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.