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Part 4: Meromorphic Functions and Residues

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Part 4: Meromorphic Functions and Residues COMPLEX ANALYSIS PART 4: MEROMORPHIC FUNCTIONS AND RESIDUES Q What are we going to discuss today ? A Before starting a new topic, let us recall what we did so far. Can you briefly describe this in your own words, sparing all details ? Q Why do we have to review at this moment? I am eager to learn new things. A In math, new things are often developed from old ones. Other subjects such as physics are all alike in this respect. To understand Einstein’s relativity theory, you have to know Newton’s mechanics. Besides, learning a math subject is like taking a long journey. From time to time we have to recall the places we have passed in order to find out where we are, to avoid being completely lost. Q I’m afraid that I can’t remember everything correctly. A Give a try. You may consult your notes if necessary. But don’t look up the notes without giving a try. In doing this, conduct a self-evaluation to see how well you can visualize the big picture. Q OK. let me go through this briefly. We begin with basics of complex numbers. Then we move to complex functions and we define a function to be analytic if it is complex differentiable at each point of its domain. This leads to CR-equations, which can be simply put as ∂f/∂z¯ = 0. Then we introduce contour integrals and we use Green’s formula to prove Cauchy’s theorem, which says Z f(z) dz = 0 ∂D for an analytic function f in a suitable set-up. This theorem is used to establish Cauchy’s integral formula, which says Z 1 f(w) f(z0) = dw (z0 ∈ D) (1) 2πi C w − z0 1 for an analytic function in an appropriate set-up. Then we use this formula to deduce several important properties of analytic functions, such as the existence of power series expansion at each point. A You are doing pretty well. Can you elaborate power series expansions a bit? Q It says that if f is an analytic function with an open set U as its domain, and if z0 is a point in U, then we can write 2 f(z) = a0 + a1(z − z0) + a2(z − z0) + ··· , (2) where the series on the right hand side converges uniformly to f(z) on any closed disk D¯(z0; R) contained in U centered at z0. Here the coefficients an can be described in two ways: (n) Z f (z0) 1 f(w) an = , and an = n+1 dw. (3) n! 2πi |w−z0|=R (w − z0) Here I simply write “|z − z0| = R” for the circle Cz0;R ≡ ∂D(z0; R) oriented in the counterclockwise manner. (n) A What happens if all derivatives f (z0) at z0, imcluding n = 0, vanish? Q In that case the above power series vanishes and hence f is zero on the entire disk D(z0; R). A This is rather amazing: what happens at a point z0 can affect the entire neighborhood D¯(z0; R). Actually, more is true: it can affect the whole domain U as long as U is connected. Let me make this precise: if f is an analytic function on a connected open set U and if (n) f (z0) = 0 (n = 0, 1, 2,... ) at some point z0 ∈ U, then f is a zero function: f ≡ 0. Q How do you prove this? A Take any point z1 in U. We have to show f(z1) = 0. Because U is connected, we can link up z0 and z1 by a curve C. Due to a topological consideration, we know that there is a positive number R > 0 such that |z − w| > R for all z on C and all w not in U. Another topological consideration allows us to find a finite sequence of open disks D(c1; R),D(c2; R),...,D(cN ; R) 2 of radius R such that c1 = z0, cN = z1 and |ck − ck+1| < R (1 ≤ k ≤ N − 1), that is, each disk contains the center of the next disk. We have seen that f vanishes on (n) the first disk D(c1; R) because c1 = z0 and f (z0) = 0 (n ≥ 0) by our assumption. (n) Since c2 belongs to D(c1; R) and since f vanishes on D(c1; R), we have f (c2) = 0 (n ≥ 0) and consequently f vanishes on D(c2; R). This argument shows in general that the property “f vanishes on D(ck; R)” passes on to the next disk. Now you can see why f(z1) ≡ f(cN ) = 0. Q From now on let us assume that f is not identically zero. Thus, in the power series expansion, there is at least one n such that an 6= 0. Now, what can we say about the local behavior of f at z0? A Let N be the smallest integer such that aN 6= 0. Thus we have N N+1 N f(z) = aN (z − z0) + aN+1(z − z0) + ··· = (z − z0) F (z) where F (z) = aN + aN+1(z − z0) + ··· with F (z0) = aN 6= 0. When N ≥ 1, we call z0 a zero of f and N the order of this zero. If f is a polynomial, then of course z0 in this case is a root of f and N is the multiplicity of this root. N From the expression (z − z0) F (z) we know that the zero z0 is isolated: there is a neighborhood N of z0 such that z0 is the only zero of f in N . Q Let me try to prove this. Since F (z0) = aN 6= 0, by the continuity of F we know that there is a neighborhood N on which F never vanishes. WE are going to show that z0 is the only zero of f in N as follows. Suppose that z1 ∈ N is a zero of f in N . Then N (z1 − z0) F (z1) = f(z1) = 0 N but F (z1) 6= 0. Hence (z1 − z0) = 0, which gives z1 = z0. A Good. Next we consider the reciprocal 1/f. What is its local behavior at z = z0? N Let us write f(z) = (z − z0) F (z) again, assuming N ≥ 1 and F does not vanish at any point in a neighborhood N of z0. (We may assume that N is an open disk, say N = D(z0; r).) Now 1/F is an analytic near z0 and hence we have a power series P∞ k expansion of 1/F at z − z0, say 1/F (z) = k=0 bk(z − z0) . Thus X∞ X 1 1 1 k−N n = N = bk(z − z0) = cn(z − z0) , f(z) (z − z0) F (z) k=0 n≥−N 3 where cn = bk in case k − N = n, or cn = bn+N . Q What are you going to do next? P n A Next, I split the sum n≥−N cn(z − z0) into two parts. The first part is a sum of partial fractions: X−1 n c−N c2 c−1 cn(z − z0) ≡ N + ··· + 2 + . (4) n=−N (z − z0) (z − z0) z − z0 P∞ n It is called the principal part. The second part is h(z) ≡ n=0 cn(z − z0) , which represents an analytic function defined on a neighborhood of z0. (Our discussion here can be repeated in the future for a meromorphic function with z0 as a pole.) Q Suppose that, instead of 1/f, we have g/f, where g is another analytic function defined on U. What happens? N M A We write down f(z) = (z−z0) F (z) again. Similarly, we write g(z) = (z−z0) G(z), where G(z0) 6= 0. Then we have g(z) G(z) = (z − z )M−N . f(z) 0 F (z) When M ≥ N, g/f is analytic at z0. When M < N, we can repeat the above discussion ... Q I know how to handle the rest. A Now let us entertain a general situation. Assume that f is a function analytic on U, except at some isolated points called singularities. Let z0 be a point of singularity for f. That z0 is isolated means that we can take a disk D(z0; R) centered at this point so that f is analytic at each point of this disk except its center z0. In a neighborhood of this singularity, we can express f by a Laurent series: X n f(z) = an(z − z0) ≡ −∞<n<∞ (5) −2 2 ··· + a−2(z − z0) + a−1(z − z0) + a0 + a1(z − z0) + a2(z − z0) + ··· which converges uniformly in an annulus {z : δ ≤ |z| ≤ R}, where δ is any positive number less than R. Q How do you prove (5)? 4 A Laurent series expansions generalize power series expansions. So you expect that the proof is similar. Alas, this is not quite the case. The first step is to guess what an in (5) are. For power series expansion at a regular point (a point is regular if it is not a singularity) we have two ways to put an: see (2) and (3) above. (2) is no good for (n) the present situation because z0 is a singularity of f and hence f (z0) does not make sense. So we guess an is given by (3). This is a correct guess. From this we can start the proof. But we are not planning to do here because it is a bit technical, and not suitable for our conversation.
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