19.5 Residue Theorem

Practice Problems: 1, 2, 4, 5, 10, 11, 13, 14, 17, 20, 24, 25, 27, 29, 31.

<aside> 🐝 The residue of a function $f(z)$ at an isolated singularity $z_0$ is the coefficient of $a_{-1}$ in the Laurent series expansion.

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• Computing the $a_{-1}$ term can be faster than computing the residue
  • $a_{-1}$ is the first term that has $(z-z_0)$ in the denominator

Classification of Singularities by Residues

• Removable Singularity: If all coefficients of negative powers in the principal part are zero, the singularity is removable.
• Pole: If there are a finite number of non-zero coefficients, $z_0$ is a pole. The order of the pole is the largest $n$ where $a_{-n} ≠ 0$
• Essential Singularity: If there are infinitely many non-zero coefficients, $z_0$ is an essential singularity.

Computing Residues

• Simple Pole

  $$ \text{Res}(f(z), z_0) = \lim_{z \to z_0} (z - z_0)f(z) $$

• Pole of Order n

  $$ \text{Res}(f(z), z_0) = \frac{1}{(n - 1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}}[(z - z_0)^nf(z)] $$

• Alternative Method for Simple Pole: If $f(z) = \frac{g(z)}{h(z)}$ where $g(z_0) ≠ 0$ and $h(z_0) = 0$ but $h’(z_0)≠0$:

  $$ \text{Res}(f(z), z_0) = \frac{g(z_0)}{h'(z_0)} $$

Cauchy’s Residue Theorem

If you integrate $f(z)$ around a closed contour $C$ that encloses singularities $z_1, z_2, \dots , z_n$, the integral is $2 \pi i$ times the sum of the residues at those singularities:

$$ \oint_C f(z) dz = 2\pi i \sum_{k=1}^{n} \text{Res}(f(z), z_k) $$

19.6 Evaluation of Real Integrals

Practice Problems: 1, 26, 28, 31, 33, 34, 35

Integrals of the Form $\int_{0}^{2\pi} F(\cos \theta, \sin \theta) d\theta$

We can convert these to complex integrals over the unit circle

$$ \int_{0}^{2\pi} F(\cos \theta, \sin \theta) d\theta = \oint_{C} F\left(\frac{1}{2}(z + z^{-1}), \frac{1}{2i}(z - z^{-1})\right) \frac{dz}{iz} $$

Improper Integrals of the Form $\int_{-\infty}^{\infty} f(x) dx$