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Counting the negative powers of a Laurent series divides isolated singularities into three kinds — removable, a pole of some order, essential — and each kind licenses a different tool, so getting the order wrong gives a wrong residue with no warning.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to classify an isolated singularity from its Laurent series, find the order of a pole by subtracting the order of the numerator's zero from the denominator's, recognise a removable singularity and repair it, recognise an essential singularity and say how differently it behaves, and say why a branch point fits none of the three.
Laurent series on a punctured disc, and the observation that the negative powers are where all the interesting behaviour is stored. This lesson reads those negative powers as a classification, and the classification is what decides which tool to use on a singularity later.
A singularity $z_0$ of $f$ is isolated when $f$ is analytic on some punctured disc $0 < |z - z_0| < \rho$. Its principal part is the negative-power half of the Laurent series there. The singularity is removable when the principal part is empty, a pole of order $m$ when the most negative power present is $-m$, and essential when infinitely many negative powers appear. A pole of order $1$ is simple. $f$ has a zero of order $m$ at $z_0$ when its Taylor series there starts at the power $m$.
Let $z_0$ be an isolated singularity, with Laurent series $\sum a_n (z - z_0)^{n}$ on a punctured disc about it. Count the terms with $n < 0$.
None: removable. $f$ has a finite limit at $z_0$, and defining $f(z_0)$ to be that limit makes $f$ analytic there. The singularity was an artefact of how the function was written. $\dfrac{\sin z}{z}$ at $0$ is the standard example; Riemann's removable singularity theorem says that merely being bounded near $z_0$ is enough to force this case.
Finitely many, down to $-m$: a pole of order $m$. Then $|f(z)| \to \infty$ as $z \to z_0$, and $(z - z_0)^{m}f(z)$ has a removable singularity with a non-zero limit. Poles are the well-behaved singularities: they are the ones residues are computed at, and the ones every rational function has.
Infinitely many: essential. Then the function does neither: it has no limit, finite or infinite. Casorati-Weierstrass says its values on every punctured disc about $z_0$ come arbitrarily close to every complex number, and Picard sharpens that to attains every complex value infinitely often, with at most one exception. $e^{1/z}$ at $0$ misses only the value $0$.
Reading the order off a quotient. If $f = g/h$ with $g$ and $h$ analytic, $g$ having a zero of order $p$ at $z_0$ and $h$ a zero of order $q$, then the singularity is
So the classification is subtraction, once the two orders are known — and the orders are read off the first surviving term of each series.
Another way: picture
Watch $|f(z)|$ as $z$ spirals in towards $z_0$. At a removable singularity it settles to a number. At a pole it climbs without limit, and climbs the same way whatever direction you come from. At an essential singularity it does neither: choose a direction and you can make the values tend to anything you like, including infinity, including zero, including $17 - 3i$. The three pictures are as different as three pictures can be, and the only thing distinguishing them is a count of terms.
Another way: steps
It is not taxonomy for its own sake. Each of the three cases licenses a different tool, and picking the wrong one wastes the work.
Removable: delete the problem. Redefine the function at the point and carry on. In a contour integral, a removable singularity inside the contour contributes nothing at all — Cauchy's theorem applies to the repaired function. A great many integrals that look hard have removable singularities and are therefore zero.
A pole: compute a residue. The next lesson's formulas are stated in terms of the order, and they need it: the simple-pole formula is a limit, and the order-$m$ formula differentiates $m - 1$ times. Getting the order wrong differentiates the wrong number of times and gives a wrong number, silently.
Essential: expand the series. No closed formula exists for the residue, because there is no finite principal part to manipulate. The only route is to write out the Laurent series and read off the coefficient of the power $-1$, which is exactly what the $e^{1/z}$ example did.
And a fourth case that is not on the list: a branch point, such as the origin for $\sqrt{z}$ or for $\log z$. It is a singularity, and it is not isolated in the sense this lesson means, because no punctured disc about it carries a single-valued analytic function. None of the three tools applies, and a cut has to be made instead. That is unit 1's ambiguity, still not finished with.
The order of a pole is a property of the function, not of the way it happens to be written. $\dfrac{\sin z}{z^{2}}$ has a simple pole at the origin despite the square, because the numerator supplies a power of $z$ that cancels one of them; and $\dfrac{z^{3}}{z^{5}}$ is a pole of order $2$, however it is written.
So the numerator always gets a vote, and the habit is to expand it before deciding. The error costs more than a wrong label: the residue formula for a pole of order $m$ differentiates $m - 1$ times, so an order that is two too large differentiates twice too often and returns a wrong residue with no warning.
The second confusion is between a pole and any old badly behaved point. A pole is where the modulus tends to infinity from every direction. $e^{1/z}$ at the origin tends to infinity along one direction and to zero along another, and is therefore not a pole of any order — no amount of multiplying by powers of $z$ will tame it.
And the third: isolated is a hypothesis. The classification covers isolated singularities, and a branch point is not one. Calling the origin a pole of $\log z$ is a category error rather than an arithmetic slip.
$\dfrac{1 - \cos z}{z^{4}}$ at $0$: the numerator is $\dfrac{z^{2}}{2} - \dfrac{z^{4}}{24} + \cdots$.
Expand the numerator.
It vanishes to order $2$, so two of the four powers cancel: $\dfrac{1}{2z^{2}} - \dfrac{1}{24} + \cdots$.
Subtract the orders.
A pole of order $2$, not $4$. Reading the exponent off the denominator would have been wrong by two.
The numerator gets a vote.
$e^{1/z}$ at $0$: substituting into the exponential series gives $1 + \dfrac{1}{z} + \dfrac{1}{2z^{2}} + \cdots$, infinitely many negative powers.
Essential.
Approach along the positive real axis: $e^{1/x} \to \infty$. Along the negative real axis: $e^{1/x} \to 0$. Along the imaginary axis the modulus is exactly $1$ the whole way.
Three directions, three completely different answers.
No limit exists, finite or infinite, so this is neither removable nor a pole — and the residue, $1$, still has to come from the series.
The third case is genuinely different.
The numerator vanishes to order $2$; $\sin z$ vanishes to order $1$.
Read off both orders.
The numerator wins by one, so the singularity is removable and the repaired function has a zero of order $1$ there.
Match each function to the kind of singularity it has at $z = 0$.
| Removable | A simple pole | A pole of order $6$ | A pole of order two | |
|---|---|---|---|---|
| $\dfrac{\sin z}{z}$ | ||||
| $\dfrac{\sin z}{z^{2}}$ | ||||
| $\dfrac{1}{z^{6}}$ | ||||
| $\dfrac{1 - \cos z}{z^{4}}$ |
What kind of singularity does $\dfrac{1}{z^{3}}$ have at $z = 0$?
Let $f(z) = \dfrac{1}{z^{2}\left(z - 3\right)^{1}}$. Fill in the table.
| Value | |
|---|---|
| The order of the pole at the origin | |
| The order of the pole at the other singularity | |
| How many singularities the function has | |
| The total of the two orders |
What is the order of the pole of $\dfrac{\sin z}{z^{6}}$ at $z = 0$?
Answer:
Select every statement that is true.
This task has no paper form; do it on a device.
$f$ has a pole of order $1$ at a point, and $g$ has a zero of order $5$ at the same point. The product $fg$ has a zero there. What is its order?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each function to the kind of singularity it has at $z = 0$.
| Removable | A simple pole | A pole of order $6$ | A pole of order two | |
|---|---|---|---|---|
| $\dfrac{\sin z}{z}$ | ||||
| $\dfrac{\sin z}{z^{2}}$ | ||||
| $\dfrac{1}{z^{6}}$ | ||||
| $\dfrac{1 - \cos z}{z^{4}}$ |
You can name the kind and the order of an isolated singularity, and you know why the exponent in the denominator is not the answer on its own. Say in your own words how a pole differs from an essential singularity. Next: the residue, which is one coefficient of the series you have been writing.
9. Your turn: classify the singularity of $\dfrac{z^{2}}{\sin z}$ at $z = 0$, step 2