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Math Help - Removable singularity

  1. #1
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    Removable singularity

    Let f be analytic in \{ z:0<|z-\alpha |<r\} and has a singularity at \alpha. Supppose there exists a neighbourhood \{ z:0<|z-\alpha |<\epsilon \} where Re(f(z))>0. Show that \alpha is a removable singularity.

    I've tried looking at e^{f(z)} in order to show that f is bounded at a neighbourhood of \alpha, but all I get is |e^{f(z)}|=e^{Re(f(z))}>1 so that dosent help, I also looked at e^{-f(z)} but got nowhere. Need some direction here please.
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  2. #2
    Senior Member Tinyboss's Avatar
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    You know it's not essential, because an analytic function attains almost every value in any neighborhood of an essential singularity. So you have to rule out a pole. So suppose there's a pole at alpha, and consider [f(z)]^{-1}. It has a removable singularity: h(z)=[f(z)]^{-1} for z\ne\alpha and h(\alpha)=0 is analytic on the original domain.

    What does its image look like?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Tinyboss View Post
    You know it's not essential, because an analytic function attains almost every value in any neighborhood of an essential singularity.

    Good, an elegant application of the Casaroti-Weierstrass theorem.
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  4. #4
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    Quote Originally Posted by Tinyboss View Post
    You know it's not essential, because an analytic function attains almost every value in any neighborhood of an essential singularity. So you have to rule out a pole. So suppose there's a pole at alpha, and consider [f(z)]^{-1}. It has a removable singularity: h(z)=[f(z)]^{-1} for z\ne\alpha and h(\alpha)=0 is analytic on the original domain.

    What does its image look like?
    I'm not sure I understand why h is analytic. We know g(z)=|z| is not analytic at z=0 ?

    It's image is the positive half of the real line, including zero. How does that contradict \alpha being a pole?
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  5. #5
    Senior Member Tinyboss's Avatar
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    I was a little sloppy--h won't be analytic on all the original domain, but only where f is nonzero. But since we're assuming f has a pole at alpha, then f is nonzero on some neighborhood. If the real part of z is positive, then so is the real part of 1/z. But zero is also in the image of h, violating the open mapping property of analytic maps.
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  6. #6
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    I actually mis-spoke earlier. Checked the book again and |z| is not analytic in the entire complex plane.

    So now I'm completely confused
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by skyking View Post
    Let f be analytic in \{ z:0<|z-\alpha |<r\} and has a singularity at \alpha. Supppose there exists a neighbourhood \{ z:0<|z-\alpha |<\epsilon \} where Re(f(z))>0. Show that \alpha is a removable singularity.

    I've tried looking at e^{f(z)} in order to show that f is bounded at a neighbourhood of \alpha, but all I get is |e^{f(z)}|=e^{Re(f(z))}>1 so that dosent help, I also looked at e^{-f(z)} but got nowhere. Need some direction here please.
    A 'counterexample': let's suppose f(z)= \ln \frac{1}{z}... setting z= r\ e^{i\ \varphi} is f(z)= - \ln r - i\ \varphi so that in 0<|z|<1 f(z) is analytic and is also \text {Re} \{f(z)\} >0 ... but , in z=0 , f(z) certainly doesn't have a 'removable singularity'...

    Kind regards

    \chi \sigma
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  8. #8
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    Quote Originally Posted by chisigma View Post
    A 'counterexample': let's suppose f(z)= \ln \frac{1}{z}... setting z= r\ e^{i\ \varphi} is f(z)= - \ln r - i\ \varphi so that in 0<|z|<1 f(z) is analytic
    Kind regards

    \chi \sigma
    But, Logz is not even defined for negative values, and in your 'counterexample' \frac {1}{z} clearly takes real negative values in 0<|z|<1 so I think this example is not valid.

    Still need help showing the above, still stuck
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by skyking View Post
    But, Logz is not even defined for negative values, and in your 'counterexample' \frac {1}{z} clearly takes real negative values in 0<|z|<1 so I think this example is not valid.

    Still need help showing the above, still stuck
    Just for curiosity: where did You read that?... perhasps in the Boys Scout Manual?... z is a complex variable that can be written, according to De Moivre's formula, as z= r\ e^{i\ \varphi} so that the [complex] logarithm \ln z = \ln r + i\ \varphi is defined in the entire \mathbb{C} with the only exception of z=0...

    Kind regards

    \chi \sigma
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  10. #10
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    Quote Originally Posted by chisigma View Post
    z is a complex variable that can be written, according to De Moivre's formula, as z= r\ e^{i\ \varphi} so that the [complex] logarithm \ln z = \ln r + i\ \varphi is defined in the entire \mathbb{C} with the only exception of z=0...
    You have to be very careful here, because the complex logarithm is not a single-valued function in general. It depends on the argument \varphi, which is only defined up to a multiple of 2\pi. In particular, the function \ln\frac1z cannot be defined in a way that makes it continuous in a (punctured) neighbourhood of the origin. So this function cannot be analytic in any such neighbourhood.
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  11. #11
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Opalg View Post
    You have to be very careful here, because the complex logarithm is not a single-valued function in general. It depends on the argument varphi, which is only defined up to a multiple of 2pi. In particular, the function lnfrac1z cannot be defined in a way that makes it continuous in a (punctured) neighbourhood of the origin. So this function cannot be analytic in any such neighbourhood.
    The question has been investigated in...

    http://www.mathhelpforum.com/math-he...st-167358.html

    ... and here it has been demonstrated, using the Cauchy-Riemann relations, that \ln z is analytic everywhere in \mathbb{C} with the only exception of z=0. Regarding the question proposed by skyking I suggest, in order to avoid problems, to modify the first line as follows...

    Let f(z) be analytic in \{z : 0<|z-\alpha|< r \}, has a singularity at \alpha and \alpha is not a branch point of f(z) ...

    For more informations see...

    Logarithmic Branch Point -- from Wolfram MathWorld

    Kind regards

    \chi \sigma
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  12. #12
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Opalg View Post
    You have to be very careful here, because the complex logarithm is not a single-valued function in general. It depends on the argument \varphi, which is only defined up to a multiple of 2\pi. In particular, the function \ln\frac1z cannot be defined in a way that makes it continuous in a (punctured) neighbourhood of the origin. So this function cannot be analytic in any such neighbourhood.
    Of course is \ln \frac{1}{z} = - \ln z so that what holds for \ln z holds for -\ln z too...

    Kind regards

    \chi \sigma
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  13. #13
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    Quote Originally Posted by chisigma View Post
    Just for curiosity: where did You read that?... perhasps in the Boys Scout Manual?...

    Kind regards

    \chi \sigma
    Very classy...
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