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Homework Help: Why this point is not essential singular point

  1. Jul 12, 2010 #1
    if we go [tex]z=0^+[/tex] f(z)=+infinity
    if we go [tex]z=0^-[/tex] f(z)=-infinity
    we dont have a limit here
    so why its a pole

    by my definition a pole is when [tex]f(z)=\frac{g(z)}{(a-z)^n}[/tex]

    if g is analitical at g(a) then 'a' in th n'th order pole

    but in my case there is no (a-z)^n representation
    and i proved that there is no limit
    Last edited: Jul 12, 2010
  2. jcsd
  3. Jul 12, 2010 #2
    in my exam i have much complicated functions
    and i cant develop it into a series every time

    i tried to solve this one strictly by the limit tests
    where am i wrong in my limit test
    where is my conclusion wrong
  4. Jul 12, 2010 #3
    i was told that there is no +infinity and -infinity in complex functions
    only infinity

    but i have a solved question from a test which is solved exactly like it you said was wrong
    and got full credit

    maybe i am missing something here

  5. Jul 12, 2010 #4
    Functions don't have a "limit" at a simple pole (like this one), they have aribitarily large norm.

    At poles, the values "wrap around" the complex plane at infinity, in other words, the function hits all complex numbers with arbitarily large norm. You always see a complex function going to the whole set of numbers [itex]re^{i\theta}[/itex] for [itex]\theta\in [0,2\pi)[/itex] and [itex]r[/itex] aribtraily large.
  6. Jul 12, 2010 #5
    i tried to develop it into a series
    if my development is correct how whould you continue to develop it
    so we will get a series with only one z variable

    i cant make a guess here its all has to be series with only one z in a certain power in order to make a conclution
  7. Jul 12, 2010 #6
    i understand why
    [MATH]\sin \left( \frac{1}{z} \right)[/MATH] has an essential singularity at [MATH]0[/MATH]

    but how we can take any function we want and say sin(f) has significant point
    at the pole of f
    like here
    [MATH]\sin ( f(z))[/MATH]
    [MATH]f(z)= \frac{z}{z+1}[/MATH]
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