Laurent Series and The Residue Calculus

Seminar
 [3:00pm] ness: ok, so let's begin [3:01pm] ness: So recall that a holomorphic function can be (locally) expanded as a power series: [3:02pm] ness: f(z) = sum_{n=0}^{infty} c_n (z-a)^n, where c_n = f^(n)(a)/n! [3:02pm] ness: This is just taylor's theorem in the complex case [3:03pm] ness: But a function described this way obviously cannot have singularities in the region (disk) of convergence [3:03pm] ness: so taylor series are of limited use if we want to study functions that contain poles but are &quot;nice&quot; otherwise [3:04pm] ness: consider therefore the following straightforward extension: [3:04pm] ness: f(z) = sum_{n=-infty}^{infty} c_n (z-a)^n, for some c_n, where it converges [3:04pm] ness: This is called a laurent series expansion. [3:05pm] ness: We will talk a bit about properties of laurent series [3:05pm] ness: Throughout, let R^* denote R_0^+ u {infty}, i.e. the non-negative reals or infinity [3:06pm] ness: this is purely symbolic, with the obvious ordering, but no more structure [3:06pm] ness: We might first of all wonder: what is the region of convergence of a particular laurent series? [3:07pm] ness: There is a relatively simple answer: there exist r_1 &lt; r_2 in R^* such that the laurent series converges inside the annullus D(a, r_1, r_2) = {z in C | r_1 &lt; |z-a| &lt; r_2}, and diverges outside of it. [3:08pm] ness: As in the taylor series case, we cannot know in advance for z on the border [3:09pm] ness: I won't prove this here, but it can be proved using much the same ideas that are used in the analogous theorem about taylor series, and there are similar lim sup formulas to determine r_1 and r_2 [3:09pm] ness: There are a few special cases: [3:10pm] ness: if c_n = 0 for all n &lt; some n_0 (where n_0 &lt; 0), then obviously r_1 = 0 [3:11pm] ness: further, if we have c_n = 0 for all n&lt;=0 we are left with a holomorphic function and thus we know that not only r_1 = 0 but also that the series converges at a at well [3:12pm] ness: Now for a bit of terminology: the negative powers part of the laurent series (i.e. the c_{-1}/z + c_{-2}/z^2 + ...) is called the prinzipal part [3:12pm] ness: er principal part [3:12pm] ness: so we can restate the above as: if the laurent series has a finite principal part, then r_1 = 0 [3:13pm] ness: We might ask: what else can we do with laurent series? [3:13pm] ness: There is one nice application: we can classify singularities [3:13pm] ness: We can say (indeed, define) the following: [3:13pm] ness: If the principal part vanishes, then we have a removable singularity [3:14pm] ness: If c_n = 0 for all n &lt; -n_0 &lt; 0, we say that the function has an n-th order pole [3:15pm] ness: and finally, if none of the above holds, the singularity is said to be essential [3:16pm] ness: Now of course there is a (superficially) more general way to classify singularities: Say we have a function f(z), not necessarily given as a laurent series expansion [3:16pm] ness: We can consider the limit lim_{z-&gt;a) (z-a)^n f(z), for n &gt;= 0 [3:17pm] ness: If the limit exists for all n (&gt;=), then we say f has a removable singularity [3:17pm] ness: if the limit exists for all n &gt;= n_0, we say that the function has an nth order pole [3:18pm] ness: and if the limit diverges for all n, we say that the function has an essential               singularity at a [3:19pm] ness: now that I use the same terminology here already suggests this: if a function has a                laurent series expansion at a with r_1 = 0, then the two ways of classifying the               singularities are *equivalent* [3:19pm] ness: We will need this later [3:20pm] ness: There is a formula to compute c_n (in fact it looks exactly like the cauchy formula expression for the c_n of a taylor series), but I won't talk about it here because we              don't need it [3:21pm] ness: So much for a detour in more or less abstract series expansions [3:21pm] ness: You may well ask how all of this is related in any direct way to the topics covered                last time [3:22pm] ness: but there is a relation, which you will see shortly [3:22pm] ness: Recall that a meromorphic function f(z) is defined as the quotient of two holomorphic               functions: f(z) = g(z)/h(z) [3:23pm] ness: We will see that all meromorphic functions have laurent series expansions with r_1 = 0               around all a (in their region of holomorphy) [3:24pm] ness: To show this, we need a &quot;classification of roots&quot; [3:24pm] ness: Let f(z) be a holomorphic function, and consider L = lim_{z-&gt;a} f(z)/(z-a)^(n+1),                for n &gt; 0 [3:25pm] ness: If L diverges for all n, we might call this a &quot;removable root&quot;, or &quot;no root at all&quot; (quotation marks because this is no standard terminology) [3:25pm] ness: If L=0 for n&lt;n_0 and l != 0 for n=n_0, we call this a root of n_0-fold multiplicity [3:26pm] ness: And finally if L=0 for all n, we might call a an &quot;essential root&quot; [3:26pm] ness: Now there is a very simple theorem about this classification: [3:27pm] ness: Non-constant holomorphic functions have *no* essential roots (this is essentially the              fundamental lemma of holomorphic continuation if you know what this is) [3:28pm] ness: We can prove this by contradiction: let f be a holomporphic function with an essential root at a. [3:28pm] ness: Then (locally) f(z) = sum_{n=0}^{infty} f^(n)(a)/n! (z-a)^n [3:29pm] ness: But also lim_{z-&gt;a} f(z)/(z-a)^n = 0 for all n (by the definition of &quot;essential root&quot;) [3:29pm] ness: so we see that all the coefficients of the power series expansion must be zero, thus f              vanishes identically [3:29pm] ness: a contradiction! [3:30pm] ness: Equipped with that, we can show the relation between meromorphic functions and laurent series: [3:30pm] ness: lot f(z)=g(z)/h(z) where g and h are holomorphic (i.e. f is meromorphic). [3:31pm] ness: Let a be in C. Then (by the above) h has at most a singularity of finite multiplicity at a, say of multiplicity n [3:31pm] ness: But then phi(z) = (z-a)^n g(z)/h(z) = (z-a)^n f(z) must be holomorphic [3:32pm] ness: So it has a (local) power series expansion around a [3:32pm] ness: and it follows that f has a local laurent series expansion (around a) with finite principle part! [3:33pm] ness: But the taylor (laurent) cofficients are arbitrary (as are g and h), so we see the following: [3:33pm] ness: Meromorphic functions (ratios of holomorphic functions) are exactly the functions that have local laurent expansions with finite principal part. [3:34pm] ness: Now we can go back to the residue theorem [3:35pm] ness: Recall that we had shown that oint_{Gamma) f(z) dz = 2 pi i sum_a ind_{Gamma}(a) res_a f [3:36pm] ness: where f is meromorphic the sum is over all the enclosed singularities (poles),              Ind_{Gamma}(a) is the &quot;winding number&quot; of Gamma around a (recall the picture! figure 6)  (read ahead for correction)[3:36pm] ness: and where we defined Res_a f = 2 pi i oint_C f(z) dz, where               C is a closed circle around a enclosing no other singularities, oriented in the                counterclockwise sense [3:37pm] ness: Now we can plug in our new knowledge on laurent series here: We know that, locally,                f(z) = sum_{n=-n_0}^{infty} c_n (z-a)^n [3:38pm] ness: We might ask: does this detailed knowledge (assume for the sake of argument we know the correct c_n) help us to evaluate the residue? [3:38pm] ness: Indeed, it does: [3:39pm] ness: First of all, we see that               f(z) = sum_{n=-n_0}^{-1} c_n (z-a)^n + sum_{n=0}^{infty} c_n (z-a)^n,                but the second sum is actually a holomorphic function! [3:40pm] ness: So wee see that all the c_n with n &gt;= 0 cannot possibly affect the residue [3:40pm] ness: Now let's look at the c_{-1} coefficient: [3:41pm] ness: oh [3:41pm] ness: I just saw I mistyped the definition of residue above [3:41pm] ness: Res_a f = 1/(2 pi i) oint_C f(z) dz is the correct version [3:42pm] ness: so the c_{-1} term contributes 1/(2 pi i) oint_C c_{-1}/z dz to the residue [3:43pm] ness: but we have evaluated this integral already, in fact it was the first contour integral we               ever evaluated! We know that oint_C dz/z = 2 pi i [3:43pm] ness: so the c_{-1} term contributes just c_{-1} to the residue (now you see why the definition has the superficially useless 1/(2 pi i) factor!) [3:44pm] ness: Now let's look at the c_n left (n &lt; -1) [3:45pm] ness: To find the contributions to the residue, we evaluate c_{-n} oint_C dz/z^n [3:46pm] ness: We can do this by parametrizing the circle as e^(it), this becomes              int_0^{2 pi} 1/e^(itn) * i e^it dt which is easily evaluated to be zero! [3:46pm] ness: So the c_{-1} term is the *only* one contributing to the residue [3:47pm] ness: in fact, we could have defined Res_a f = c_{-1}, where                f(z) = sum_{n=-n_0}^{infty} c_n (z-a)^n in the first place [3:48pm] ness: So we see that evaluating integrals using the residue theorem boils down to finding                the c_{-1} coefficient [3:48pm] ness: &quot;tricks&quot; for doing so are often summarized as calculus of residues [3:49pm] ness: Let's start with the obvious: Res_a (f+g) = Res_a f + Res_a g, and Res_a (c*f) = c*Res_a f where c is a constant [3:50pm] ness: Now of course we can always find Res_a f by expanding f into a laurent series (perhaps               only partially, so that we see the c_{-1} coefficient), probably using clever expansions of the functions f is built up from [3:51pm] ness: But there is one more, particularly useful formula for evaluating Res_a f, in the case that we know f to have an nth order pole at a: [3:51pm] ness: assume thet f has an nth order pole at a, i.e. locally f(z) = sum_{k=-n} c_k (z-a)^k [3:52pm] ness: But then (z-a)^n f(z) is holomorphic, so (again locally) (z-a)^n f(z) = sum_{k=0}^{infty} b_k (z-a)^k, where b_k = 1/k! lim_{z-&gt;a} d^k/dz^k [(z-a)^n f(z)] [3:53pm] ness: (The formula for b_k might look daunting, but it is just the usual c_n = f^(n)/n! for              this case) [3:53pm] ness: (er f^(n)(a)/n!) [3:54pm] ness: But this gives us another laurent expansion of f(z), by dividing the obtained expansion of (z-a)^n f(z) by (z-a)^n [3:54pm] ness: So f(z) = sum_{k=-n}^{infty} b_k (z-a)^{k-n} [3:55pm] ness: and, as the regions of convergence must overlap, we can match coefficients to yield c_{-1} = b_{n-1} [3:55pm] ness: So we get [3:55pm] ness: Res_a f = 1/(n-1)! lim_{z-&gt;a} d^{n-1}/dz^{n-1} [(z-a)^n f(z)] [3:56pm] ness: Again, this might look daunting at first, but it is a very convenient way to evaluate certain integrals [3:56pm] ness: To demonstrate this (and other things I talked of in the other two seminars), let me show you how the machinery developed can be used to evaluate complicated real integrals [3:57pm] ness: Let's try to evaluate I = int_{-infty}^{infty} dx/(x^2+x+1) [3:57pm] ness: This looks quite hard if we try it using traditional techniques (seems to be an arctangent              in somewhere, but nothing obvious, at least to me) [3:57pm] ness: We extend this to a contour integral: [3:58pm] ness: Consider I' = oint_C dz/(z^2+z +1), where the contour is a semicircle as indicated in               figure 7  [3:59pm] ness: The semicircle is oriented counterclockwise and the radius should tend to infinity [3:59pm] ness: I claim that I' = I [4:00pm] ness: Indeed, this is because lim_{|z| =&gt; infty} z*f(z) = 0 [4:00pm] ness: We can clearly see that our integrand satisfies this relation, but how does this help? [4:01pm] ness: Now, it can be shown that |int_C f(z) dz| &lt;= int_C |f(z)| |dz| &lt;= L * M, where L is the length of the contour and M is a bound of |f(z)| on the contour. This is known as the standard estimate [4:03pm] ness: Applying this to a (semi-circle) we find that |int_C f(z) dz| &lt;= pi * R * M. But R*M goes to zero as R (=|z|) goes to infinity (by assumption) [4:03pm] ness: so we see: I = I' [4:03pm] ness: We have thus &quot;reduced&quot; a real integral to an integral around a closed contour in the complex plane, and we can use the residue theorem [4:04pm] ness: To do so, we must first find the poles of the integrand [4:04pm] ness: we solve z^2+z+1 = 0, find z_{1,2} = (-1 +- isqrt(3))/2 [4:04pm] ness: but only z_1 is inside the contour of integration (effectively the upper half plane) [4:05pm] ness: so I = 2 pi i Res_{z_1} 1/(z^2+z+1) [4:05pm] ness: But 1/(z^2+z+1) = 1/(z-z_1)*1/(z-z_2), so the pole is simple, and by the above formula for residues, [4:06pm] ness: Res_{z_1} 1/(z^2+z+1) = lim_{z-&gt;z_1} (z-z_1)/(z^2+z+1) = 1/(z-z_2) evaluated at               z=z_1, = 1/(i sqrt(3)) [4:07pm] ness: Thus we find I' = I = 2pi/sqrt(3), almost without thinking! [4:07pm] ness: Ok. [4:07pm] ness: So much for the residue calculus [4:07pm] ness: I will stop here [4:08pm] ness: There is still plenty of stuff to explore in the wonderful world of complex analysis [4:08pm] ness: Are there any questions? [4:09pm] ness: Hehe, all bored away