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Math Help - complex analytic functions

  1. #1
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    complex analytic functions

    Prove there exists no function F(z) analytic in the annulus D: 1<|z|<2 such that F'(z) = 1/z.
    I assumed there was such a function.
    let F(z)=Log(z) + c
    F'(z)= 1/z
    I don't know how to do this.
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  2. #2
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    If f(z) is analytic on an (non-trivial) open set S and e^{f(z)} = z for all point z\in S then f(z) = \ln |z| + i\arg(z) for some particular definition of \arg z. Thus, our goal is to prove that F(z) as you defined it above has the property that e^{F(z)} = z but then that would mean F(z) = \ln |z| + i \arg(z) for some particular definition of of the argument, but then this contradicts the fact that F is analytic on 1<|z|<2 because the branch of \arg z will have to hit the annulus as it leaves the origin. There is just one detail I do not know how to show, maybe it is not needed: there is a point z_0 on the annulus so that e^{F(z_0)} = z_0. With that we can complete the proof. Suppose that F'(z) = 1/z and consider g(z) = ze^{-F(z)} then g'(z) = e^{-F(z)} - z(1/z)e^{-F(z)} = 0 and this means g(z) = k for some complex number k because the annulus is a region (open connected set). This means, ze^{-F(z)} = k for all point on the annulus. In particular for z_0, which yields. z_0e^{-F(z_0)} = k \implies k=1. Thus, ze^{-F(z)} = 1 \implies e^{F(z)} = z. We have show F(z) is the inverse function of e^z and this completes the proof using the begining comments.
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