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Math Help - True or false about analycity

  1. #1
    MHF Contributor arbolis's Avatar
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    True or false about analycity

    True or false? Justify.
    Let z_0 be an isolated singularity of f(z) such that Res(f,z_0)=0. Then z_0 is a removable singularity and thus f(z) is analytic in a neighborhood of z_0.
    Attempt: True. I know that if z_0 is a removable singularity, then f can be extended to an analytic function by giving a value to f(z_0). Thus I also believe that f is bounded if its singularity is removable.
    Though formally I don't know how to prove it. If f(z)=\sum _{n=-\infty}^{+\infty} a_n (z-z_0)^n (I know I assume that f has a Laurent series representation, I'm not sure I can assume that) then I'm told that \lim _{z \to z_0} \sum _{n=-\infty}^{+\infty} a_n (z-z_0)^{n+1}=0. I'm not going far with this.
    Any idea?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by arbolis View Post
    True or false? Justify.
    Let z_0 be an isolated singularity of f(z) such that Res(f,z_0)=0. Then z_0 is a removable singularity and thus f(z) is analytic in a neighborhood of z_0.
    Attempt: True. I know that if z_0 is a removable singularity, then f can be extended to an analytic function by giving a value to f(z_0). Thus I also believe that f is bounded if its singularity is removable.
    Though formally I don't know how to prove it. If f(z)=\sum _{n=-\infty}^{+\infty} a_n (z-z_0)^n (I know I assume that f has a Laurent series representation, I'm not sure I can assume that) then I'm told that \lim _{z \to z_0} \sum _{n=-\infty}^{+\infty} a_n (z-z_0)^{n+1}=0. I'm not going far with this.
    Any idea?
    What about the function  \displaystyle f(z)=e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n} .
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by chiph588@ View Post
    What about the function  \displaystyle f(z)=e^{1/z}-\frac1z = 1+\sum_{n=2}^{\infty} \frac{1}{n!z^n} .
    Oh, that would be a counter example?
    I realize that e^{1/z}=\sum _{n=0}^{+\infty} \frac{1}{z^n n!}=1+ \sum _{n=1}^{+\infty} \frac{1}{z^n n!}.
    But I'm not sure how to reach e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}.
    Also, Res(e^{1/z},z=0)=1 while  Res(-1/z,z=0)=-1 so indeed Res(f,z=0)=0. So the singularity of e^{1/z} is essential (there's an infinity of negative powers of z in the series expansion of the function) and so should be the one of f... So that after all, z isn't bounded nor analytic in a neighborhood of z=0?
    Should I use Cauchy-Riemann equations to show that f isn't analytic? How would you do it?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by arbolis View Post
    Oh, that would be a counter example?
    I realize that e^{1/z}=\sum _{n=0}^{+\infty} \frac{1}{z^n n!}=1+ \sum _{n=1}^{+\infty} \frac{1}{z^n n!}.
    But I'm not sure how to reach e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}.
    Also, Res(e^{1/z},z=0)=1 while  Res(-1/z,z=0)=-1 so indeed Res(f,z=0)=0. So the singularity of e^{1/z} is essential (there's an infinity of negative powers of z in the series expansion of the function) and so should be the one of f... So that after all, z isn't bounded nor analytic in a neighborhood of z=0?
    Should I use Cauchy-Riemann equations to show that f isn't analytic? How would you do it?
     \displaystyle e^{1/z} = \sum_{n=0}^\infty \frac{1}{n!z^n} = 1+\frac1z+\sum_{n=2}^\infty \frac{1}{n!z^n}

    Therefore  \displaystyle e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}
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    MHF Contributor Bruno J.'s Avatar
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    Or just look at 1/z^2... it has a vanishing residue at 0.
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Or just look at 1/z^2... it has a vanishing residue at 0.
    Ok thank you. And to show it's not analytic, I could either show that f'(z) doesn't exist or show that C-R equations aren't satisfied?
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Bruno J. View Post
    Or just look at 1/z^2... it has a vanishing residue at 0.
    But that's a removable singularity...
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by chiph588@ View Post
    But that's a removable singularity...
    And I think it's analytic in a neiborhood of 0. Is that right? I've shown that f'(z)=\frac{-2}{z^3} for any z \neq 0.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    But that's a removable singularity...
    No, it's not! It's a second-order pole.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by arbolis View Post
    And I think it's analytic in a neiborhood of 0. Is that right? I've shown that f'(z)=\frac{-2}{z^3} for any z \neq 0.
    No, it's not analytic at the origin. It's analytic in a punctured neighbourhood of the origin, but not at the origin. (Unless you consider poles to be regular points, which is sometimes done.)
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  11. #11
    MHF Contributor chiph588@'s Avatar
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    Whoops! I forgot my definitions!

    I thought a removable singularity was when you could multiply by a polynomial of an arbitrary degree to remove the singularity.
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  12. #12
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Whoops! I forgot my definitions!

    I thought a removable singularity was when you could multiply by a polynomial of an arbitrary degree to remove the singularity.
    No worries!
    What you describe is a pole. A removable singularity is just camouflaged regular point!
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  13. #13
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Bruno J. View Post
    No, it's not analytic at the origin. It's analytic in a punctured neighbourhood of the origin, but not at the origin. (Unless you consider poles to be regular points, which is sometimes done.)
    Ah I see. I'm not sure it would fit as a counter example due to the "punctured" neighborhood.
    By the way, last question before my exam: if f(z)=g(z)+h(z), does Res(f,z_0)=Res(g,z_0)+Res(h,z_0) assuming that f (and maybe g and h) has a pole at z_0.
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  14. #14
    MHF Contributor Bruno J.'s Avatar
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    Yup, by linearity of the integral!
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