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I Complex analysis - removable singular points

  1. Sep 12, 2018 #1

    dyn

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    Hi. I have 2 questions regarding removable singular points.
    1 - the residue at a removable singularity is always zero so by the residue theorem the integral around a closed simple contour is zero. Cauchy's theorem states the integral around a simple closed contour for an analytic function is zero so am I right in thinking that the integral being zero does not imply the function is analytic ? As it is also zero for functions with a removable singularity.

    2 - I have seen the following example in a book where it is shown that (z-sinz)/z3 has a removable singularity at z=0 but then it states that the series converges for all values of z but surely the function is undefined at z=0 ?
    Thanks
     
  2. jcsd
  3. Sep 12, 2018 #2

    mathwonk

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    1. there is no difference in integrating an analytic function versus a function with a removable singularity. more interesting is the case of a function which does not have a removable singularity but still has integral zero due to there being a pole there of higher order and residue zero. i.e. integral = 0 just means the term 1/z has coefficient zero, but therte could still be terms of order 1/z^2 or 1/z^3 etc with non zero coefficients but the integral will not detect that.

    2. in this example sin has only odd degree terms of positive degree, and the linear term equals z, so subtracting off z reduces it to beginning in degree 3. thus dividing by z^3 gives an analytic function (i.e. one with a removable singularity). do not obsess over whether a function has or has not yet been stated as defined at a point when it is perfectly capable of being well defined there. all functions with removable singularities should have them removed, before studying the function.
     
  4. Sep 12, 2018 #3

    dyn

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    Thanks for your reply.
    1- so if an integral around a closed loop is zero ; that doesn't tell me if the function is analytic or not ?
    2- You say that (z-sinz)/z3 is an analytic function ie. one with a removable singularity. Aren't they 2 different things ? A singular point is a point where a function is not analytic.
     
  5. Sep 13, 2018 #4

    mfb

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    A removable singularity just means you chose a poor way to define a function that could be analytic at that point. Like ##f(z)=\frac{z^2+z}{z}##. It has a removable singularity at z=0. You could have defined f(z)=z+1 and you wouldn't have had that issue.
     
  6. Sep 13, 2018 #5

    dyn

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    Thanks. So for the function f(z) = (z-sinz)/z3 ; it only converges if f(0) is defined and so it is analytic ? If f(0) is not defined it does not converge and is not analytic ?
     
  7. Sep 14, 2018 #6

    mfb

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    f(z) converges for z->0 which is independent of what happens at exactly 0.

    ##f(z)=\frac 1 6 - \frac{z^2}{120} + \frac{z^4}{5040}\pm ...##
     
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