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Math Help - why this point is not essential singular point

  1. #1
    MHF Contributor
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    why this point is not essential singular point

    f(x)=\frac{e^{\frac{1}{z-1}}}{e^z-1}
    if we go z=0^+ f(z)=+infinity
    if we go z=0^- f(z)=-infinity
    we dont have a limit here
    so why its a pole
    ??
    by my definition a pole is when f(z)=\frac{g(z)}{(a-z)^n}
    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

    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
    ??
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  2. #2
    Super Member
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    México
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    Quote Originally Posted by transgalactic View Post
    f(x)=\frac{e^{\frac{1}{z-1}}}{e^z-1}
    if we go z=0^+ f(z)=+infinity
    if we go z=0^- f(z)=-infinity
    we dont have a limit here
    so why its a pole
    ??
    by my definition a pole is when f(z)=\frac{g(z)}{(a-z)^n}
    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

    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
    ??
    If f(z) has a zero of order m at z_0 then \frac{1}{f(z)} has a pole of order m at z_0. Now use this with f(z)=e^z-1. Note however that the function does have an essential singularity at z=1. Another thing, your argument is wrong since there is no +\infty or -\infty when dealing with complex functions (only a single point \infty).
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  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    thanks
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