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Thread: isolated singularity

  1. #1
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    isolated singularity

    Locate each of the isolated singularities of the function:

    (e^z - 1) / z

    How do I do this? Thanks in advance...
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  2. #2
    MHF Contributor chisigma's Avatar
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    The answer is very easy: the function...

    $\displaystyle f(z)= \frac{e^{z}-1}{z}$ (1)

    ... is analytic on the whole complex plane...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Quote Originally Posted by jzellt View Post
    Locate each of the isolated singularities of the function:

    (e^z - 1) / z

    How do I do this? Thanks in advance...
    z = 0 is a removable singularity: Removable singularity - Wikipedia, the free encyclopedia
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  4. #4
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    Quote Originally Posted by chisigma View Post
    The answer is very easy: the function...

    $\displaystyle f(z)= \frac{e^{z}-1}{z}$ (1)

    ... is analytic on the whole complex plane...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$

    I'm new to this stuff, but I wanted to check myself. I think it goes something like this:

    $\displaystyle f(z)$ clearly has a singularity at $\displaystyle z=0$. I mean, it's undefined there so $\displaystyle f(z)$ itself is not what is holomorphic, right? But because 0 is a removable singularity it is possible to define a function $\displaystyle \hat{f}(z) $ where the new function is everywhere equal to the original except at the singularity, where we can define $\displaystyle \hat{f}(0)$ to be the limit of $\displaystyle f(z)$ as $\displaystyle z\to 0$, which is legitimate because this limit exists and is finite. Have I understood this correctly?

    Also, what is the justification that 0 is a removable singularity for this function? Is it, in fact, because $\displaystyle \lim_{z\to 0} f(z)$ exists and is finite? If so, would this need to be shown for completeness, or could it be taken as obvious enough?
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