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Complex Analysis Question

  1. Mar 11, 2006 #1
    I'm glad to see that the physics forum website is back online.

    Suppose you have a function with double poles somewhere on the complex plane. Are there complex analysis techniques that can be used to split the double pole into two single isolated poles?

    Some example functions might be

    f(z) = 1/(z - 2)^2


    g(z) = csc(pi*z)*csc(2*pi*z) = 1/sin(pi*z)*1/sin(2*pi*z)
    of the complex variable z = a + bi.

    The 2nd function g(z) has doubles at all the integers, and single poles at all points 1/2 + n and 1/2 - n where n is an integer 1, 2, 3...etc.

    Are there any techniques that can allow us to split the double poles at the integers of the 2nd function into isolated singularities off the real axis, while leaving the single poles on the real axis?


    Edwin G. Schasteen
  2. jcsd
  3. Mar 11, 2006 #2


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    You can't turn a double pole into two isolated single poles. Why would you need to though?

    You can't really move poles either, you can do a change of variables but if the pole was on your path of integration before (for ecample) it still will be in the new variable.
  4. Mar 11, 2006 #3

    I was thinking along the lines of multiplying g(z) through by another function, or otherwise altering the function g(z). Multiplying the function changes the location where poles exist. It wouldn't actually be 'moving the poles,' it would generate two new isolated poles, while canceling the double poles.
    Last edited: Mar 11, 2006
  5. Mar 11, 2006 #4
    Oh, the reason is that I have another function of a similar form that has interesting information that is yielded by doing a contour integral around the double pole. Non-interesting values occurs at single poles. Problem is that the double poles, and uninteresting single poles all exist on the real line. The hope was to find a way to find a similar function that yields poles in the complex plane with coordinates real part the same as the old function's real part, but also with some imaginary part, so that one can do a contour integral in the complex plane real close to the real line and contain the "interesting point."
  6. Mar 11, 2006 #5


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    Look into the Weierstrass product, which lets you can form an analytic function with zeros at given locations (subject to some assumptions). This is more complicated than with polynomials where you have a finite number of zeros, but morally the same idea. You can then multiply by this function and remove your simple poles. This of course changes your residues at the double-poles and I have absolutely no idea whether whatever you have planned will work or not as your description was sufficiently vague that I have no other clue what to suggest.
  7. Mar 11, 2006 #6
    Thankyou. I'll read up on the Weierstrass product and try it out to see if it works.
  8. Mar 11, 2006 #7


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    You could also write your function as a limit of functions with two single poles. e.g.

    \lim_{d \rightarrow 0} \frac{1}{(z - 2 - di)(z - 2 + di)} = \frac{1}{(z - 2)^2}

    or maybe even study

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

    itself, as a function of two variables, z and d. (And then look at what happens near d = 0) Or, to be more general, to study

    [tex]\frac{1}{(z - a)(z - b)}[/tex]

    as a function of three variables. And then look at what happens on (and near) the space a = b, or a = b = 2.
  9. Mar 12, 2006 #8


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    hurkyl has explained how to do it. just consider a function like
    sin(piz-ti)sin(piz+ti) where t is a parameter, then multipy your function

    g by sin^2(piz), to kill all the ples at the integers, and then divide it by

    sin(pi(z-ti))sin(pi(z+ti)), to introduce simple poles at n ± ti.

    the resulting function is a "deformation" of your original function, which converges to g as t approaches zero.

    this sort of deformation technique is basic in all advanced mathematics, but for some reason is not taught much in beginning courses.
  10. Mar 18, 2006 #9
    Hurkyl, Shmoe, and Mathwonk,

    Thank you very much for your assistance. I think those techniques will all work!

    Best Regards,

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