms of the kind [itex](z-i)[/itex]. For this, we will have to use the following very well known series

    [tex]\frac{1}{1-z}=1+z+z^2+z^3+z^4+...[/tex]

    Can you use the above series to write

    [tex]\frac{1}{2i(z+i)}[/tex]

    in the form of a series?
     
  8. Jul 14, 2011 #7
    Hmmm I can't see how to do it that way.

    I tried a Taylor series expansion about z=i of that expression and I think it is wrong but I found that....

    [tex]\frac{1}{2i(z+i)}=-\frac{1}{4}-\frac{i}{8}(z-i)-\frac{1}{4}(z-i)^{2}+.....[/tex]

    Now that fits, but I don't fully understand my own reasoning for getting there so I don't feel comfortable. Also I would like to see how using the well know series works.

    Cheers
     
  9. Jul 14, 2011 #8

    micromass

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    For the well-known series approach, do

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

    Now apply the well-known series

    [tex]\frac{1}{1-a}=1+a+a^2+a^3+a^4+...[/tex]

    with a=zi.
     
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