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using Plots, ComplexPhasePortrait, ApproxFun
gr();
Dr. Sheehan Olver
s.olver@imperial.ac.uk
This lecture we cover
In this course, we will always think of Laurent series living on a circle $ \gamma_r(z_0) = \{z : |z-z_0| = r \}$. That is, $$ f(z) \approx \sum_{k=-\infty}^\infty f_k (z-z_0)^k $$
for $z \in \gamma_r(z_0)$.
Proposition (Residue on a circle) Suppose the Laurent series is absolutely summable on $\gamma_r$. Then $$ \oint_{\gamma_r} f(z) dz = 2 \pi i f_{-1} $$ We refer to $f_{-1}$ as the _residue over $\gamma_r$_.
Example For all $0 < r < \infty$,
$$ \oint_{\gamma_r} {1 \over z} dz = 2 \pi i $$
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phaseplot(-2..2, -2..2, z-> 1/z)
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γ = Circle(0.0, 1.0)
z = Fun(γ)
sum(1/z)
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Example This works for functions not analytic: $$ \oint_{\gamma_1} (\sqrt{z-1}\sqrt{z+1})^3 dz $$
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f = z -> (sqrt(z-1)*sqrt(z+1))^3
phaseplot(-2..2, -2..2, f)
plot!(Circle(1.1); color=:black, label="contour", linewidth=1.5, arrow=true)
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@show sum(Fun(f, Laurent(Circle(1.1)))) # integral over circle
f₋₁ = Fun(f, Laurent(Circle(1.1))).coefficients[2] # numerical Laurent coefficient
@show 2π*im*f₋₁;
When $f$ is holomorphic in a neighbourhood of the circle, we can extend it to an annulus (like Taylor series and disks):
Proposition (Laurent series in an annulus) Suppose $f$ is holomorphic in an open annulus $A_{\rho R}(z_0) = \{z : \rho < | z - z_0| < R\}$. Then the Laurent series converges uniformly in any closed annulus inside $A_{\rho R}$
Proof Exercise. Hint: use the decay in the Laurent coefficients $f_k$ from last lecture.
Proposition (Residue on a circle) holds true regardless of the radius.
Definition (isolated singularity) $f$ has an isolated singularity at $z_0$ if it is holomorphic in an open annulus with inner radius 0: $$ A_{0R}(z_0) = \{z : 0 < |z - z_0| < R \}. $$
Definition (Removable singularity) $f$ has a removable singularity at $z_0$ if it has an isolated singularity at $z_0$ and all negative terms in the Laurent series in $A_{0R}(z_0)$ are zero: $$ f(z) = f_0 + f_1 (z-z_0) + f_2 (z-z_0)^2 + \cdots $$
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f = z -> (exp(z)-1)/z
f(0.0)
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phaseplot(-1..1, -1..1, f) # no singularity appears because
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Proposition (Removing a removable singularity) If $f$ has a removable singularity at $z_0$, then $$ \tilde f(z) = \begin{cases} f_0 & z = z_0 \\ f(z) & 0 < |z-z_0| < R \end{cases} $$ is analytic in the disk $B_R(z_0) = \{ z : |z-z_0| < R \}$, with a convergent Taylor series. Hence the name.
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f̃ = z -> z ≈ 0 ? 1 : f(z)
phaseplot(-1..1, -1..1, f̃)
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Definition (simple pole) $f$ has a simple pole at $z_0$ if it is holomorphic in $$ A_{0R}(z_0) = \{z : 0 < |z - z_0| < R \} $$ with only one negative term in the Laurent series in $A_{0R}(z_0)$: $$ f(z) = {f_{-1} \over z - z_0} + f_0 + f_1 (z - z_0) + \cdots $$ where $f_{-1} \neq 0$.
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phaseplot(-1..1, -1..1, z -> exp(z)/z)
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Definition (higher order pole) $f$ has a pole of order $N$ at ${z_0}$ if it is holomorphic in $$ A_{0R}(z_0) = \{z : 0 < |z - z_0| < R \} $$ with only $N$ negative coefficients in the Laurent series: $$ f(z) = {f_{-N} \over (z - z_0)^N} + {f_{1-N} \over (z - z_0)^{N-1}} + \cdots + {f_{-1} \over z-z_0} + f_0 + f_1 (z-z_0) + \cdots $$ where $f_{-N} \neq 0$.
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phaseplot(-1..1, -1..1, z -> exp(z)/z^3)
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Definition (essential singularity) $f$ has an essential singularity at $z_0$ if it is holomorphic in $A_{0R}(z_0)$ and has an infinite number of negative Laurent coefficients.
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phaseplot(-1..1, -1..1, z -> exp(1/z))
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sum(Fun(z -> exp(1/z), Circle())),
2π*im*Fun(z -> exp(1/z), Laurent(Circle())).coefficients[2]
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Definition (Residue at a point) Suppose $f$ has an isolated singularity at $z_0$, and is analytic in the annulus $A_{0R}(z_0)$ for some $R > 0$. Then we define the residue at $z_0$ as $$ {\underset{z = z_0}{\rm Res}}\, f(z) = f_{-1} $$ where $f_{-1}$ is the first negative coefficent of the Laurent series in $A_{0R}(z_0)$.
Proposition (Residue of ratio of analytic functions with simple pole) Suppose $$ f(z) = {A(z) \over B(z)} $$ and $A$, $B$ are analytic/holomorphic in a disk of radius $R$ around $z_0$ and that $B$ has only a single zero at $z_0$: \begin{align*} A(z) = A_0 + A_1(z-z_0) + \cdots \cr B(z) = B_1(z-z_0) + \cdots \end{align*} Then ${\underset{z = z_0}{\rm Res}}\, f(z) = {A_0 \over B_1}$
Exercise (Residue of ratio of analytic functions with higher order poles) What is the residue at $z_0$ if $B$ has a higher order zero: $B(z) = B_N (z-z_0)^N + \cdots$?
We still have the contour integral over a circle, and so Proposition (Residue on a circle) still holds true for $r > 2$. But we can also deform the contour into three contours:
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f = z -> sqrt(z-1)sqrt(z+1)/(z^2+4)
Γ = Circle(1.1) ∪ Circle(2.0im,0.1) ∪ Circle(-2.0im,0.1)
phaseplot(-2..2, -3..3, f)
plot!(Γ; color=:black, label=:contour, arrow=true, linewidth=1.5)
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sum(Fun(f, Circle(2.1))), sum(Fun(f, Γ))
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Thus we can sum over three residues.
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f = z -> 1/(z*(z+2))
phaseplot(-3..3, -3..3, f)
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sum(Fun(f, Circle(3.0)))
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f = z -> exp(z)/(z*(z+2))
phaseplot(-3..3, -3..3, f)
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sum(Fun(f, Circle(3.0)))
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2π*im*(1/2 - exp(-2)/2)
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sum(Fun(z -> exp(z)/(z^2*(z+2)), Circle(3.0)))
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2*pi*im * (1/4 + exp(-2)/4)
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We can calculate integrals of the form $$\int_0^{2 \pi} R(\cos \theta, \sin \theta) d \theta$$ where $R(x,y)$ is rational by doing the change of variables $z = e^{i \theta}$ to reduce it to $$\oint_{\gamma_1} R\left({z + z^{-1} \over 2}, {z - z^{-1} \over 2 i} \right) {d z \over i z}$$
Example Consider
$$\int_0^{2\pi} {d \theta \over 1 - 2\rho \cos \theta + \rho^2}$$for $0 < \rho < 1$.
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ρ = 0.5
plot(Fun(θ -> 1/(1-2ρ*cos(θ) + ρ^2), 0 .. 2π))
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phaseplot(-2..2, -2..2, z -> 1/(1-ρ*(z+(z^(-1))) + ρ^2) * 1/(im*z))
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sum(Fun(θ -> 1/(1-2ρ*cos(θ) + ρ^2), 0 .. 2π)), 2π /(1-ρ^2)
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