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Singularities at ∞
Adrian Down
November 10, 2005
1 Outline
Today, we will discuss meromorphic functions. This idea is an extension of the notion of differentiability. With this idea, we will investigate singularities at ∞. We will also see some examples. This will conclude the discussion of singularities. Next, we will move on to calculating integrals using Cauchy’s residue theorem. The proof will be deferred, and we will mainly discuss applications.
2 Singularities
2.1 Review
Last lecture, we saw that if f has an isolated singularities at z 0 with Laurent coefficients an (expanded in powers of (z − z 0 )), then for some r > 0 such that f is differentiable in the disk of radius r, ∫
Cr
f = 2πı · a− 1
A better way to see this is to consider ∫
Cr
f (z)dz =
Cr
f (z) (z − z 0 )n+^
dz
If we choose n = −1, then this is a true statement. However, we already encountered the integral on the right. It is simply 2πı · an.
2.2 Meromorphic functions
2.2.1 Definition
Definition (Meromorphic). Let S ⊂ C be open. f is meromorphic in S if f is bounded in an open set S˜ of S and f has only isolated singularities in S, all of which are poles.
Note. • S˜ = S with a subset deleted, where this deleted subset has no limit points in S.
- Meromorphic functions are an extension of the idea of differentiable functions. Meromorphic functions differ from differentiable functions in that we allow meromorphic functions to have poles.
- We allow that f has removable singularities in S˜, since we assume that we can immediately remove any removable singularities.
2.2.2 Examples
Example.
f (z) =
z
S = C S˜ = C r { 0 }
Example.
f (z) =
P (z) Q(z)
where P, Q are polynomials and Q is not identically 0. Then,
S = C S^ ˜ = { z : Q(z) 6 = 0 }
The set of points where Q(z) = 0 is a finite set since Q is a polynomial.
Example.
f (z) = csc z =
sin z
f (z) is defined and differentiable where sin z 6 = 0 ⇔ z /∈ Z · π. The function has infinitely many singularities, but they are isolated. We show that f has a pole at 0.
Note. The point of this example is that we can easily calculate the leading coefficient of a Taylor expansion for a function g, a 0 = g(z 0 ), without obtain- ing explicit expressions for the rest of this series. This expansion can then be used to calculate the order of poles. This method is quite general and is often useful.
We have only found the behavior of sin z at z = 0. However, we previously proved that sin z is periodic with period 2π. Thus once we have found that sin z has a pole of order 1 at the origin, we know that all other singularities of sin z are also of order 1. Analytically, we would write, 1 sin z
sin w
where w = z − π. Then we could write the power series for (^) sin^1 w. Thus csc z is meromorphic.
Example. Thus far, all of the examples have been meromorphic on the entire complex plane. However, a function need not be meromorphic on the entire complex plane, as this example shows.
f (z) =
Logz z − 1
S = Cπ
The only singularity in S is at z = 1. To find the behavior at z = 1, observe that since Log is differentiable in a neighborhood about z = 1, it can be expanded in a Taylor series.
f (z) =
Log(1) z − 1
Since Log(1) = 0, the singularity at z = 1 is actually removable. The example could be made more complicated by considering,
f (z) =
Logz (z − 1)^5
The resulting Taylor expansion of f (z) will then have a pole of order 4 at z = 1.
Note. Any differentiable function divided by (z − α) to a power will have a pole at z = α.
2.3 Singularities at ∞
2.3.1 Isolated singularities
Definition (Isolated singularities at ∞). Suppose
- f is defined ∀z such that |z| > R
- 0 ≤ R < ∞
- f is differentiable for |z| > R
Then we say f has an isolated singularity at ∞.
Example.
f (z) = z f (z) = 0
2.3.2 Examples
Example.
csc z =
sin z
csc z does not have an isolated singularity at ∞, since sin z has a 0 for |z| > R for any R.
Example.
f (z) = z g(w) = f
w
w
This function has a pole at 0.
2.3.3 Types of singularities at ∞
Suppose f has an isolated singularity at ∞ (f is differentiable in { z : | z| > R } where R > 0). Consider
g(z) = f
z
2.5 Meromorphic functions and ∞
Theorem. Suppose
- f is a meromorphic function on C
- f has an isolated singularity at ∞
Then f is a rational function. I.e. f takes the form
f (z) =
P (z) Q(z)
for certain polynomials P and Q, where all zeros of Q are singularities of f.
Note. • Think of this as a generalization of Loiville’s theorem.
- In the interest of exploring some applications, we defer the proof of this theorem until next lecture.
3 Computing integrals
3.1 Cauchy’s Residue theorem
3.1.1 Preliminary definitions
Definition (Simple contour). Let γ be a closed contour. γ is called simple if for any z /∈ γ,
w(γ, z) ∈ { 0 , 1 }
Note. • Simple contours are similar to Jordan curves. Any Jordan curve is simple, but simple curves need not be a Jordan curve.
- We will soon have to choose curves along which to perform integrals, and we will often try to choose simple curves.
Definition (Inside/outside of a simple curve). If γ is a simple contour (or path), a point z is inside of γ if w(γ, z) = 1. z is said to be outside of γ if w(γ, z) = 0.
3.1.2 Theorem
Theorem (Cauchy’s Residue Theorem). Suppose
- D is a domain
- f is defined on D except for isolated singularities.
- γ is a simple contour in D, so that every point inside γ lies in D
- No singularity f lies in the range of γ
Then ∫
γ
f =
∑^ N
j=
(2πı) · res(f, zj )
where { z 1 , z 2 ,... , zN } is the set of all the singularities of f which are inside γ.
Note. • Recall the definition of the residue of a function f at an isolated singularity, res(f, zj ) = coefficient of (z−zj )−^1 in the Laurent expansion of f around zj
- We defer the proof of this theorem until next time in favor of some applications.
- At first glance, it looks as if this theorem has nothing to do with inte- gration. Integrals from −∞ to ∞ correspond to an infinite path, not a closed path. It is important to recognize this disconnect before seeing how the theorem does indeed relate to integrals.
3.2 Applications
−∞
dx x^2 +
Real analysis method ∫ (^) ∞
−∞
dx 1 + x^2
= lim N →∞
∫ N
−N
dx 1 + x^2 = lim N →∞ (arctan(N ) − arctan(−N ))
=
π 2
π 2
= π
We have obtained the right answer, but we have done the wrong problem. The integral is not along the infinite real line, since we have considered only a finite portion of the real line, and we have introduced an extra semi-circular path. We will continue next lecture and demonstrate how to connect the prob- lem that we just solved with the actual integral that we would like to cocm- pute.
