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Complex analysis: Terminology

  1. Jun 12, 2009 #1
    Hi all

    We look at [itex]f(z)=\sqrt z [/itex]. Here the point z0=0 is a branch point, but can/is z0=0 also regarded as a zero?
  2. jcsd
  3. Jun 12, 2009 #2


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    Yes, the square root of 0 is 0.
  4. Jun 14, 2009 #3
    Thanks. It is kind of you to help me. I have another question related to complex analysis, which I hope you will help me with.

    Is it correct that [itex]z_0=\infty[/itex] is an essential singularity for [itex]f(z)=z\cos z[/itex]?
  5. Jun 14, 2009 #4
    Yes, any analytic function that is not a polynomial has an essential singularity at infinity.
  6. Jun 14, 2009 #5
  7. Jun 17, 2009 #6
    I have a question similar to my previous one, so I thought I could just ask it here, instead of creating a new thread.

    In my book it says: "A function f which is not analytic - or even necessarily defined - at a point z is said to have a singularity at z ...". In the case of [itex]f(z)=z\cos(1/z)[/itex], then the function is not defined for [itex]z_0=0[/itex], but the limit does exist, and it is 0.

    Does this mean that there is a singularity there or not? I believe that there is, but I would very much like for someone to confirm this.

    I really appreciate your help. Thank you very much in advance.
    Last edited: Jun 17, 2009
  8. Jun 17, 2009 #7
    The limit only exists if it doesn't depend from which way you approach z = 0. If you approach it from the real axis, then the limit is zero. But if you approach it from the imiaginary axis, the limit is not zero. So, you don't have a limit that is independent on the way you approach the limit point so "the" limit for z = 0 does not exist.
  9. Jun 17, 2009 #8
    I see. Would you advise me just to look for the points where the function is not defined, just like my book says? This way (if you do advise me to do this), then I don't have to worry about limits, and which way they are approached from.

    Thanks for replying.
  10. Jun 17, 2009 #9

    That won't work, becuase if the function is sin(z)/z, then the function is not defined at z = 0 either, yet the limit does exist and is equal to 1.

    So, you have a "removalbe singularity", which means that you can define the value at the undefined point z = 0 to be equal to 1, and then the function is continuous and even analytic.

    Now, when you are dealing with functions that are analytic except for possible exception points, the only tyupes of sungularities are poles, branch point singularites or essential singularities.

    Poles are singularties around which the expansion starts with a negative power. If the expnsion around z = z_0 starts like:

    a (z-z_0)^(-n) + b (z-z_0)^(-n+1) + ...

    then we say that the pole is of order n. Then mltiplying the function by (z-z_0)^n would yield an analytic function (you can then define it at the point z = z_0 such that it become analytic there).

    A function f(z) has an essential singularity at z = z_0 if there does not exist a finite n such that multiplying the function by (z-z_0)^n will make the function analytic. So for all n > 0 the function (z-z_0)^n f(z) is singular at z = z_0.

    I case of your example, you can see from the series expansion of
    cos(1/z) that it has an essential singularity for z = 0.
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