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A typical problem related to Laurent series is this:

For the function $\frac 1{(z-1)(z-2)}$, find the Laurent series expansion in the following regions: $\\(a) |z|<1, \\ (b) 1<|z|<2, \\(c)|z|>2.$

Now, while I know how to solve such problems (using algebraic manipulations and geometric series), I don't understand why the function can have Laurent series in so many regions. Isn't a Laurent series meaningful only within an open annulus centred at each singularity? Why does the given function have Laurent series for all points except at circles containing its singularities? I feel that there's a fundamental gap in my understanding, so please forgive me if my question seems fuzzy.

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The main idea is that all of these Laurent series that you're finding are centered at the same point (namely zero). A Laurent series expansion exists in any annulus in which the function is analytic, so if there are any singularities (at the center of the Laurent series or at any other point), this will create the ring about the center where there isn't an expansion.

If you wanted to center at a different point, then you would have different expansions for each annulus about that point that avoided the singularities.

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    $\begingroup$ Just like with Taylor series, Laurent series have fixed centers. The difference is that the Taylor series must be analytic at its center, but the Laurent series does not have to be (if it is, then it has a Taylor series expansion in some open disc about its center). The function will have a Laurent series in any open annulus about its center where it's analytic, so you have to view everything in circles around the fixed center (which may or may not be a singularity for the function). In your example, zero is the center of each of the annuli, and it's not a singularity for the function. $\endgroup$ – William Mar 27 '13 at 3:28
  • $\begingroup$ Where you said "the centre where there isn't an expansion", did you mean "each singularity"? If so, then I totally agree with everything, except that I'm no closer to understanding why, if there's only a ring around each singularity whereby the function has a Laurent series, is it that the given function has Laurent series everywhere except at the circles containing its singularities? Am I making sense? $\endgroup$ – Ryan Mar 27 '13 at 3:48
  • $\begingroup$ Oh I just realised that for isolated singularities, the annulus is actually a punctured neighbourhood that is possibly infinite. So I think I've resolved my confusion. $\endgroup$ – Ryan Mar 27 '13 at 6:37
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I think some points of confusion can be clarified by looking at power series first.

Let $B(z_0,r) = \{z | |z-z_0|<r\}$, $\operatorname{ann}(z_0,r_1,r_2)= \{ z | r_1 < |z-z_0| < r_2\}$.

A power series centred at $z_0$ given by $z \mapsto \sum_{n=0}^\infty a_n (z-z_0)^n$ defines an analytic function for $z \in B(z_0,R)$, and diverges for $|z-z_0|>R$, where $\frac{1}{R} = \limsup_n \sqrt[n]{|a_n|}$. (Hence $R$ is the largest number such that the power series is analytic on $z \in B(z_0,R)$).

If the function defined by the power series has a non-removable singularity at some $z_1$, then clearly $R\le |z_1-z_0|$ (otherwise the function would be analytic at $z_1$). This is also what limits the radius of convergence of the Laurent series.

The power series coefficients (and hence the radius of convergence) depend on the function and the choice of $z_0$. For example, take $f(z) = \frac{1}{z}$. If we take $z_0=1$, then $f(z) = 1 - (z-1)+(z-1)^2-\cdots$ on $z \in B(1,1)$, whereas if we take $z_0=2$, we have $f(z) = 1-\frac{z-2}{2}+(\frac{z-2}{2})^2-\cdots$ on $z \in B(2,2)$. So $f$ has different power series in different regions. In this example, the radius of convergence of the power series will always be the distance from the centre to the singularity at $z=0$.

If a function $f$ is analytic on $\operatorname{ann}(z_0,r_1,r_2)$, then it has a Laurent series that converges uniformly & absolutely on any compact subset of the annulus. As with the power series, we must have ${r_2} \le {1 \over \limsup_{n \to \infty} \sqrt[n]{|a_n|}}$ and ${r_1} \ge { \limsup_{n \to \infty} \sqrt[n]{|a_{-n}|}}$.

By separating the negative and non-negative coefficient index parts of the Laurent expansion, we see that $f$ can be written as $f(z) = g_1(\frac{1}{z-z_0})+g_2(z-z_0)$, where $g_1,g_2$ are functions that are analytic on $B(0,\frac{1}{r_1})$ and $B(0,r_2)$ respectively. Hence the above remarks regarding power series can by applied mutatis mutandis to the Laurent series, taking into account the two-sided nature of things.

Note that all that is required is that $f$ be analytic on $\operatorname{ann}(z_0,r_1,r_2)$. There is no requirement that the center of the annulus lie on a singularity. For example, with $f(z) = \frac{1}{z}$ above, $f$ is analytic on $\operatorname{ann}(2,1,2)$, and the Laurent expansion is precisely the power series expansion in this case.

As with the power series, the Laurent expansion depends on the chosen annulus.

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  • $\begingroup$ Thanks. (a) is an open disc, why do you call it an annulus? Also, I don't see any removable singularity at z=0, nor do I know what B(0,1) means. $\endgroup$ – Ryan Mar 27 '13 at 3:07
  • $\begingroup$ The function is defined at $z=0$, so how is $0$ a removable singularity? I feel that I'm fundamentally misunderstanding your point. $\endgroup$ – Ryan Mar 27 '13 at 3:20
  • $\begingroup$ A bit late, but I have rewritten my answer to address what I think you were asking. $\endgroup$ – copper.hat Mar 27 '13 at 5:40
  • $\begingroup$ Thank you so much for taking the time to write this up. It has helped further my understanding. $\endgroup$ – Ryan Mar 27 '13 at 13:24
  • $\begingroup$ @copper.hat Did you mean to say $$\frac{1}{r_2} \ge \limsup_{n \to \infty} (|a_{n}|^{1/n})$$ instead of $\le$? Or perhaps I'm mistaken? $\endgroup$ – Muno May 10 '17 at 16:38
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Note that the given regions do not contain the singularities. A Laurent series is meaningful as long as the regions of definition do not contain the singularities. If the annuli are centered at a singularity the annuli will not contain the singularity.

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