$$ \def\E{{\rm e}} $$

M3M6: Methods of Mathematical Physics

Dr. Sheehan Olver
s.olver@imperial.ac.uk

Website: Blackboard

Lecture 6: Analytic functions at infinity

This lecture we cover

1. Riemann sphere and analyticity at infinity
   • Cauchy's integral formula exterior to a contour
   • Exterior Residue theorem
2. Reperesenting functions by their behaviour near singularities
   • Application: Partial fraction expansion

Riemann sphere and analyticity at infinity

Definition (Riemann sphere) The Riemann sphere is the compactification of ${\mathbb C}$: $$ \bar {\mathbb C} = {\mathbb C} \cup \{\infty\} $$

Without delving on the details, we can define an open set $D \subset \bar{\mathbb C}$ on the Riemann sphere, where $\infty \in D$ implies that there exists an $R$ such that $\{ z : |z| > R\} \subset D$.

Definition (Analytic at infinity) A function $f(z)$ defined on an open set $D \subset \bar {\mathbb C}$ such that $\infty \in D$ is analytic at ∞ if $f(z^{-1})$ is analytic at zero.

Proposition (Taylor series at infinity) If $f$ is analytic at infinity, then there exists an $R$ such that for all $|z| > R$ we have $$ f(z) = \sum_{k=-\infty}^0 f_k z^k $$ The coefficients $f_k$ are defined by $$ f_k = {1 \over 2 \pi i} \oint_\gamma {f(z) \over z^{k+1}} dz $$ where $\gamma$ is any simple closed positively oriented contour such that $f$ is analytic outside of.

Demonstration $f(z) = \E^{1/z}$ is not analytic at zero, but is analytic at infinity because $f(1/z) = \E^z$ is analytic at zero:

We therefore have a convergence "Taylor" series in inverse powers of $z$: $$ \E^{1/z} = 1 + {1 \over z} + {1 \over 2 z^2} + {1 \over 3! z^3} + \cdots $$

These coefficients can be calculated as integrals:

Theorem (Cauchy's integral theorem near infinity) Suppose $f$ is analytic outside and on a positively oriented, simple, closed contour $\gamma$, and $$f(\infty) = 0.$$ Then we have $$ f(z) = -{1 \over 2 \pi i} \oint {f(\zeta) \over \zeta - z} d \zeta $$

Example The function $f(z) = \E^{1/z} - 1$ vanishes at $\infty$ and so can thence be recovered as a Cauchy integral:

The decay at infinity is required:

Exterior Residue theorem

Definition (Residue at infinity) Suppose $f$ is analytic in the annulus $A_{R\infty} = \{z : R < |z| < \infty \}$. Then the residue at infinity is $$ {\underset{z = \infty}{\rm Res}} \, f(z) = -f_{-1} $$ where $f_{-1}$ is (again) the Laurent coefficient for any circle in $A_{R\infty}$.

Theorem (Exterier Residue Theorem) Let $f$ be holomprohic outside and on a simple closed, positively oriented contour $\gamma$ except at isolated points $z_1, \ldots, z_r$ outside $\gamma$. Then

$$\oint_\gamma f(z) dz = -2 \pi i \sum_{j=1}^r {\underset{z = z_j}{\rm Res}}\, f(z)$$

Let's return to our simple examples from before:

Because it is analytic outside a circle of radius 3 and decays like $O(z^{-2})$, its residue at infinity is zero:

Here's another example with singularities at $0$ and $-2$:

This has a residue at infinity:

On a smaller circle of radius 1 we pick up another term:

Here's a more complicated example:

This example has an essential singularity at zero so the classical Residue theorem is not much use, but we can use the residue theorem at infinity:

Representing analytic functions by their behaviour near singularities

A key theme in complex analysis is representing functions by their behaviour near singularities. A rather simple example of this is a side-effect of Cauchy's integral representation:

Corollary (Cauchy's integral representation around holes) Let $D \subset {\mathbb C}$ be a domain with $g$ holes (i.e., genus $g$). Suppose $f$ is holmorphic in and on the boundary of $D$. Given $g$ simple closed negatively oriented contours surrounding the holes $\gamma_1, \ldots, \gamma_g$ and a simple closed positively oriented contour $\gamma_{0}$ surrounding the outer boundary of $D$, we have $$ f(z) = {1 \over 2 \pi i} \sum_{k=0}^{g} \oint_{\gamma_k} {f(\zeta) \over \zeta - z} d \zeta $$

Here are two examples:

Application: Partial fraction expansion

Suppose we have a rational function $$ r(z) = {p(z) \over q(z)} $$ where $p,q$ are both polynomials. This is analytic everywhere apart for the roots of $q$, which we enumerate $\lambda_1,\ldots,\lambda_g$. If we draw negatively oriented circles around each root, the previous result applies:

But now we can use the residue theorem to simplify the integrals! Note the following:

For example, near the $j$th root we have the Laurent series $$ r(z) = r_{-N_j}^j (z-\lambda_j)^{-N_j} + \cdots + r_{-1}^j (z-\lambda_j)^{-1} + r_0 + r_1 (z-\lambda_j) + \cdots $$ where $N_j$ is the order of the zero of $q(z)$ at $\lambda_j$.

Then it follows that $${1 \over 2 \pi i} \oint_{\gamma_j} {r(\zeta) \over z - \zeta} d\zeta = r_{-N_j}^j (z-\lambda_j)^{-N} + \cdots + r_{-1}^j (z-\lambda_j)^{-1}$$ for $z$ outside the contour $\gamma_j$.

Similarly, for the contour around infinity $\gamma_0$, if we have the Laurent series $$ r(z) = \cdots + r_{-1}^0 z^{-1} + r_0^0 + r_1^0 z + \dots + r_{N_0}^0 z^{N_0} $$ where $N_0$ is the degree of $p(z)$ divided by the degree of $q(z)$. Then we have $${1 \over 2 \pi i} \oint_{\gamma_{g+1}} {r(\zeta) \over z - \zeta} d\zeta = r_0^0 + r_1^0 z + \cdots + r_{N_0}^0 z^{N_0}.$$

Thus we have the expansion summing over the behaviour near each singularity that holds for all $z$:

$$ r(z) = \sum_{k=0}^{N_0} r_k^0 z^k + \sum_{j=1}^d \sum_{k = -N_j}^{-1} r_k^j (z - \lambda_j)^k $$

Example When we only have simple poles and no polynomial growth at $\infty$, this has a simple form in terms of residues:

$$ r(z) = r(\infty) + \sum_{j=1}^d (z - \lambda_j)^{-1} \underset{z = \lambda_j}{\rm Res}\, r(z) $$