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) ≡ −∞

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.

Q But can you give a rough outline of this?

A OK, a very rough outline. I let you fill in the detail. Define Z 1 f(w) an = n+1 dw 2πi |w−z0|=r (w − z0)

where r is any number satisfying 0 < r < R. First we check that an is independent of the choice of r > 0. This can be achieved by applying Cauchy’s theorem to the n+1 function f(z)/(z − z0) on the annulus {z : r1 ≤ |z − z0| ≤ r2} for arbitrary

r1, r2 with 0 < r1 < r2 < R. Next, fix a point z in the open disk D(z0; R) with

z 6= z0. Take any r1, r2 > 0 such that 0 < r1 < |z − z0| < r2 < R and let D be

the annulus {w : r1 < |w − z0| < r2}. By Cauchy’s formula, Z ÃZ Z ! 1 f(w) 1 f(w) dw f(z) = dw = − . 2πi ∂D w − z0 2πi |w−z0|=r2 |w−z0|=r1 w − z0

If |w − z0| = r1, then |w − z0| < |z − z0| and hence

X∞ k 1 1 1 1 (w − z0) − = − = = k+1 . w−z0 (w−z0) − (z−z0) z−z0 1 − (w−z0)/(z−z0) (z − z0) k=0

Consequently à ! Z X∞ Z 1 f(w) 1 k 1 − dw = f(w)(w − z0) dw k+1 2πi w − z0 2πi (z − z0) |w−z0|=r1 k=0 |w−z0|=r1 à Z ! X∞ 1 X∞ = f(w)(w − z )−n−1dw (z − z )n = a (z − z )n. 2πi 0 0 n 0 n=−1 |w−z0|=r1 n=−1 R P∞ n In the same way we get = an(z − z0) . Done. |w−z0|=r2 n=0

5 Q How do you classify singularities?

A Back to (5), consider two cases: the case that there exits an integer n0 such that

an = 0 for all n < n0 and the case that such n0 does not exist. In the latter z0 is called an essential singularity of f. For example, the Laurent expansion of e1/z at z = 0 is given by 1 1 e1/z = 1 + + ··· + + ··· z n!zn and hence e1/z has an essential singularity at z = 0. In the first case we can pick

n0 such that an0 6= 0 and an = 0 for n < n0. Then, as before, we may write

n0 f(z) = (z − z0) F (z), where F is an analytic function with F (z0) 6= 0. Here n0 is

negative; otherwise z0 would not be a singular point. In the first case, z0 is called a pole of f.

Q I’ve heard that meromorphic functions are those for which all singularities are poles. Why do we bother to give a name for them?

A They are important for several reasons. First, complex functions encountered in prac- tice are mostly meromorphic functions, such as tan z. Second, all meromorphic

functions defined on something called Riemann surface S form a field MS: if f, g are meromorphic, then so are f + g, fg and f/g (assuming g 6≡ 0). This gives a link between field extensions and branched coverings of Riemann surfaces, one from the area of algebra and the other from topology. It relates field automorphisms to covering transformations. But we are not going to touch upon this interesting but advanced aspect of meromorphic functions: we keep our discussion elementary. For our purpose here, meromorphic functions are important because of the residue the- orem. The residue calculus allows us to manipulate many intriguing integrals, most of which are inaccessible by elementary calculus. So it is considered to be a major application of complex analysis.

Q How should we think of meromorphic functions?

A One may think that analytic functions generalize polynomials and meromorphic func- tions generalize rational functions. We have seen that the quotient of two analytic

functions is meromorphic. The converse holds at a local level. Indeed, if z0 is a

n0 pole of a meromorphic function f, we can write f(z) = (z − z0) F (z), where n0

is a negative integer and F is analytic on a neighborhood N of z0, which is a

6 N quotient of f and (z − z0) , where N = −n0 > 0. This indicates that the relation of meromorphic functions to analytic functions is like that of rational functions to

polynomials. There is another way to think of meromorphic functions: at a pole z0 a meromorphic function is an analytic function plus a sum of partial fractions, say

P (z; z0), which is an expression of the form given in (4) above, called the principal

part of f at z0. Suppose that f has only finitely many poles, say z0, z1, . . . , zN . Let PN R(z) = k=0 P (z; zk), the sum of all principal parts of f. Let F (z) = f(z) − R(z). Then g has no poles and hence is analytic. rewrite F = f − R as f = R + F , we see that the meromorphic function f with finitely many poles is the sum of a rational function R and an analytic function F .

Q In the first year calculus you use partial fractions to compute integrals of rational functions. But why one should put partial fractions in a particular pattern is rarely explained. Now I can tell that we are getting close to a good explanation.

A Oh, yes. Suppose that f is rational. Clearly f has only finitely many poles in the complex plane C. Write f = R + F as before. Then F = f − R is also a rational function. But F has no poles in C. So F must be a polynomial. Thus f = F + R gives us the partial fraction decomposition of f used in the first year calculus.

Q You have mentioned the residue theorem. I am eager to know this important thing. What is it about? R A It is about a contour integral ∂D f(z) dz, of a meromorphic function f with finitely many poles in D and no poles on ∂D. Again write z0, z1, . . . , zN for these poles and let f = R + F be the decomposition as before. Since F is analytic, R we have F (z) dz = 0 in view of Cauchy’s theorem. So it remains to consider R ∂D ∂D R(z) dz. Here R is a sum of principle parts. Let us take one of them, say PN −k P (z; z0) = a−k(z − z0) . Thus it remains to consider integrals of the form R k=1 −k ∂D (z − z0) dz. When k = 1, the answer is 2πi, obtained by applying Cauchy’s −k formula (1) to f ≡ 1. When k 6= 1, we can write (z − z0) as the derivative of −k+1 (z − z0) /(−k + 1) and hence the integral is zero. (If C is a path running R 0 from z0 to z1, then g (z) dz = g(z1) − g(z0) for any function g analytic R C R 0 along C. In particular, g (z) dz = 0 if C is closed.) So we have P (z; z0) dz = R C C −1 a−1 (z−z0) dz = 2πi a−1. The number a−1 is called the residue of f at z0 and C R will be denoted by Res(f; z0). We have shown that ∂D P (z; z0) dz = 2πi Res(f; z0).

7 Now we put things together: Z Z Z XN XN f(z) dz = R(z) dz = P (z; z0) dz = 2πi Res(f; zk). ∂D ∂D k=0 ∂D k=0

Thus we obtain the following identity, which concludes the residue theorem: Z 1 XN f(z) dz = Res(f; z0). 2πi ∂D k=0 R That is, the integral (1/2πi) ∂D f(z) dz of the meromorphic function f is equal to the sum of all its residues in D.

Q After Cauchy’s formula the residue theorem is the next important theorem, I suppose.

A Quite so. But Cauchy’s theorem can be regarded as a special case of the residue

theorem. Indeed, the integrand g(z) ≡ f(z)/(z − z0) in (1) is a meromorphic

function with z0 as its only pole. We can compute its residue as follows. Write down P∞ n the power series expansion of f at z0: f(z) = n=0 an(z − z0) . Then we have the

Laurent expansion at z = z0:

f(z) a0 2 g(z) = = + a1 + a2(z − z0) + a3(z − z0) + ··· z − z0 z − z0

So Res(g; z0) = a0 = f(z0). Thus, applying the residue theorem to g, we get Cauchy’s

formula. By the way, the method of figuring out the residue of g at z0 here is applicable in many situations.

Q I think now I understand the residue theorem well. Tell me its application to evaluation of integrals.

A When we try to use the residue theorem to evaluate an integral, we have to take three major steps: find a suitable meromorphic function, choose an appropriate contour, and calculate residues. The second step is often more tricky. In applying to some improper integrals, often we have to move parts of contour to infinity or to shrink some semicircle to a point. Many examples can be found in standard textbooks.

Q OK. Let us stop now.

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