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Describing Singularities

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    2cxamo.jpg

    3. The attempt at a solution

    Both [itex]\displaystyle \frac{\cos(z)-1}{z^2}[/itex] and [itex]\displaystyle \frac{\sinh(z)}{z^2}[/itex] have 1 singular point at [itex]z=0[/itex].

    For (a):

    z=0 is a removable singularity since defining f(0)=1 makes it analytic at all [itex]z\in\mathbb{C}[/itex].

    z=0 is isolated since f(z) is analytic for 0<|z|<1. But z=0 is not a pole since cos(0)-1 =0, and so z=0 is an essential singularity.

    For (b):

    z=0 is a removable singularity since defining f(0)=1 makes it analytic at all [itex]z\in\mathbb{C}[/itex].

    z=0 is isolated since f(z) is analytic for 0<|z|<1. But z=0 is not a pole since sinh(0)=0, and so z=0 is an essential singularity.

    Is this correct?
     
  2. jcsd
  3. Jan 21, 2012 #2

    Dick

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    Parts of it might be true. You said basically the same thing about both functions and you didn't prove anything you said. Give some arguments. If f(z)=sinh(z)/z^2, why does defining f(0)=1 make it analytic on C?
     
  4. Jan 21, 2012 #3
    Probably because I'm not understanding the definitions correctly!

    These are my set of definitions:
    1zfn5kz.jpg

    I think for both (a) and (b), z=0 is an isolated singularity but not a pole, so an essential singularity. But they probably aren't removable.
     
  5. Jan 21, 2012 #4

    Dick

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    The definitions will be clearer to you if you look at a power series expansion of each function around z=0.
     
  6. Jan 21, 2012 #5
    I don't like how some of these definitions are given so if I use this definition of pole:

    11c8nlc.jpg

    Clearly [itex]z_0=0[/itex] is an isolated singularity since it is the only singularity for both (a) and (b).

    (a) [itex]\displaystyle \lim_{z\to 0} \;(z-0)^N f(z) = \lim_{z\to 0} \; z^{N-2} (\cos(z)-1) = 0 \;\; \forall \;N>0[/itex] so [itex]z_0=0[/itex] is not a pole. Hence it is an essential singularity.

    (b) If N=1 then [itex]\displaystyle \lim_{z\to 0} \;(z-0) f(z) = \lim_{z\to 0} \frac{\sinh(z)}{z} = 1 \neq 0[/itex] so [itex]z_0=0[/tex] is a simple pole (of order 1). What would be the strength of the pole? It is not an essential singularity.

    I'm not understanding how to see if 0 is a removable singularity in each case?
     
  7. Jan 21, 2012 #6

    Dick

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    You know how to expand cos(z) and sinh(z) in a power series around z=0. Put those expansions into the two functions and simplify. See what you think. Then look back at the definitions.
     
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