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Thread: why this point is not essential singular point

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

    $\displaystyle f(x)=\frac{e^{\frac{1}{z-1}}}{e^z-1}$
    if we go $\displaystyle z=0^+$ f(z)=+infinity
    if we go $\displaystyle z=0^-$ f(z)=-infinity
    we dont have a limit here
    so why its a pole
    ??
    by my definition a pole is when $\displaystyle f(z)=\frac{g(z)}{(a-z)^n}$
    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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    Quote Originally Posted by transgalactic View Post
    $\displaystyle f(x)=\frac{e^{\frac{1}{z-1}}}{e^z-1}$
    if we go $\displaystyle z=0^+$ f(z)=+infinity
    if we go $\displaystyle z=0^-$ f(z)=-infinity
    we dont have a limit here
    so why its a pole
    ??
    by my definition a pole is when $\displaystyle f(z)=\frac{g(z)}{(a-z)^n}$
    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 $\displaystyle f(z)$ has a zero of order $\displaystyle m$ at $\displaystyle z_0$ then $\displaystyle \frac{1}{f(z)}$ has a pole of order $\displaystyle m$ at $\displaystyle z_0$. Now use this with $\displaystyle f(z)=e^z-1$. Note however that the function does have an essential singularity at $\displaystyle z=1$. Another thing, your argument is wrong since there is no $\displaystyle +\infty$ or $\displaystyle -\infty$ when dealing with complex functions (only a single point $\displaystyle \infty$).
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  3. #3
    MHF Contributor
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    thanks
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