Complex Analysis: #21 Logarithmic Derivative

• Complex Analysis: #21 The Logarithmic Derivative

Let f : G → ℂ be a function. Thinking about the rules for derivatives, we can combine what we know about the derivative of a logarithm and the chain rule to arrive at the interesting observation that

This is an interesting equation, particularly so in the field of analytical number theory. But the theorem we will look at here concerns path integrals. For simplicity, we will assume that G = ℂ, and that the function f is meromorphic, so that it is analytic everywhere in ℂ, except possibly for some set of isolated singularities. At each of these singularities, f has a pole of some order.

So let us say that f has a pole of order k at the point a ∈ ℂ. Then we can write

Combining this with the residue theorem, we have a method of counting the zeros and poles. Namely:


Theorem 40
Let f be a meromorphic function, defined in a region G ⊂ ℂ, and let Ω be some cycle which avoids all zeros and poles of f and which is such that all points of ℂ \ G have index zero with respect to Ω. Let N_Ω be the number of zeros of f which have non-vanishing index with respect to Ω, counted according to their orders, and let P_Ω be the poles, again counted with their orders. Then

Another way to look at this is to remember that f is, after all, just a mapping of a region of ℂ back into ℂ. Thus if γ is a closed path in the region G (again, with winding number zero with respect to all points in the compliment of G in ℂ), it follows that f ◦ γ, given by f ◦ γ(t) = f(γ(t)), is itself another closed path in ℂ. If we assume that γ passes through no zero or pole of f, then the path f ◦ γ avoids both 0 ∈ ℂ and also the special point ∞. Therefore we can think about the index of 0 with respect to this path.


Theorem 41
ν_f◦γ(0) = N_γ − P_γ, where the numbers N_γ and P_γ have been defined in the previous theorem (theorem 40).

Proof
As we saw in theorem 25, we have

where we use the result of the previous theorem. Here, we have just imagined that the path γ is defined on some interval of the form [a, b]. Of course it is a trivial matter to see that we could replace the single path γ with a cycle Ω.


Theorem 42 (Roché)
Let f, g : G → ℂ be analytic functions, and let Ω be a cycle in G. Assume A = {w ∈ ℂ : ν_Ω(w) ≠ 0} ⊂ G, and assume furthermore that |g(z)| < |f(z)| for all z which lie directly on Ω. Then both functions f and f + g have the same number of zeros in A.

Proof
Let z be a point of Ω. Then since 0 ≤ |g(z)| < |f(z)|, we certainly do not have f(z) = 0. But also, for all τ ∈ [0, 1], we have

|f(z) + τg(z)| ≥ |f(z)| − τ|g(z)| > 0. 

Therefore the cycle (f + g) ◦ Ω is homotopic to the cycle f ◦ Ω in ℂ \ {0}, and the result then follows from theorem 41.

This theorem gives us a very quick proof of the Fundamental Theorem of Calculus. For let

P(z) = zⁿ + a_n−1zⁿ + · · · + a₁z + a₀

be some polynomial of degree n ≥ 1 in ℂ. Let R = |a_n−1| + · · · + |a₀| + 1. Then on the circle given by |z| = R, we have

0 ≤ |a_n−1zⁿ + · · · + a₁z + a₀| < |zⁿ|. 

But then theorem 42 implies that P(z) must have n zeros (counted with their multiplicities) within the circle of radius R.

• Complex Analysis: #20 Symmetric Real Functions

For example, how do we calculate the integral

On the one hand, there is no singularity at 0, but on the other hand, how are we to calculate the integral other than by using the calculus of residues? [For example, partial integration only seems to make things more complicated here. Nevertheless, I see that the computer algebra system MuPAD does give the correct answer with little fuss!] Let us begin by noting that

But the residue of exp(ix)/x at 0 is simply 1. Therefore, theorem 38 shows that

This example shows that if we have a meromorphic function f which is defined in ℂ, such that f(x) = f(−x) for all x ∈ ℝ, and such that the conditions of theorem 38 are satisfied, then it does make sense to calculate the integral  ∫₀^∞ f(x)dx, using the Cauchy principle value technique.

• Complex Analysis: #19 Integrating Out From a Pole

Maybe we are dissatisfied with this “Cauchy principle value” technique. After all, it is rather like cheating! So let’s see what we can do with an integral like

where 0 is a pole, and ∞ is a zero of f. For example, look at the function f(x) = 1/x. But here we see big problems! Both of the integrals ∫₁^∞ f(x)dx and  ∫₀¹ f(x)dx are divergent.

Thinking about this, we see that the problems with the function 1/x stem from the fact that, first of all, the zero at ∞ is simple, and second of all, the pole at 0 is simple. This leads us to formulate the following theorem.


Theorem 39
Again, let R be a rational function defined throughout ℂ, but this time with a zero of order at least 2 at ∞. Furthermore, R has no poles in the positive real numbers (x > 0), and at most a simple pole at 0. Then we have

for all 0 < λ < 1.

Proof
As before, the function z^λR(z) has at most finitely many poles in ℂ. Let γ_{r, φ}, where 0 < r < 1 and 0 < φ < π, be the following closed curve. It starts at the point re^φi and follows a straight line out to the point Te^φi, where T = 1/r. Next it travels counter-clockwise around the circle of radius T, centered at 0, till it reaches the point Te^(2π−φ)i. Next it travels along a straight line to the point re^(2π−φ)i. Finally it travels back to the starting point, following the circle of radius r in a clockwise direction. Lets call these segments of the path L₁, L₂, L₃ and L₄. By choosing r and φ to be sufficiently small, we ensure that the path γ_{r, φ} encloses all poles of the function. In the exercises, we have seen that the path integrals along the segments L₂ and L₄ tend to zero for r → 0. Thus we have

• Complex Analysis: #18 Integrating across a Pole

For example, consider the "integral"

Obviously this is nonsense, since the integrals from −1 to 0, and from 0 to 1 of the function 1/x diverge. Specifically, for 0 < ∈ < 1 we have

Thus, if we agree to abandon the principles we have learned in the analysis lectures, and simply say that

then we have the Cauchy principle value of the integral, and we see that in this case it is simply zero. One could make an important-looking definition here, but let us confine our attention to integrals along closed intervals [a, b] ⊂ ℝ of complex-valued functions, where there might be poles of the function in the given interval. Assume for the moment there is a single pole at the point p ∈ (a, b). Then we will define the principle value of the integral (if it exists) to be

Then the generalization to having a finite number of poles of f along the interval (but not at the endpoints) is clear.


Theorem 38
Let R be a rational function, defined throughout ℂ (together with it’s poles). Assume that it has a zero at infinity, so that there can only be finitely many poles. Let p₁ < . . .  < p_m be the poles of R which happen to lie on ℝ. Assume that each of these poles is simple; that is, of order 1. We distinguish two cases:

• If R has a simple zero at infinity (that is, of order 1), then we take f(z) = R(z)e^iz. 
• Otherwise, R has a pole of order at least 2 at infinity, and in this case we take f(z) = R(z). 

Then we have

Proof
In either case, we can have only finitely many poles of the function f; therefore only finitely many poles along the real number line. Let γ_δ be the path along the real number line from −∞ to ∞, but altered slightly, following a semi-circle of radius δ above each of the poles on the real line. Furthermore, δ is sufficiently small that no other pole of f is enclosed within any of the semi-circles. Then, according to our previous theorems, we have

To simplify our thoughts, let us first consider the case that there is only one single pole p ∈ ℝ on the real number line. And to simplify our thoughts even further, assume that p = 0. Then we have the path γ_δ coming from −∞ to −δ, then it follows the path δe^iπ(1−t), for t going from 0 to 1, and then finally it goes straight along the real number line from δ to ∞.

Let us now consider the Laurent series around 0. We can write

say. But then we can just define the new function g throughout ℂ (leaving out the finite set of poles of f) by the rule

Obviously g has no pole at 0, but otherwise it has the same set of poles as the original function f. Since the function c₋₁/z is analytic at all these other poles, the residue of g is identical with that of f around each of these poles. Therefore

Now, to get the principle value of the integral for f, we use the path γ_δ, but we must remove the semi-circle part of it. That is, let

• Complex Analysis: #17 Residues Around the Point at "Infinity"

In many applications, one speaks of the properties of a function “at infinity”. For this, we take the “compactification” of the complex number plane. This is the set ℂ ∪ {∞}, where “∞” is simply an abstract symbol. Then the open sets of ℂ ∪ {∞} are, first of all the familiar open sets of ℂ, then in addition, we say that any set of the form {z ∈ ℂ : |z| > R} ∪ {∞} is open, for R > 0. Finally, the union of all these sets of sets gives the topology of ℂ ∪ {∞}. It turns out that ℂ ∪ {∞} is homeomorphic to the standard 2-sphere. It is often called the “Riemann sphere”.

Definition 13
Let G ⊂ ℂ ∪ {∞} be a region containing ∞. (Thus G is open and connected.) [Note that we must then have ℂ \ G being a closed and bounded set, thus compact.] We will say that f : G → ℂ is analytic at ∞ if the function given by f(1/z) has a removable singularity at 0. Similarly, f has a zero of order n, or a pole of order n at ∞ if the respective property is true of the function f(1/z) at 0. That is to say, if ∞ is a zero of order n of f then we would like to have the function wⁿf(w) having a removable, non-zero singularity at ∞. However this is the same as looking at the limit as z → 0 of the function which is given by substituting w = 1/z. That is, the function

So f has a zero of order n at infinity if g has a removable, non-zero singularity at 0.


Theorem 36
Let g, h : ℂ → ℂ be entire, not constant, functions (thus they are analytic at all points of ℂ). Let f be the meromorphic function f = g/h. Assume that there is no zero of h on the real number line. Assume furthermore that f has a zero at infinity of order at least 2. Then we have

(Here the sum is over all poles of the function f in the “upper” half plane of ℂ. That is, the set of all complex numbers with positive imaginary parts.)

Proof
Because f has a zero at infinity, we have f being bounded outside of a compact disc of the form D = {z ∈ ℂ : |z| ≥ r}, for some sufficiently large r > 0. In particular, all of the poles of f must be within D. Since the zeros of h are isolated, there are only finitely many of them, thus the sum over the poles is finite. Therefore, the residue theorem shows that

where γ is the closed curve which consists of two segments: first the segment along the real number line from −r to r, then the segment consisting of the semi-circle of radius r around zero, traveling upwards from r and around through ir, then coming back to −r. That is to say,

We are assuming that f has a zero at infinity of order at least 2. That means, for all ∈ > 0 there exists some δ > 0 such that for all w ≠ 0 in ℂ with |w| < δ, we have

Since ∈ can be taken to be arbitrarily small, giving a corresponding δ, we can choose our r to be greater than 1/δ, and thus the integral around the half-circle α_r is small. The limit r → ∞ gives the formula of the theorem.

So this gives a method of calculating an integral along the real number line without actually having to do the integral at all! We only need to know the residues of the poles of the function in the upper half-plane of ℂ; no poles are allowed to be on the real number line; and the function should tend to zero sufficiently quickly at infinity.

But the assumption that the zero at infinity is of order 2 or more might be too restrictive. Perhaps the function we happen to be looking at only has a simple zero at infinity. For example consider the function

The integral along the real number line does not converge, and so this shows that we cannot expect the integral to exist if the zero at infinity is only of the first order. But perhaps the following theorem, where f(x) is multiplied with the “rotating” function e^ix, thus mixing things up nicely, might be useful. However, in contrast to the case where the zero at infinity is of at least second order, here we cannot expect that the integral over the absolute value of the function also converges.


Theorem 37
The same assumptions as in the previous theorem, except that the zero of f at infinity is only of the first order. Then we have

Proof
Again, we only have finitely many singularities in the upper half-plane. This time take a closed path γ consisting of four straight segments. The first segment is the straight line from the point −r to +r, along the real number line. The second segment goes from r to r + ir. The third from r + ir to −r + ir, and the fourth from −r + ir back to the starting point at −r. Let’s call these segments γ₁(r), . . . , γ₄(r). Following the ideas in the proof of the previous theorem, we see that it is only necessary to show that

for j = 2, 3, 4. Our assumption implies that for all ∈ > 0 there exists an r₀ > 0 such that |f(z)| < ∈ for all z with |z| > r₀. We will show that the absolute value of the integral along the path γ₂(r₀) is less than ∈ . The calculation for the other paths is similar. We have

Complex Analysis: #16 Residues Calculus

• Complex Analysis: #16 The Calculus of Residues

Let’s begin by thinking about a function f with a pole of order m at the point a ∈ ℂ. That is, in a sufficiently small neighborhood of a we can write h(z) = f(z)(z−a)^m, and after filling in the removable singularity of h at a, we have h(a) ≠ 0. Let

So this is a formula for the residue of a function with a pole at a.

A rather special case is the following. Let us assume that G ⊂ ℂ is a region, and g, h are both analytic functions defined on G. Assume that a ∈ G is a simple zero of h. (That is h(a) = 0, but h'(a) ≠ 0.) Assume furthermore that g(a) ≠ 0. Then let f = g/h. (Note that a function such as f, which is defined to be the ratio of two analytic functions, is called a rational function.) Therefore f is meromorphic in G. What is the residue of f at a? Writing g and h as power series around a, we have

Of course, going in the other direction, if a is a zero of g and h(a) ≠ 0, then the residue of f at a is simply zero. This is trivial.


Theorem 35 (The Residue Theorem)
Let the function f be defined and analytic throughout the region G ⊂ ℂ, except perhaps for a set S ⊂ G of isolated, not removable singularities. Let Ω be a cycle in G which avoids all these singularities and which is such that the winding number of Ω around all points of the compliment of G is zero. Then only finitely many points of S have non-vanishing index with respect to Ω and we have the residue formula

Proof
Nothing is lost if we assume that Ω simply consists of a single closed path γ. So it is contained in a compact disc in ℂ which must contain all points of ℂ having a non-vanishing index with respect to γ. If there were infinitely many such points, then they must have an accumulation point, which must lie in the compliment of G. But such a point has index zero with respect to γ. Since that point does not lie on γ, it must have a neighborhood which contains only points with index zero with respect to γ. Thus we have a contradiction. The residue theorem now follows from Cauchy’s theorem (theorem 26).

Complex Analysis: #15 Laurent Series

• Complex Analysis: #15 The Laurent Series

Since a pole of order n involves a function which looks somewhat like (z − z₀)⁻ⁿ, at least near to the singularity z₀, it seems reasonable to expand our idea of power series into the negative direction. This gives us the Laurent series. That is, a sum which looks like this:

There is a little problem with this notation. After all, if the series is not absolutely convergent, then we might get different sums if we start at different places in the doubly infinite sequence. Let us therefore say that the sum from 0 to +∞ is the positive series, and that from −∞ to −1 is the negative series. So the whole series is absolutely convergent if both the positive and negative series are absolutely convergent. [The negative series is called the “Hauptteil” in German, whilst the positive series is the “Nebenteil”.] One can obviously imagine that the negative series is really a positive series in the variable 1/(z − z₀).

So given some collection of coefficients c_n for all n ∈ ℤ, let R ≥ 0 be the radius of convergence of the series

That is to say, if 1/|z − z₀| < 1/r, or put another way, if |z − z₀| > r, then the negative series

converges.

Therefore, there is an open ring (or annulus) of points of the complex plane, namely the region {z ∈ ℂ : r < |z − z₀| < R}, where the Laurent series is absolutely convergent. In this ring, the function defined by the Laurent series

is analytic. Note however that it is not necessarily true that f has an antiderivative. We see this by observing that the particular term c₋₁(z − z₀)⁻¹ has no antiderivative in the ring. On the other hand, if we happen to have c₋₁ = 0 then there is an antiderivative, namely the function in the ring given by the Laurent series

Theorem 29
Assume the Laurent series converges in the ring between 0 ≤ r < R ≤ ∞. Let r < ρ < R. The coefficients of the Laurent series are then given by

Proof
Since the series is uniformly convergent around the circle of radius ρ, we can exchange sum and integral signs to write

However, if we look at the individual terms, we see that if k ≠ n, then each term has an antiderivative,
and thus the path-integral is zero for that term. We are left with the n-term, and this is then simply

Conversely, we have:


Theorem 30
Given 0 ≤ r < ρ ≤ R, let G = {z ∈ ℂ : r < |z − z₀| < R} and f : G → ℂ be an analytic function. Then we have f(z) = ∑ c_nzⁿ, (with n = −∞, . . ., ∞) , where

Proof
For simplicity, choose z₀ = 0. Let z be given with r < |z| < R, and take ∈ > 0 such that ∈ < min{R − |z|, |z| − r}. Thus, according to theorems 5 and 6, we have

(Note that theorem 5 shows that the integrals for |z| = R − ∈, and for |z| = r + ∈ are equal to the integrals, taken along the path |z| = ρ.)


Theorem 31
Again, the same assumptions as in theorem 30. Assume further that there exists some M > 0 with |f(z)| ≤ M for all z with |z − z₀| = ρ. Then |c_n| ≤ M/ρⁿ for all n.

Proof

Theorem 32 (Riemann)
Let a ∈ G ⊂ ℂ be an isolated singularity of an analytic function f : G\{a} → ℂ such that the exists an M > 0 and an ∈ > 0 with |f(z)| ≤ M for all z₀ ∈ G with z ≠ a and |z − a| < ∈ . Then a is a removable singularity.

Proof
For then |c_n| ≤ M/rⁿ for all 0 < r < ∈ , and therefore, for the terms with n < 0 we must have c_n = 0, showing that in fact f is given by a normal power series, and thus it is also analytic in a.


Theorem 33 (Casorati-Weierstrass)
Let a be an essential singularity of the function f : G \ {a} → ℂ. Then for all ∈ , δ > 0 and w ∈ ℂ, there exists a z ∈ G with |z − a| < ∈ such that |f(z) − w| < δ. (Which is to say, arbitrarily small neighborhoods of a are “exploded” through the action of f throughout ℂ, so that they form a dense subset of ℂ!)

Proof
Otherwise, there must exist some w₀ ∈ ℂ such that there exists an ∈ > 0 and |f(z) −w₀| ≥ δ for all z ∈ G with |z − a| < ∈ . Let B(a, ∈) = {z ∈ ℂ : |z − a| < ∈ } and define the function h : (B(a, ∈ )\{a})∩G → ℂ to be

Clearly h is analytic, with an isolated singularity at the point a. Furthermore.

Therefore, according to theorem 32 we must have a being removable. Thus, writing

we see that since the function given by 1/h(z) has at most a pole at a, we cannot have a being an essential singularity of f. This is a contradiction.

As an example of a function with an essential singularity, consider the function f(z) = exp(1/z). Clearly f is defined for all z ≠ 0, and not defined for the single point 0. In fact, 0 is an essential singularity. To see this, consider the exponential series

The singularity at 0 obviously cannot be a pole of the function, since the negative series is infinite. In fact we can make a theorem out of this observation.


Theorem 34
Let a function f be defined by a Laurent series ∑ c_n(z − z₀)ⁿ, (with n = −∞, . . ., ∞), around a point z₀ ∈ ℂ. Assume that the series converges in a “punctured disc” {0 < |z − z₀| < R}. If infinitely many of the terms c_n, for n < 0, are not zero, then z₀ is an essential singularity of f.

Proof
Obviously z₀ is not a removable singularity. If it were a pole of order n, then all the terms c_m, for m < −n must vanish. The only remaining possibility is that z₀ is an essential singularity.