I have the following functions:

a) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{e^{\frac{1}{z}}-1}$

b) $\displaystyle f:\mathbb{C}\backslash\{0,2\}\rightarrow\mathbb{C},\ f(z)=\frac{\sin z ^2}{z^2(z-2)}$

c) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\cos\left(\frac{1}{z}\right)$

d) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{1-\cos\left(\frac{1}{z}\right)}$

e) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{\sin\left(\frac{1}{z}\right)}$

What would the quickest approach to determine if $f$ has a removable singularity, a pole or an essential singularity? What would be the thinking $behind$ the approach?


What I know/ What I have tried:

I know that if we have an open set $\Omega \subseteq \mathbb{C}$, then we call an isolated singularity, a point, where $f$ is not analytic in $\Omega$ ($f \in H(\Omega \backslash \{a\}$).

The functions in (a)-(e) are not defined on some values. So I suspect, that these are the first candidates for singularities. For instance in (a), it would be 0. In (b), it would be 0 and 2.

Question: Could there be any other points where these functions are not analytic?

Let's call our isolated singularity $a$. Furthermore I know that we have 3 types of singularities:

1) removable

This would be the case when $f$ is bounded on the disk $D(a,r)$ for some $r>0$.

2) pole

There is $c_1, ... , c_m \in \mathbb{C},\ m\in\mathbb{N}$ with $c_m \neq 0$, so that:

$$f(z)-\sum\limits_{k=1}^m c_k\cdot\frac{1}{(z-a)^k},\ z \in \Omega \backslash \{a\})$$

has a removable singularity in $a$, then we call $a$ a pole.

We also know that in this case:

$|f(z)|\rightarrow \infty$ when $z\rightarrow a$.

3) essential

If the disk $D(a,r) \subseteq \Omega$, then $f(D(a,r)\backslash\{a\})$ is dense in $\mathbb{C}$ and we call $a$ essential singularity.

The books that I have been using (Zill - Complex Analysis and Murray Spiegel - Complex Analysis) both expand the function as a Laurent series and then check the singularities. But how do I do this, if I use the definitions above? It doesn't seem to me to be so straight forward...

What I would want to learn a method which allows me to do the following:

I look at the function and the I try approach X to determine if it has a removable singularity. If not continue with approach Y to see if we have a pole and if not Z, to see if we have an essential singularity. An algorithmic set of steps so to speak, to check such functions as presented in (a) to (e).

Edit 2: This is not homework and I would start a bounty if I could, because I need to understand how this works by tommorow. Unfortunately I can start a bounty only tommorow...

Edit 3: Is this so easy? Because using the definitions, I am getting nowhere in determing the types of singularities...

  • 4
    $\begingroup$ You have to stop throwing questions around like that and start answering the comments/answers that were left on your other questions. MSE is a community, and as such, there has to be some exchange between the different parties. You can't just ask questions without leaving feedback. $\endgroup$
    – M Turgeon
    Jun 6, 2012 at 15:15
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    $\begingroup$ Of course, you are free to do what you like. But one thing which is certain: if you leave feedback, if you accept answers, people will feel more inclined to answer your future questions. My comment comes from the exasperation of seeing too many of your questions without feedback, and I will venture to say that I am not the only one who dislikes such behaviour. $\endgroup$
    – M Turgeon
    Jun 6, 2012 at 15:45
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    $\begingroup$ I will leave feedback on all of them today. I appreciate all the given help tremendously and am very honored that I may use this great platform. $\endgroup$
    – Chris
    Jun 6, 2012 at 15:51
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    $\begingroup$ It appears that all others who left comments felt this question was so easy, that there should be no need to give a detailed answer, but instead the inductive steps and thinking. From my point of view, nevertheless, this approach takes too much time to answer such a question. So I might post an answer, while I am really not good at it. Hence could I suggest someone to post an answer? In any case, this is not a homework, is it? $\endgroup$
    – awllower
    Jun 7, 2012 at 4:01
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    $\begingroup$ Have you seen Laurent series? $\endgroup$ Jun 7, 2012 at 22:33

3 Answers 3


a) $\displaystyle{f(z)=\dfrac{1}{e^{1/z}-1}}$.

It says $f:\mathbb C\setminus\{0\}\to\mathbb C$, but this is incorrect, because $f$ has a simple pole at $z=\dfrac{1}{2\pi ki}$ for each nonzero integer $k$, and $z=0$ is not even an isolated singularity. If you change the codomain to $\mathbb C\cup\{\infty\}$ and think of $f$ as a meromorphic function, then it has an essential singularity at $0$.

In fact, you can show that $f(D(0,r)\setminus\{0\})=(\mathbb C\cup\{\infty\})\setminus\{0,-1\}$ for all $r>0$, using elementary properties of the exponential function.

b) $\displaystyle f:\mathbb{C}\backslash\{0,2\}\rightarrow\mathbb{C},\ f(z)=\frac{\sin z ^2}{z^2(z-2)}$

Evaluate $\lim\limits_{z\to 0}f(z)$ and $\lim\limits_{z\to 2}f(z)$. One is finite, the other is $\infty$, so you have a removable singularity and a pole.

c) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\cos\left(\frac{1}{z}\right)$

In this case, you should be able to show, even just using real variables, that $\lim\limits_{z\to 0}f(z)$ does not exist in either a finite or infinite sense. Sketch a graph of $y=\cos(1/t)$ close to $0$. Another useful tool is the Laurent series, which in this case is obtained from the power series expansion of $\cos$ by substitution of $1/z$. And similarly to a), you could use elementary properties of the exponential function along with the identity $\cos(z)=\frac{1}{2}(e^{iz}+e^{-iz})$ to find the image of a small punctured disk at $0$.

d) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{1-\cos\left(\frac{1}{z}\right)}$

Similarly to a), this is incorrect. Either the domain or the codomain should be changed. If you don't change the codomain, then $f$ is undefined where $\cos(1/z)=1$, and there is not an isolated singularity at $0$. If you allow meromorphic functions, then it is an essential singularity at $0$. (And again you could even explicitly find the range, or you could more simply show that no limit exists by choosing special values.)

e) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{\sin\left(\frac{1}{z}\right)}$

See a) and d).

  • $\begingroup$ For d) What if we change the domain to: $\mathbb{C}\backslash\{0,\frac{1}{2k\pi}\}$ ? In e) We should change it to $\mathbb{C}\backslash\{k\pi\}$ right? $\endgroup$
    – Chris
    Jun 7, 2012 at 23:31
  • $\begingroup$ @Chris: For d), do you actually mean $\mathbb C\setminus(\{0\}\cup\{\frac{1}{2k\pi}:k\in\mathbb Z\setminus\{0\}\})$? If you change the domain to that, then you do not have an isolated singularity at $0$, but you have a pole at $\frac{1}{2k\pi}$ for each nonzero integer $k$. For e), no, but $\mathbb C\setminus(\{0\}\cup\{\frac{1}{k\pi}:k\in\mathbb Z\setminus\{0\}\})$ would work as a domain if you want it to still be complex valued. Again, $0$ is not an isolated singularity in that case, and you have a pole at the new removed points. $\endgroup$ Jun 7, 2012 at 23:33
  • $\begingroup$ @Chris: FYI I will not be responding further (at least for a while), but perhaps others will chime in if you have other questions about my answer, or someone will clarify things with their own answer, or I will respond to further questions in time. $\endgroup$ Jun 7, 2012 at 23:41

Thank you for all your feedback. I've decided to simplify things and use the method from Schaum's Outline on Complex Analysis. Otherwise, I am getting nowhere.

They do it like this:

(i) If $\lim_{z\rightarrow a} f(z)$ exists then we have a removal singularity.

(ii) If $\lim_{z\rightarrow a} (z-a)^n f(z) = A \neq 0$, then $z=a$ is a pole of order $n$.

If we don't have (i) or (ii), then the singularity is essential.

Question: Why are these 3 options, the only ones for isolated singularities? A short explanation in words would be nice!

So, using this we have:

a) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{e^{\frac{1}{z}}-1}$

We must check $\lim_{z\rightarrow 0} z^n \frac{1}{e^{\frac{1}{z}}-1}$

We have $\lim_{z\rightarrow 0} z^n \frac{1}{e^{\frac{1}{z}}-1}=0$ for any natural number $n$.

So, this means that 0 is an essential singularity here.

b) $\displaystyle f:\mathbb{C}\backslash\{0,2\}\rightarrow\mathbb{C},\ f(z)=\frac{\sin z ^2}{z^2(z-2)}$

$\lim_{z\rightarrow 0} z^n \frac{\sin z ^2}{z^2(z-2)}=0$

$\lim_{z\rightarrow 2} z^n \frac{\sin z ^2}{z^2(z-2)}=-\infty$

So, we have again essential singularities, I believe...

c) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\cos\left(\frac{1}{z}\right)$

$\lim_{z\rightarrow 0} z^n \cos\left(\frac{1}{z}\right)=0$

Uhem... essential again...

d) $\displaystyle f:\mathbb{C}\backslash\{0,\frac{1}{2k\pi}\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{1-\cos\left(\frac{1}{z}\right)}$

$\lim_{z\rightarrow 0} z^n \frac{1}{1-\cos\left(\frac{1}{z}\right)}$

0 is odd here... might it be that 0 is no singularity?

For $2k\pi,\ k\neq 0$, the limit can be evaluated to something. I believe these values are the poles then. I think we have $n$ of them.

e) $\displaystyle f:\mathbb{C}\backslash\{0,\frac{1}{k\pi}\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{\sin\left(\frac{1}{z}\right)}$

$\lim_{z\rightarrow 0} z^n\frac{1}{\sin\left(\frac{1}{z}\right)}$

Same as d, 0 is no singularity?

For $n = 1$, the limit is $1$. So, we got a pole of order $1$ at $z=0$.

Do you agree with the above?

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    $\begingroup$ This is mostly very incorrect. How are you computing these limits? $\endgroup$ Jun 7, 2012 at 23:25
  • $\begingroup$ I evaluated them with Mathematica. Sometime I've used certain values for n, so that I would get a result. $\endgroup$
    – Chris
    Jun 7, 2012 at 23:26
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    $\begingroup$ I appreciate your attempt. Regarding your new question on why those are the only three options, it really depends on your definitions. In some sense it is a tautology that those are the only three options, because essential singularities can be defined simply as those that are not removable or poles. However, with the definition you gave in your question, you need to use the Casorati-Weierstrass theorem to see that those are the only options. $\endgroup$ Jun 7, 2012 at 23:27
  • $\begingroup$ Something went wrong with your Mathematica attempts. They are not correct. $\endgroup$ Jun 7, 2012 at 23:28
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    $\begingroup$ $@$Chris: To give an example of something that is wrong aside from the incorrect evaluation of limits, note that if $f$ has an essential singularity at $z=a$ then $\lim\limits_{z\to a}(z-a)^nf(z)$ will never exist for any nonnegative integer $n$. If it is ever $0$, then you have a pole or a removable singularity. If it is always zero, then you have a removable singularity. $\endgroup$ Jun 7, 2012 at 23:45

So I can't give you a nice tool and I'm no pro by all means, but let me share you my approach. First, for isolated singularities, we can look at the Laurent series to determine the type of the singularity. We know that the Laurent series consists of two parts: The principal part and the analytical part. Now we further know:

Removable: Degree of the principal part is zero: We have a Taylor The principal part series. E.g. $\frac{sin(z)}{z}$

Pole: Degree of the principal part is finite: The degree of the principal part corresponds to the degree of the pole. E.g. $\frac{\sin(z)}{z^2}$

Essential: Degree of the principal part is infinite. E.g $\sin(\frac{1}{z})$

Now what I do is: I look at the given function $f$. I check the Taylor series of the function which my $f$ consists of. You also look at the argument of these functions and basically check if the argument reduces the degree of the Taylor series into the negative or not. You also consider the how the denominator influence the degree. Then you use the statements above.

Example: Let's consider the examples above. We know that $sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-...$

So if we have $\sin(z)/z$ we see, that we get $1-\frac{z^2}{3!}+... $, so the principal part has a degree of 0. So it's a removable singularity.

If we look at $\sin(z)/z^2$ we see, that we now do get one negative term. So we have a simple pole.

If we look at $\sin(1/z)$ we see that the degree of the principal part is infinite. So we have an essential pole.

I hope that helps.

Another thing to note is that if you have functions that share a root, you have to look closer at it. E.g. if you have $\sin(\pi z)/(z-1)$ you have a problem point at $z=1$, which first looks like a simple pole but you also see that $\sin(\pi \cdot 1)=0$, so $z=1$ is a root of $\sin(\pi z)$. The safest bet here is to check $\lim_{z\to 1}\frac{\sin(\pi z}{1-z}$. If that limit exists you found a continuation of the function at $z=1$, making it a removable singularity.


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