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Complex Analysis: Poles and Singularities

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    According to my book, a pole [itex]z_0[/itex] of a function f(z) is defined as

    [tex]
    \mathop {\lim }\limits_{z \to z_0 } f(z) = \infty.
    [/tex]

    Now lets look at e.g. f(z) = exp(z). Thus we have a singularity for z = infinity, since the limit in this case is infinity.

    This is what I don't understand: Definitions aside, f(z) = exp(z) is still analytic when it is infinite, so how can there be a singularity there?
     
  2. jcsd
  3. May 9, 2009 #2

    Defennder

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    Infinity isn't a complex number. It's not a number, so you can't say there's singularity at infinity. Usually a singularity is identified by noting that the denominator of f(z) approaches 0 when z -> z_0, if f(z) may be written as a fraction of 2 functions.
     
  4. May 9, 2009 #3

    HallsofIvy

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    While it is true that "infinity" is not a complex number, texts on complex analysis often talk about "poles at infinity" or "singularities at infinity". To quote Complex Analysis by Theodore Gamelin, "We say that a function, f(z), has an isolated singularity at infinity if it is analytic outside some bounded set." In other words, ez has a "singularity at infinity" precisely because it is analytic for large |z|.
     
    Last edited: May 9, 2009
  5. May 9, 2009 #4
    The functional behavior of [tex]f(z)[/tex] at infinity is determined by the function's behavior at the point [tex]z=1/w[/tex] at [tex]w=0[/tex]. [tex]f(z)[/tex] then is analytic, has a pole, essential singularity if [tex]f(1/w)[/tex] has the same property at [tex]w=0[/tex]. Then [tex]e^{z}[/tex] has an essential singularity at the point of infinity. In the latter case, we encounter Picards's First Theorem: Any entire function that is not a polynomial has an essential singularity at infinity.

    May I recommend "Basic Complex Analysis" by Marsden and Hoffman. This is my favorite introductory text on Complex Analysis and if you look into it I hope you agree :).
     
  6. May 12, 2009 #5
    What about e.g.

    [tex]
    f(z) = \frac{\sin z}{z^2}.
    [/tex]

    This is not an entire function, but it is not defined for z going to infinity (since we by the above metod would have 1/0 in the sine). Is infinity then also an essential singularity?
     
  7. May 12, 2009 #6

    Dick

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    Look at the limit of sin(z)/z^2 as you approach infinity via z=x real, and z=iy imaginary. The real limit is 0 and the imaginary limit is infinity. It's pretty clearly an essential singularity, isn't it?
     
  8. May 13, 2009 #7
    So I just have to find the limit when z goes to the singularity from different ways. If the limits are not the same, then it is an essential singularity? (Is this equivalent of the limit not existing?)

    If you can confirm the following example, then I think I have understood it (I used your above method):

    We look at that [itex]f(z) = \tan (1/z)[/itex] for z -> 0. We have the following limits:

    [tex]
    \mathop {\lim }\limits_{z \to 0} \frac{{\sin (1/z)}}{{\cos (1/z)}} = \mathop {\lim }\limits_{z \to \infty } \frac{{\sin (z)}}{{\cos (z)}}.
    [/tex]

    If we let z = x, then we get an "oscillating limit". If we let z = iy, we get infinity, which is not oscillating. Thus z = 0 is an essential singularity (Am I correct?).

    Thank you very much.
     
    Last edited: May 13, 2009
  9. May 13, 2009 #8

    Dick

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    Check on the oscillating behavior for real z, but if z=iy you have basically sinh(y)/cosh(y) which does have a finite limit. But the real behavior means it doesn't have a limit and doesn't behave like a pole. What's left?
     
  10. May 13, 2009 #9
    Then its an essential singularity. But this analysis seems rather long: Most of the functions we look at in class are hardly this simple. From http://en.wikipedia.org/wiki/Essential_singularity it says:

    "If neither [itex]\mathop {\lim }\limits_{z \to a} f(z)[/itex] nor [itex]\mathop {\lim }\limits_{z \to a} 1/f(z) [/itex] exists, then a is an essential singularity of both f and 1/f."

    Couldn't we just say, then, that since [itex]f(z) = \tan (1/z)[/itex] is not defined for z=0, then the limit does not exist. Thus it must be an essential singularity. Is this reasoning valid?
     
  11. May 13, 2009 #10

    Dick

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    'Not defined' doesn't imply 'limit does not exist'. 1/z isn't defined at z=0, but it does have a 'limit' as z->0 (in the extended complex plane). It goes to infinity. It's a pole. Not an essential singularity. tan(1/z) has oscillatory behavior as z->0. It's all about the limiting behavior as you approach the singularity.
     
  12. May 13, 2009 #11
    Are you talking about 1/z or tan(1/z)? Because tan(1/z) was essential, correct?

    So my approach when classying an essential singularity is the following (will you be so kind to confirm if this is correct, please?):

    Find the limit for z -> z0 (where z0 is the singularity) from different directions. If they are not the same, then it is an essential singularity.
     
  13. May 13, 2009 #12
    May I make a suggestion: look at the Laurent series as one way to determine if a function has an essential singularity. By definition, a function with an essential singularity, has an infinite singular component (the 1/(z-za)^n terms in its Laurent expansion). Now consider an analytic function that is not a polynomial like sin(z). It's Laurent series at zero is an infinite Taylor series of z^n terms. However, when you substitute 1/z into that series, it then becomes an infinite series of 1/z^n terms thus making sin(1/z) have an essential singularity at zero or equivalently, sin(z) having an essential singularity at infinity.
     
  14. May 13, 2009 #13

    Dick

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    With the proviso that 'infinity' is a valid limit, yes. That's conceptually what is 'essential' about the singularity. There are easier criteria for finding this out than checking all possible directions, as squidsoft has just pointed out. And, yes, I was talking about 1/z.
     
    Last edited: May 13, 2009
  15. May 13, 2009 #14
    Ok, I will do some more problems on this topic. Thanks to everybody who participated and for your patience.
     
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