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9. Week 9: Laurent Series and Singularities

≪ 8. Week 8: Orders of Zeroes | Table of Contents | 10. Week 10: Contour Integration ≫

Last week, we talked about finding orders of zeroes of analytic functions, and how this property could be “read off” from the Taylor series of a function centred at said zero.

Unfortunately, analytic functions are not the only functions we will ever encounter in our lives, and we will often encounter functions that perform some kind of “division by zero” at one point or another in their domain. For instance, there’s the function \(\frac{e^z}{z^2+1}\), which is dividing by zero whenever \(z = \pm i\). However, as discussed many weeks ago, we can still use this function to compute things like \(\int _{-\infty}^{\infty} \frac{\cos x}{x^2+1}dx\), and we did so by appealing to the Cauchy integral formula.

Sometimes, we’ll encounter computations such as \[\oint _{\left\lvert z \right\rvert = 1}\frac{1}{\sin z} dz,\] which don’t at first glance look like they can be handled by any of the theorems we’ve learned so far. However, if we recognise that \(\sin z\) has a simple zero at \(z=0\), we can write \(\sin z = z g(z)\) for some analytic function \(g\) such that \(g(0) \neq 0\). Then, \[\oint _{\left\lvert z \right\rvert = 1} \frac{1}{\sin z } dz = \oint _{\left\lvert z \right\rvert = 1} \frac{g(z) ^{-1}}{z} dz.\] Now the numerator is holomorphic and we can use Cauchy’s integral formula again!

These computations can quickly grow out of hand because it’s hard to get a handle on what \(g(z)\) is explicitly. Sure we can write it as a power series in this case, but there may be scenarios in which this is less feasible, particularly when we don’t have the luxury of a simple zero.

Moreover, such arguments don’t work if we end up dealing with functions like \(\exp \left( \frac{1}{z} \right)\). Can I still “divide out” the zeroes in the denominator? How do I know how many zeroes to “divide out”?

There is a unified approach to addressing all of these computational considerations, and that’s encapsulated by the theory of Laurent series. Ultimately, this will help us develop a computationally feasible approach to handling almost any contour integral under the sun in a theorem called the residue theorem.

Laurent Series and Singularities

A power series centred at \(z_0\) is simply something of the form \(\sum _{n=0}^{\infty} a_n \left( z-z_0 \right)^n\). A Laurent series centred at \(z_0\) is exactly a power series, but \(n\) is allowed to be negative: \[\sum _{n=-\infty}^{\infty} a_n \left( z - z_0 \right)^n = \sum _{n=1}^{\infty} a _{-n} \left( z-z_0 \right) ^{-n} + \sum _{n=0}^{\infty} a_n \left( z-z_0 \right)^n.\] Here, we’ve separated our Laurent series into two pieces: one containing only the nonnegative terms and one containing only the negative terms. From our theory of power series, the nonnegative series will converge on some disc (or possibly all of \(\mathbb{C} ^n\)) centred at \(z_0\). The negative series is actually a power series at \(\infty\) of some function! Thus it will converge on \(\mathbb{C} ^n \) minus some disc centred at \(z_0\). The Laurent series converges on the intersection of these two domains, an annulus (or donut-shaped region).

Let’s consider an example: take \(f(z) = \frac{1}{1-z}\). This has a power series expansion at \(z = 0\) given by the usual geometric series: \(f(z) = \sum _{n=0}^{\infty} z^n\). This converges on the disc \(\left\lbrace \left\lvert z \right\rvert < 1 \right\rbrace\) since \(f\) has a singularity at \(z=1\). The power series is just a Laurent series for \(f\) on the disc centred at \(0\).

However, we can also write a power series for \(f\) at infinity: \[f(z) = \frac{1}{z}\cdot \frac{1}{1 - \frac{1}{z}} = \frac{1}{z} \sum _{n=0}^{\infty} \left( \frac{1}{z} \right)^n = \sum _{n=1}^{\infty} z ^{-n}.\] This converges whenever \(\left\lvert \frac{1}{z} \right\rvert < 1\), or when \(\left\lvert z \right\rvert > 1\). But this is also a Laurent series centred at \(z=0\), the only diffence being where the series converges.

What we did was we took the domain of \(f\) — \(\mathbb{C}\setminus \left\lbrace 1 \right\rbrace\) — and cut it up into two annular regions, \(\left\lbrace \left\lvert z \right\rvert < 1 \right\rbrace\) and \(\left\lbrace \left\lvert z \right\rvert > 1 \right\rbrace\). We’re missing a very thin region, but this is the best we can do. For each of these disjoint regions, there will be a Laurent series for \(f\)!

Solution

The domain of \(f\) is \(\mathbb{C} \setminus \left\lbrace \pm 1, \pm 2 \right\rbrace\), which we can cut up into three annular regions: \(\left\lbrace \left\lvert z \right\rvert < 1 \right\rbrace\), \(\left\lbrace 1 < \left\lvert z \right\rvert < 2\right\rbrace\), and \(\left\lbrace \left\lvert z \right\rvert > 2 \right\rbrace\).

Now we’ll use partial fractions to rewrite \[f(z) = \frac{1}{3\left( z^2-1 \right)} - \frac{1}{3\left( z^2-4 \right)}.\]

To get the Laurent series for the first region \(\left\lbrace \left\lvert z \right\rvert < 1 \right\rbrace\), we’ll expand both parts of \(f\) as power series centred at \(0\): \[\frac{1}{3\left( z^2-1 \right)}= -\frac{1}{3} \frac{1}{1-z^2} = -\frac{1}{3}\sum _{n=0}^{\infty} z ^{2n}\] and \[\frac{1}{3\left( z^2-4 \right)} = -\frac{1}{12} \frac{1}{1 - \left( \frac{z}{2} \right)^2} = -\frac{1}{12} \sum _{n=0}^{\infty} \left( \frac{z}{2} \right) ^{2n}.\] The first power series converges on \(\left\lbrace \left\lvert z \right\rvert < 1 \right\rbrace\) while the second converges on \(\left\lbrace \left\lvert z \right\rvert < 2 \right\rbrace\), hence their sum \[f(z) = \sum _{n=0}^{\infty} \left( -\frac{1}{3} - \frac{1}{12\cdot 4^n} \right) z ^{2n}\] converges on \(\left\lbrace \left\lvert z \right\rvert < 1 \right\rbrace\).

For the “middle region”, we’ll expand the first term at \(\infty\) to get \[-\frac{1}{3 \left( z^2-1 \right)} = -\frac{1}{3z^2} \frac{1}{1 - \left( \frac{1}{z} \right)^2} = -\frac{1}{3z^2} \sum _{n=0}^{\infty} \left( \frac{1}{z} \right) ^{2n} = -\frac{1}{3}\sum _{n=1}^{\infty} z ^{-2n}. \] This converges whenever \(\left\lvert z \right\rvert > 1\). Thus, the Laurent series \[f(z) = -\frac{1}{3}\sum _{n=1}^{\infty} z ^{-2n} - \frac{1}{12} \sum _{n=0}^{\infty} \left( \frac{z}{2} \right) ^{2n}\] converges on the annulus \(\left\lbrace 1 < \left\lvert z \right\rvert < 2 \right\rbrace\).

And finally, for the third region, we’ll expand both terms in \(f\) as power series at \(\infty\). We already did the first term, and for the second term, we have \[-\frac{1}{3 \left( z^2-4 \right)} = -\frac{1}{3z^2} \frac{1}{1 - \left( \frac{2}{z} \right)^2} = -\frac{1}{3z^2} \sum _{n=0}^{\infty} \left( \frac{2}{z} \right) ^{2n} = -\frac{1}{3} \sum _{n=}^{\infty} \left( \frac{2}{z} \right) ^{2n}.\] This converges whenever \(\left\lvert z \right\rvert > 2\). Putting these two series at \(\infty\) together yields \[f(z) = -\frac{1}{3}\sum _{n=1}^{\infty} \frac{1+2^n}{ z ^{2n}},\] which converges whenever \(\left\lvert z \right\rvert > 2\). \(\square\)

One really nice application of Laurent series is the following: suppose one has a Laurent series \[f(z) = \sum _{n=-\infty}^{\infty} a_n \left( z-z_0 \right)^n\] converging on the annulus \(\left\lbrace r < \left\lvert z \right\rvert < R \right\rbrace\). Then, for any constant \(r < c < R\), one has \[\frac{1}{2 \pi i}\oint _{\left\lvert z \right\rvert = c} f(z) dz = a _{-1}.\] This should resemble Cauchy’s integral formula, and in fact, it is a vast generalisation of Cauchy’s integral formual! This is because the Laurent series converges uniformly on the circle \(\left\lvert z \right\rvert = c\). While one would normally have to be very careful with swapping series with integrals, in this situation it’s okay, and we get \[\oint _{\left\lvert z \right\rvert = c } f(z) = \sum _{n=-\infty}^{\infty} \oint _{\left\lvert z \right\rvert = c} a_n \left( z-z_0 \right)^n dz.\] Whenever \(n\neq -1\), the summand becomes zero: for \(n > -1\) the integrand is holomorphic everywhere and Cauchy’s theorem does the trick. For \(n < -1\), use Cauchy’s generalised integral formula. You pick up some derivatives of the numerator, but since the numerator is constant, those all become zero! For \(n = -1\), you just get the numerator right back (and the factor of \(2 \pi i\)).

This coefficient \(a _{-1}\) is very special, and we’ll see it come up in the near future (or near past?) when you discuss the residue theorem.

Special Types of Singularities

Based on what we now know about Laurent series, \(f\) has an isolated singularity at \(z_0\) if and only if it has a Laurent series expansion centred at \(z_0\) that converges on a punctured disc! For instance, the function \[\exp \left( \frac{1}{z} \right) = \sum _{n=0}^{\infty} \frac{1}{n! z ^{n}}\] has an isolated singularity at \(z=0\). Most interestingly, using our formula from earlier, we can easily compute things like \[\oint _{\left\lvert z \right\rvert = 1} e ^{\frac{1}{z}} dz = 2 \pi i.\] What!? I didn’t have to do anything — no parameterisations, no factoring out zeroes, no Cauchy’s formula, nothing — all I needed to know was one coefficient from the Laurent series of \(\exp\) near this isolated singularity.

The function \(\frac{1}{\sin z}\) has a much nicer Laurent series at \(z = 0\), which is also an isolated singularity: \[\frac{1}{\sin z} = \frac{1}{z} + \frac{1}{6} z + \frac{7}{360}z^3 + \cdots\] It’s not important to know where I got this series from. It’s nice because there’s only one nonnegative term, and if we multiply by \(z\) we get a nice holomorphic function back, since the Laurent series no longer has any negative terms! We still know that \[\oint _{\left\lvert z \right\rvert = 1} \frac{1}{\sin z} dz = 2 \pi i.\]

But what about something nastier like \[\oint _{\left\lvert z \right\rvert = 1} \frac{\cos\left( z \right) - 1}{\sin\left(z\right)^2\left( z+4 \right)} dz?\] There’s no shot that I can compute the Laurent series of this thing. However, there is a way to compute just the \(a _{-1}\) term of the Laurent series using things we already know how to do.

Returning to the example of \(\frac{1}{\sin z}\), we know that \(\sin z\) has a simple pole at \(z=0\), hence \[z\cdot \frac{1}{\sin z} = a_0 + a_1 z + a_2 z^2 + \cdots\] should be a holomorphic function at \(z=0\)! We have “divided out” the zero of \(\sin z\). If we go backwards, we see that \[\frac{1}{\sin z} = \frac{a_0}{z} + a_1 + a_2 z + a_3z^2 + \cdots\] is our Laurent series, which converges exactly when our original Taylor series converges, except at \(z=0\), the centre of the Laurent series.

This is special because there’s only finitely many negative terms, in contrast to \(\exp \left( \frac{1}{z} \right)\)! This will eventually really simplify our computational lives.

Now suppose \(f(z)\) has a pole of order \(n\) at \(z_0\). We need a good way to compute its residue without having to do a bunch of garbage power series computations. We know \[f(z) = a _{-n} \left( z-z_0 \right) ^{-n} + \cdots + a _{-1} \left( z-z_0 \right) ^{-1} + a_0 + a_1 \left( z-z_0 \right) + \cdots\] Multiplying by \(\left( z-z_0 \right) ^{n}\), we have \[\left( z-z_0 \right) ^{n} f(z) = a _{-n} + a _{-n+1} \left( z-z_0 \right) ^{1} +\cdots + a _{-1} \left( z- z_0 \right) ^{n-1} + \cdots\] But this is just the Taylor series of \(\left( z-z_0 \right) ^{n} f(z)\) centred at \(z_0\)! So the residue, i.e. \(a _{-1}\), is equal to the Taylor coefficient of the \(\left( z-z_0 \right) ^{n-1}\) term, hence \[a _{-1} = \operatorname{Res} \left( f; z_0 \right) = \frac{1}{n!}\left. \frac{d ^{n-1}}{dz ^{n-1}} \left( z-z_0 \right) ^{n} f(z)\right\rvert_{z_0}.\]

You do not need to compute the Laurent series of this guy at each of the singularities. You just need to determine what the order of the pole is, then use the residue formula!