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I have a Complex Analysis exam in 2 days. The last exam had, among other exercises, the following:

Let $f$ be a function holomorphic in $\mathbb{D}\smallsetminus\{0\}$ that does not have a removable singularity ad the origin.

  1. What kind of singularities can it have? Why?

  2. Show that the origin is an essential singularity for $e^f$.

Point one is fairly easy: it can either have a pole or an essential singularity, because those are the only possible isolated singularities except for a removable one, which is excluded by hypothesis, and that singularity is surely isolated since the function is holomorphic in the rest of the disk $\mathbb{D}$. But how do I go about the second one? I thought of trying to prove the conclusions of the Casorati-Weierstrass theorem, because that would exclude a removable singularity since the function would not be bounded, and a pole since the modulus would wildly oscillate and therefore not tend to infinity, which is equivalent to 0 being a pole. But I'm still stuck. I mean, if $f$ has real part going to $-\infty$ and imaginary part doing whatever it wants, then $e^f$ has a removable singularity at 0, yet $f$ still has a pole, since whatever the imaginary part does, the modulus will tend to infinity. Am I missing something or is this really an impossible exercise right out of an exam? And if not - which I guess is most likely - what am I missing, and how do I solve this?

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  • $\begingroup$ Luckily the real part of $f(z)$ does not tend to $-\infty$ as $z\to0$. Whether $f$ has a pole or an essential singularity you can show there exist $z_n\to0$ with $\Re f(z_n)\to-\infty$ and also $w_n\to0$ with $\Re f(w_n)\to\infty$. $\endgroup$ – David C. Ullrich Jul 12 '15 at 19:53
  • $\begingroup$ @David , How can you show that if $f$ has pole ? $\endgroup$ – d_e Jul 12 '15 at 20:24
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    $\begingroup$ If you're close enough to the pole then the highest-order term in the Laurent series is twice as large as the sum of all the other terms... $\endgroup$ – David C. Ullrich Jul 12 '15 at 20:32
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Hints: If $f$ has a pole at $0,$ then $f(\{0<|z|<r\})$ contains $\{|z|>R\}$ for some large $R.$ What can you then say about $\exp (\{|z|>R\})?$

If $f$ has an essential singularity at $0,$ then what can you say about $f(\{0<|z|<r\})?$ And then what can you say about $\exp (f(\{0<|z|<r\}))?$

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  • $\begingroup$ If $f$ has an essential singularity, that image is dense in $\mathbb{C}$, so it contains complex numbers with arbitrarily large real part, and then so must the image under the exponential. But how do I show that in the pole case I have all of $\{|z|>R\}$ inside the image? If I do, then I also have numbers with arbitrarily large and positive real part, so the removable singularity is excluded, and since I have arbitrarily big and negative real parts I can exclude a pole as well. But why is it impossible for the real part of $f$ to tend to $-\infty$ if $f$ has a pole? $\endgroup$ – MickG Jul 12 '15 at 20:16
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    $\begingroup$ For the case of a pole, I think David Ullrich's way is easier. There is a leading term for the principal part in the Laurent expansion, and that $\text {Re} f$ has arbitrarily large positive and negative values in any $\{0<|z|<r\}$ follows from that. To get the idea of how to prove it, first think about the case where $f(z) = 1/z^N + g(z),$ where $g$ is entire. $\endgroup$ – zhw. Jul 13 '15 at 3:23

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