I am working on some exam practice questions (and do have the answers to these questions) but I am having trouble developing proper 'insight' as to how these answers are derived:

(a) $$\frac{\sin z}{z^2}, \frac{\cos z-1}{z^3}$$ In both these cases, we have a pole of order 1 at $z=0$, and in evaluating these, we evaluate the limits: $$\lim_{z \to 0} \frac{\sin z}{z}, \frac{\cos z-1}{z^2}$$

When I did these questions myself, my first thought was to set: $$g(0)=\lim_{z \to 0}(z^2f(z))=\lim_{z \to 0} \sin z = 0 $$ which obviously would be incorrect. I'm assuming the solution is actually to do $$g(0)=\lim_{z \to 0}(zf(z))=\lim_{z \to 0} \frac{\sin z}{z} = 1 $$ So my question is this: How do we know to multiply by $z$ rather than $z^2$? Is this just by intuition and knowing that this would 'fix' the function to make it analytic?

(b) For the function $$ z^4\sin \left(\frac{1}{z}\right)$$ we have a singularity at $z=0$, my intuition is that this is an essential singularity because there is no way to 'fix' the function to make it analytic, but the solutions suggest writing it as a Laurent expansion, to which it then simply says that it is clear $z=0$ is an essential singularity. Is this because in the Laurent expansion none of the coefficients are $0$?

(c) For the function $$ \frac{1+z}{1-z^4}$$ we have singularities at $z=1,-1,i,-i$, If I evaluate the limits (for example) $(z-1)f(z)$ as $z$ approaches 1, then they all give something non-zero, except $z=-1$, then I get simple poles at $z=1,i,-i$ and an essential pole at $z=-1$. But the solutions says that $z=-1$ is a removable pole? Why is this?

I would greatly appreciate some help in order to clear up my confusions.

Many thanks!


2 Answers 2


There are a lot of quick tricks to figure out the location and nature of singularity without ever having to deal with Laurent expansion.

Method 1: determine how singularity transform through composition and arithmetic.

Consider 2 analytic non constant function $f,g$. Assume that you already determine zeros and isolated singularity and that you know (somehow) that $f$ have no essential singularity. For pole and zero, assume you know their order as well. Define a pole of order $k$ to be a zero of order $-k$. Every other normal point is assumed to be zeros of order $0$. Then:

  1. $f\pm g$. Singularity point only appear at singularity of at least 1 of them. If $g$ have essential singularity the sum is still essential singularity. If they have different zeros order, then the lower one is the zero order of the singularity. For example: $\frac{1}{e^{x^{2}}-1}+\frac{1}{z}$. $\frac{1}{e^{x^{2}}-1}$ have pole order 2 (so zero order $-2$) and $\frac{1}{z}$ have zero order $-1$, so the sum have zero order $-2$ ie. pole order $-2$. If the zero order is the same, this won't work.

  2. $fg$. Singularity point only appear at singularity of at least 1 of them. If $g$ have essential singularity, this is still essential singularity. Otherwise, simply take the sum of the order. For example, $z\frac{1}{e^{x^{2}}-1}$. $z$ have zero order $1$ and $\frac{1}{e^{x^{2}}-1}$ have zero order $-2$ so this product is a zero of order $-1$ (pole of order 1).

  3. $\frac{f}{g}$. Singularity point only appear at singularity of at least 1 of them. If $g$ have essential singularity, this is still essential singularity. Otherwise, take zero order of $f$ minus zero order of $g$.

  4. $f\circ g$. Singularity appear only where $g$ have one, or at preimage through $g$ of singularity of $f$. If $g$ is essential singularity then the composition might produce non-isolated singularity (so yeah it can be pretty bad). If $g$ have actual zero (zero with positive order), then the zero order of the composition is just the product of the order. If $g$ have actual pole (negative zero order), then look for the power of the highest power term in Laurent series of $f$, and multiply it with zero order of $g$ to get the zero order of the composition (if $f$ do not have highest power term, this will produce essential singularity). This is not as simple if $g$ have neither pole nor zero.

We just assumed at least 1 function do not have essential singularity. If both function have essential singularity, these trick do not work, except for composition: composition will produce essential singularity.

(a quick-and-dirty rule of thumb for remembering the above tricks is that essential singularity is infectious but harder to come up, pole and zero dominated by pole when add, cancel when multiply, and multiply when composed (for zero only))

Method 2: look up tables for information on well known function.

In particular, for polynomial, they have no singularity, and zero are the root, with zero order being multiplicity of the root.

Method 3: construct a sequence approaching the singularity that satisfy certain property:

Consider an analytic $f$. If you figure out that a singularity is at $z$ and you can construct a sequence $z_{n}\rightarrow z$ such that the sequence $f(z_{n})$ have the property:

-The modulus oscillate: then must be essential singularity.

-Do not converge: then cannot be removable singularity.

-Do not diverge to infinity: then cannot be pole.

Unfortunately, you cannot rule out essential singularity with this trick.

Now back to example in your case. We can apply these tricks.

$\frac{\sin z}{z^{2}}$. Look up to find $\sin z$ have zero order $1$. $z^{2}$ have zero order $2$. So this is pole order $1$ ($1-2=-1$).

$\frac{\cos z-1}{z^{3}}$. Unfortunately, the trick won't help you to find the zero of the numerator. Simply look up Taylor series expansion to find $\cos z-1$ have zero order $2$. $x^{3}$ have zero order $3$. So this become pole order $1$.

$z^{4}\sin\frac{1}{z}$. $\frac{1}{z}$ is pole order $1$, but $\sin$ have infinite Taylor expansion, so $\sin\frac{1}{z}$ is essential singularity. $z^{4}$ do not have essential singularity, so the product is still essential singularity.

$\frac{1+z}{1-z^{4}}$. $1-z^{4}$ have zero order $1$ at these $4$ point. $1+z$ have zero order $1$ at only $z=-1$. So at $z=-1$ this cancel, so is not a pole. At other point there are no zero order in the numerator to cancel out that in the denumerator, so they are pole of order $1$.

  • $\begingroup$ Wow, thanks for your comprehensive answer. Very helpful. I'm still a bit confused for the last one. You have said that $z=-1$ is not a pole, so how does that lead to it being a removable singularity? $\endgroup$
    – JJJ
    Commented Jun 9, 2014 at 22:11
  • 1
    $\begingroup$ @user153663: it is not an essential singularity ($\frac{f}{g}$ can only have essential singularity where at least $f$ or $g$ have one). In fact, using the above tricks you can simply do this: $1+z$ have a zero of order $1$ at $z=-1$, $1-z^{4}$ have a zero of order $1$ at $z=-1$. So $\frac{1+z}{1-z^{4}}$ have a zero of order $1-1=0$ at $z=-1$. A zero of order $0$ is not a pole (which have negative zero order). So it is either not singularity, or is a removable singularity. It is clearly a singularity. $\endgroup$
    – Gina
    Commented Jun 9, 2014 at 22:56
  • $\begingroup$ ahh right. Very helpful. Thank you! $\endgroup$
    – JJJ
    Commented Jun 9, 2014 at 23:02

a) exactly because the limit is $0$ tells you that the factor $z^2$ has a too large degree. So the next step is to explore the factor $z$.

b) Because infinitely many coefficients for negative degrees are non-zero.

c) $(1-z^4)=(1-z^2)(1+z^2)=(1-z)(1+z)(1+z^2)$.

  • $\begingroup$ Thanks, for (c) I'm still confused. Could you elaborate a bit more? I noticed that if I simply evaluate the limit as $z$ approaches each of the respective singularities (using L'Hopital's rule) of the function without multiplying it by anything, they all give me something non-zero! So this leads me to conclude that they should all be removable singularities? $\endgroup$
    – JJJ
    Commented Jun 9, 2014 at 22:16
  • 1
    $\begingroup$ You can only apply l'Hopital if both numerator and denominator have the value zero at the point of interest. The only root of the numerator is $z=-1$. But there you can just cancel the terms algebraically, no need for limits. $\endgroup$ Commented Jun 9, 2014 at 22:39
  • $\begingroup$ Yep that makes perfect sense. Thanks! $\endgroup$
    – JJJ
    Commented Jun 9, 2014 at 23:02

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