Notebook

In [1]:

using Plots, ComplexPhasePortrait, ApproxFun
gr();

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

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

Riemann sphere and analyticity at infinity
- Cauchy's integral formula exterior to a contour
- Exterior Residue theorem
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:

In [2]:

f = z -> exp(1/z)
phaseplot(-6..6, -6..6, f)

Out[2]:

We therefore have a convergence "Taylor" series in inverse powers of $z$ :

$\E^{1/z} = 1 + {1 \over z} + {1 \over 2 z^2} + {1 \over 3! z^3} + \cdots$

In [6]:

z = 2.0+2im
sum([z^k/factorial(-1.0k) for k=-100:0]) - exp(1/z)

Out[6]:

2.220446049250313e-16 - 5.551115123125783e-17im

These coefficients can be calculated as integrals:

In [7]:

k=-5;
sum(Fun(z -> f(z)/z^(k+1), Circle()))/(2π*im) - 1/factorial(-1.0k)

Out[7]:

-3.4867941867133823e-16 + 1.13303848301229e-16im

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:

In [8]:

f = z -> exp(1/z) - 1

ζ = Fun(Circle())
-sum(f.(ζ)/(ζ-z))/(2π*im) - f(z)

Out[8]:

2.220446049250313e-16 + 5.551115123125783e-17im

The decay at infinity is required:

In [143]:

f = z -> exp(1/z) 

ζ = Fun(Circle())
-sum(f.(ζ)/(ζ-z))/(2π*im) - f(z)

Out[143]:

-0.9999999999999997 + 5.551115123125783e-17im

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:

In [3]:

f = z -> 1/(z*(z+2))
phaseplot(-3..3, -3..3, f)

Out[3]:

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

In [146]:

sum(Fun(f, Circle(3.0)))

Out[146]:

9.191214874231192e-18 - 8.870531755418961e-17im

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

In [4]:

f = z -> 1/z + 1/(z+2)
phaseplot(-3..3, -3..3, f)

Out[4]:

This has a residue at infinity:

In [12]:

f₋₁ = 2
res∞ = -f₋₁
sum(Fun(f, Circle(3.0))), -2π*im*res∞

Out[12]:

(-1.7038443268502702e-15 + 12.566370614359174im, 0.0 + 12.566370614359172im)

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

In [13]:

res∞ = -2
res₋₂ = 1
sum(Fun(f, Circle(1.0))), -2π*im*(res∞ + res₋₂)

Out[13]:

(-1.032906916559743e-15 + 6.283185307179586im, 0.0 + 6.283185307179586im)

Here's a more complicated example:

In [14]:

f = z -> exp(1/z)/(z*(z+2))
portrait(-3..3, -3..3, f)

Out[14]:

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:

In [15]:

sum(Fun(f, Circle(3.0)))

Out[15]:

1.2241051404579888e-18 - 1.191336987305697e-16im

In [16]:

sum(Fun(f, Circle(1.0))), -2π*im * exp(-1/2)/(-2)

Out[16]:

(-4.3477951067425517e-16 + 1.9054722647301798im, 0.0 + 1.9054722647301798im)

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:

In [7]:

f = z -> (exp(1/z) + exp(z))/(z*(z+2))
Γ = Circle(0.0, 4.0) ∪ Circle(0.0,0.5,false) ∪ Circle(-2.0,0.1,false)
phaseplot(-5..5, -5..5, f)
plot!(Γ; color=:black, label=:contour, arrow=true, linewidth=1.5)

Out[7]:

In [18]:

ζ = Fun(Γ)
z = 2.0+1.0im
sum(f.(ζ)/(ζ - z))/(2π*im)

Out[18]:

0.8671607060038516 + 0.10261889457156087im

In [19]:

f(z)

Out[19]:

0.8671607060038514 + 0.10261889457156062im

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:

In [9]:

n = 7
m = 5
p = Fun(Taylor(), randn(n))
q = Fun(Taylor(), randn(m))
λ = complexroots(q)

Out[9]:

4-element Array{Complex{Float64},1}:
  0.6574583015092604 + 0.8969475017786706im
  0.6574583015092604 - 0.8969475017786706im
 -0.6950370227626129 + 0.0im               
 -1.5131182364477396 + 0.0im

In [11]:

Γ = Circle(0.0, 5.0)
for λ in λ
    Γ = Γ ∪ Circle(λ, 0.1, false)
end
r = z -> extrapolate(p,z)/extrapolate(q,z)

phaseplot(-5..5, -5..5, r)
plot!(Γ; color=:black, label=:contour)

Out[11]:

In [29]:

ζ = Fun(Γ)
z = 2.0+2.0im
sum(r.(ζ)/(ζ - z))/(2π*im) - r(z)

Out[29]:

-1.6653345369377348e-16 + 2.6645352591003757e-15im

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:

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