In [25]:
using Plots, ComplexPhasePortrait, ApproxFun
gr();

M3M6: Methods of Mathematical Physics

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

Lecture 5: Residue Theorem

This lecture we cover

  1. Contour integrals and Laurent coefficients
  2. Isolated singularities
    • Residue at a point
  3. Contour integrals in domains with multiple holes
    • The residue theorem
  4. Calculated integrals
    • Application: Trigonometric integrals with rational functions

Contour integrals and Laurent coefficients

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 $0 < r < \infty$,

$$ \oint_{\gamma_r} {1 \over z} dz = 2 \pi i $$

In [26]:
phaseplot(-2..2, -2..2, z-> 1/z)


Out[26]:
-2 -1 0 1 2 -2 -1 0 1 2

In [3]:
γ = Circle(0.0, 1.0)
z = Fun(γ)
sum(1/z)


Out[3]:
0.0 + 6.283185307179586im

Example This works for functions not analytic: $$ \oint_{\gamma_1} (\sqrt{z-1}\sqrt{z+1})^3 dz $$


In [42]:
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)


Out[42]:
-2 -1 0 1 2 -2 -1 0 1 2 contour

In [29]:
@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;


sum(Fun(f, Laurent(Circle(1.1)))) = -1.2751605303035503e-16 + 2.3561944901923444im
(2π) * im * f₋₁ = -1.1592368457305e-16 + 2.141994991083949im

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.

Isolated singularities

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: $$ f(z) = f_0 + f_1 (z-z_0) + f_2 (z-z_0)^2 + \cdots $$


In [31]:
f = z -> (exp(z)-1)/z
f(0.0)


Out[31]:
NaN

In [32]:
phaseplot(-1..1, -1..1, f) # no singularity appears because


Out[32]:
-1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0

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.


In [33]:
 = z -> z  0 ? 1 : f(z)
phaseplot(-1..1, -1..1, )


Out[33]:
-1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0

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 $f_{-1} \neq 0$.


In [34]:
phaseplot(-1..1, -1..1, z -> exp(z)/z)


Out[34]:
-1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0

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 $f_{-N} \neq 0$.


In [35]:
phaseplot(-1..1, -1..1, z -> exp(z)/z^3)


Out[35]:
-1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0

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.


In [36]:
phaseplot(-1..1, -1..1, z -> exp(1/z))


Out[36]:
-1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0