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Math Help - f and e^f have no common pole?

  1. #1
    ynj
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    f and e^f have no common pole?

    Prove that an isolated sigularity of f(z) cannot be a pole of e^{f(z)}.
    I know the key is to prove that fand e^fhave no common pole, but how??
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    MHF Contributor Bruno J.'s Avatar
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    Hello! Here is an idea.

    It is easy to see that this is true if z_0 is an essential or removable singularity of f(z).

    In the case of an n-th order pole, then I claim that z_0 is an essential singularity of g(z)=e^{f(z)}. Clearly it is not a regular point, so it suffices to show that |g(z)| does not tend to infinity as z \rightarrow z_0. To do this, it suffices to show that every neighbourhood of z_0 contains points \zeta such that \mbox{Re } f(\zeta)<M, for some fixed M.
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  3. #3
    ynj
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    Quote Originally Posted by Bruno J. View Post
    Hello! Here is an idea.

    It is easy to see that this is true if z_0 is an essential or removable singularity of f(z).

    In the case of an n-th order pole, then I claim that z_0 is an essential singularity of g(z)=e^{f(z)}. Clearly it is not a regular point, so it suffices to show that |g(z)| does not tend to infinity as z \rightarrow z_0. To do this, it suffices to show that every neighbourhood of z_0 contains points \zeta such that \mbox{Re } f(\zeta)<M, for some fixed M.
    Hmm...Would you please give some more details about your last step?
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    MHF Contributor Bruno J.'s Avatar
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    Hi Ynj!

    I know this is a really late answer but I had forgotten about this problem and I remembered it today and solved it. So here it is.

    Let f(z) have a pole at z_0. I claim that g(z)=e^{f(z)} has an essential singularity at z_0.

    The problem is solved if we can show that there exists a sequence \{z_n\} \rightarrow z_0 such that |e^{f(z_n)}| is bounded as n \rightarrow \infty; this is because if z_0 were a pole of g(z) then we would have g(z_n)\rightarrow \infty as n \rightarrow \infty. This is equivalent to finding a sequence such that \mbox{Re } f(z_n) remains bounded, since |e^{x+iy}|=|e^x|. Since we know that |f(z_n)| necessarily blows up to infinity as as n \rightarrow \infty, it is the imaginary part which must become very big.

    Consider F(z)=\frac{1}{f(z)}. This is clearly holomorphic at z_0 and F(z_0)=0. Let N_1\supset N_2 \supset \dots be a sequence of bounded open neighbourhoods of z_0 whose intersection is z_0. Then \{F(N_n)\} is a sequence of neighbourhoods of 0; in particular, each of the F(N_n) is bounded, and thus S_n=\{1/F(z) : z \in N_n\} is a sequence of neighbourhoods of infinity. So there are points of the imaginary axis in each of the S_n; pick z_n \in N_n such that 1/F(z) \in i\mathbb{R}. Then the sequence z_n converges to z and \mbox{Re } 1/F(z_n)=\mbox{Re } f(z_n) =0. Thus |e^{f(z_n)}|=1 for n \geq 1 and we are done.
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