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I am working on the following question:

Suppose $z_0 \in \mathbb{C}$ is an isolated singular point of the function f of a given type (removable, pole of order N, essential). Show that $z_0$ is an isolated singular point of

  • $g(z)=1/f(z)$
  • $h(z)=f^2(z)$

and find its type.

My approach:

First, to show that $z_0$ is still an isolated singular point I thought I could apply the chain rule and then get an expression with the derivative of $f$ and since we know $f$ is not analytic at the isolated singular points then $g$ and $h$ must not be analytic at the isolated singular points.

However, when I looked at the example $f(z)=\frac{e^z}{1+z^2}$ and $g(z)=\frac{1+z^2}{e^z}$ it is clear $f(z)$ has isolated singular points at $i$ and $-i$ but these are not isolated singular points of $g(z)$.

To determine the types of the singular points I attempted to use the limit definitions. So if $z_0$ is a pole of $f(z)$ then $\lim_{z\to z_0}f(z)= \infty$. Therefore, $\lim_{z\to z_0}g(z)= 1/\infty=0$ which is finite and therefore $z_0$ is a removable singularity of $g(z)$. Is this approach valid?

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  • $\begingroup$ That must be a mistake, I think: for $\;g\;$ that'd be an isolated zero, not pole. $\endgroup$
    – DonAntonio
    Apr 25, 2016 at 17:32
  • $\begingroup$ For removable and pole of order $N$, I would suggest using the definition of each type of pole to rewrite $f$, then use that expression for $f$ to rewrite $g$ and $h$. $\endgroup$ Apr 25, 2016 at 17:33
  • $\begingroup$ @Joanpemo that is the source of my confusion, the question makes it seems like all isolated singular points of $f$ must be isolated singular points of $g$ and $h$ $\endgroup$
    – klib
    Apr 25, 2016 at 17:35
  • $\begingroup$ @John Martin for removable I think I was still able to use a similar approach if $z_0$ is a removable singularity then the limit is finite and therefore it is a removable singularity of $g$ unless the limit is 0 in which case it is a pole of some order that we cannot determine $\endgroup$
    – klib
    Apr 25, 2016 at 17:38
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    $\begingroup$ @klib I see, and the simple function $\;f(z)=\frac1z\implies g(z)=z\;$ is a simple counter example. May you had to decide whether the claim is true or false? $\endgroup$
    – DonAntonio
    Apr 25, 2016 at 17:39

2 Answers 2

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For example, if "$g(z) = 1/f(z)$" is taken as the definition of $g$, it implies the domain of $g$ is all points $z$ in the domain of $f$ where $f(z) \ne 0$. It could be, as in your example, that you can take a formula $f(z) = \ldots$ and simplify $g(z) = 1/\ldots$ to an expression that is defined at some points where $f$ is not defined. That will just say that the singularity of $g$ at such a point is removable.

EDIT: In fact, part of the answer to this problem will say that if $f$ has a pole at $z_0$, $g$ will have a removable singularity there. On the other hand, if $f$ has a removable singularity at $z_0$ your answer will depend on whether, after removing the removable singularity, you get a value of $0$ there.

However, the statement of the question is wrong for a different reason: if $f$ has an essential singularity at $z_0$, it is quite possible that $z_0$ is a limit point of a sequence of zeros of $f$, and then the singularity of $g$ at $z_0$ is not isolated. Consider e.g. $f(z) = \sin(1/z)$.

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  • $\begingroup$ Ok, this makes sense to me. Then to show that the points are still isolated singular points is using the chain rule approach I mentioned still valid? How can I show that the isolated singular points of $f$ are still isolated singular points of $g$ and $h$ in general? $\endgroup$
    – klib
    Apr 25, 2016 at 17:44
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If $f$ is defined and analytic in a punctured neighborhood $\dot U$ of $0$ then either there is a $k\in{\mathbb Z}$ and an $f_1$ analytic in a full neighborhood $V$ of $0$ such that $f_1(0)\ne 0$ and $f(z)=z^k f_1(z)$ in $\dot U\cap V$, or $f$ has an essential singularity at $0$. In the second case $f$, as well as $1/f$, or $f^2$, behave very strange near $0$.

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