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Let $z_0$ be a non-removable isolated singularity of $f$. Show that $z_0$ is then an essential singularity of $\exp(f)$.

Hello, unfortunately I do not know how to proof that. To my opinion one has to consider two cases:

  1. $z_0$ is a pole of order $k$ of $f$.

  2. $z_0$ is an essential singularity of $f$.

Stefan Hamcke
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3 Answers3

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We can also look at it from the other side.

If $z_0$ is a removable singularity of $e^f$, then $\lvert e^{f(z)}\rvert < K$ in some punctured neighbourhood of $z_0$. Since $\lvert e^w\rvert = e^{\operatorname{Re} w}$, that means $\operatorname{Re} f(z) < K'\; (= \log K)$ in a punctured neighbourhood $\dot{D}_\varepsilon(z_0)$ of $z_0$, and that implies that $z_0$ is a removable singularity of $f$. (Were it a pole, $f(\dot{D}_\varepsilon(z_0))$ would contain the complement of some disk $D_r(0)$; were it an essential singularity, each $f(\dot{D}_\varepsilon(z_0))$ would be dense in $\mathbb{C}$ by Casorati-Weierstraß; in both cases $\operatorname{Re} f(z)$ is unbounded on $\dot{D}_\varepsilon(z_0)$.)

If $z_0$ were a pole of $e^f$, it would be a removable singularity of $e^{-f}$, hence $z_0$ would be a removable singularity of $-f$ by the above, hence $z_0$ would be a removable singularity of $f$, and therefore a removable singularity of $e^f$ - contradiction.

Daniel Fischer
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  • I can't understand that if $z_0$ is a removable singularity of $e^f$, why $z_0$ is a removable singularity of $f$? Why can't it be a pole? Please explain. –  Sep 28 '13 at 20:22
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    That should have been a new question, this part is for answers to the question. But to answer it: if $z_0$ is a pole of $f$, then $f({ z : 0 < \lvert z-z_0\rvert < \varepsilon})$ contains the complement of a disk ${ \lvert z\rvert \leqslant K}$, in particular a half plane ${\Im z > K}$. Since $e^z$ has the period $2\pi i$, the image of that half plane under the exponential function is the entire plane minus $0$, so $e^f$ has an essential singularity in $z_0$. – Daniel Fischer Sep 28 '13 at 20:28
  • @DanielFischer I just wanted to add something as this sentence is wrong: "If $Re(f(z))$ is bounded by some $K'$ in a punctured neighbourhood of $z_0$ it implies that $z_0$ is a removable singularity." As $|f(z)|^2=Re(f(z))^2+Im(f(z))^2$ we could have that $Im(f(z))$ grows arbitrarily big and that $|f(z)|$ isn't bounded in some punctured neighbourhood of $z_0$. You have to argue otherwise – Thomas Produit Feb 04 '14 at 12:48
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    @ThomasProduit The point is that that cannot happen. If $\operatorname{Re} f(z)$ is bounded on $D_r(z_0)\setminus {z_0}$, and $f$ is holomorphic there, then it follows that $\operatorname{Im} f(z)$ is bounded on $D_\rho(z_0)\setminus {z_0}$ for some $0 < \rho \leqslant r$. For in a punctured neighbourhood of a pole or an essential singularity, the real part of a holomorphic function is unbounded. – Daniel Fischer Feb 04 '14 at 12:55
  • I just find it not so straightforward then as I didn't know that it followed directly. I will search about it. Thanks – Thomas Produit Feb 04 '14 at 12:58
  • @ThomasProduit The part in parentheses after it explains it. One needs some results of complex analysis to show it, it does not follow directly. – Daniel Fischer Feb 04 '14 at 13:00
  • Mmmh yes indeed.. – Thomas Produit Feb 04 '14 at 13:02
  • As you're 'here' can you just explain this implication ? $z_0$ pole of $e^f \Rightarrow z_0$ removable singularity of $e^{-f}$ – Thomas Produit Feb 04 '14 at 13:05
  • @ThomasProduit If $z_0$ is a pole of $g$, then it is a removable singularity of $\dfrac{1}{g}$. Apply that to $e^{-f} = \dfrac{1}{e^f}$. – Daniel Fischer Feb 04 '14 at 13:07
  • yes of course thanks for everything ! – Thomas Produit Feb 04 '14 at 13:09
  • Regarding your explanation that if $z_0$ is a pole of then it's an essential singularity of $e^f$, you claim that the image of some open disk contains the complement of a disk. The definition of being a pole implies that the image of some disk is contained in the complement of a disk, I fail to see the inclusion in the other direction. How do you find a disk, such that any $z$ outside has a preimage? – leo Jun 23 '15 at 02:06
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    @leo It's the open mapping theorem. If $z_0$ is a pole of $f$, then $g = 1/f$ is non-constant and $g(z_0) = 0$. By the open mapping theorem, for every neighbourhood $U$ of $z_0$, $g(U) \supset D_\varepsilon(0)$ for some $\varepsilon > 0$. Then $f(U\setminus {z_0}) \supset \mathbb{C}\setminus \overline{D_{1/\varepsilon}(0)}$. – Daniel Fischer Jun 23 '15 at 08:31
  • @Daniel Fischer♦,can you please explain your last line in your last comment! – A learner Jun 20 '20 at 15:34
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The above proof is very nice, here I provide another idea. The real question is that if $f$ has a pole, then $e^f$ has an essential singularity.

Assume that the conclusion is false, then there exist an integer $k$ and an analytic function $g$ in some disc $|z|<\delta$, for example, $g(0)\neq0$, such that $e^f=z^kg$,

$$f'=\frac{(e^f)'}{e^f}=\frac{k}{z}+\frac{g'}{g}$$

integrating it on the contour $|z|=\frac{\delta}{2}$ we find $k=0$, since $g$ is analytic. Using the fact that $g(0)\neq0$ we can find another analytic function $h$ in the disc $|z|<\delta$(actually we should shrink this disc), such that $e^f=g=e^h$, so $f=h+2n\pi i$, $f$ has a removable singularity, contradiction!

Kira Yamato
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Under aforementioned conditions the function $f(z)$ may be expanded about $z = z_0$ up to a certain radius (radius of convergence) namely \begin{equation*} f(z) = \sum_{k= -\infty}^{+ \infty} a_k (z-z_0)^k = \sum_{k= 1}^{+\infty} \frac{a_{-k}}{ (z-z_0)^k} + \sum_{k= 0}^{+ \infty} a_k (z-z_0)^k = f_0(z) + f_1(z); \end{equation*} here we note the principal part $f_0(z)$ and the non-principal holomorphic part $f_1(z)$.

Because the isolated singularity of $f(z)$ at $z=z_0$ is not removable, it means that we have either a pole of order $n \in \mathbb{N}$ or an essential singularity. In case of a pole at least $a_{-n} \ne 0$, in case of an essential singularity infinitely many coefficients $a_{-k},\ k \in \mathbb{N}$ are non-zero - in both cases the principal part $f_0(z)$ is not trivially equivalent to zero.

We may inspect $g(z) = e^{f(z)}$ and its expression about $z=z_0$ namely \begin{equation*} g(z) = e^{f(z)} %= e^{ f_0(z) + f_1(z)} = e^{f_0(z)} \cdot e^{f_1(z)} = = \sum_{j=0}^{+\infty} \frac{(f_0(z))^j}{j!} \cdot e^{f_1(z)} = \sum_{j=0}^{+\infty} \frac{1}{j!} \left( \sum_{k= 1}^{+\infty} \frac{a_{-k}}{ (z-z_0)^k} \right)^j \cdot e^{f_1(z)} = g_0(z) \cdot g_1(z); \end{equation*} here we note the non-trivial multiplicand $g_0(z)$ and the leftover $g_1(z) = e^{f_1(z)}$ is a holomorphic non-vanishing function in a neighborhood of $z=z_0$. However it is clear that the principal part of $g_0(z)$ exhibits an infinitude of summands in the obtained expression - it is equivalent then to $g(z)$ having an essential singularity at $z=z_0$.

Pranasas
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  • Isn't the expansion you used for $e$ (in which you substituted $f(z)$), the expansion of $e$ around $0$, and not around $z_0$? – Batrachotoxin Dec 13 '20 at 12:47