residues

Problem-solving strategies for residues in complex analysis

422 stars

Best use case

residues is best used when you need a repeatable AI agent workflow instead of a one-off prompt.

Problem-solving strategies for residues in complex analysis

Teams using residues should expect a more consistent output, faster repeated execution, less prompt rewriting.

When to use this skill

  • You want a reusable workflow that can be run more than once with consistent structure.

When not to use this skill

  • You only need a quick one-off answer and do not need a reusable workflow.
  • You cannot install or maintain the underlying files, dependencies, or repository context.

Installation

Claude Code / Cursor / Codex

$curl -o ~/.claude/skills/residues/SKILL.md --create-dirs "https://raw.githubusercontent.com/vibeeval/vibecosystem/main/skills/math/complex-analysis/residues/SKILL.md"

Manual Installation

  1. Download SKILL.md from GitHub
  2. Place it in .claude/skills/residues/SKILL.md inside your project
  3. Restart your AI agent — it will auto-discover the skill

How residues Compares

Feature / AgentresiduesStandard Approach
Platform SupportNot specifiedLimited / Varies
Context Awareness High Baseline
Installation ComplexityUnknownN/A

Frequently Asked Questions

What does this skill do?

Problem-solving strategies for residues in complex analysis

Where can I find the source code?

You can find the source code on GitHub using the link provided at the top of the page.

SKILL.md Source

# Residues

## When to Use

Use this skill when working on residues problems in complex analysis.

## Decision Tree


1. **Computing Residues**
   - Simple pole at z0:
     * Res(f, z0) = lim_{z->z0} (z - z0)f(z)
     * `sympy_compute.py limit "(z - z0)*f(z)" --var z --at z0`
   - Pole of order n:
     * Res(f, z0) = (1/(n-1)!) * lim d^{n-1}/dz^{n-1}[(z-z0)^n f(z)]
     * `sympy_compute.py diff "((z-z0)**n)*f(z)" --var z --order n-1`
   - L'Hopital shortcut for f = g/h with simple pole:
     * Res(f, z0) = g(z0)/h'(z0)

2. **Identify Pole Order**
   - Simple pole: (z - z0)f(z) has finite limit
   - Order n: (z - z0)^n f(z) has finite limit, but (z - z0)^{n-1} f(z) doesn't
   - `sympy_compute.py limit "(z - z0)**n * f(z)" --var z --at z0`

3. **Essential Singularities**
   - Neither pole nor removable (e.g., e^{1/z} at z=0)
   - Compute residue via Laurent series
   - `sympy_compute.py series "exp(1/z)" --var z --at 0`

4. **Apply Residue Theorem**
   - oint_C f(z)dz = 2*pi*i * (sum of residues inside C)
   - Count only poles INSIDE the contour
   - `z3_solve.py prove "pole_inside_contour"`


## Tool Commands

### Sympy_Residue
```bash
uv run python -m runtime.harness scripts/sympy_compute.py residue "1/((z-1)*(z-2))" --var z --at 1
```

### Sympy_Limit
```bash
uv run python -m runtime.harness scripts/sympy_compute.py limit "(z - z0)*f(z)" --var z --at z0
```

### Sympy_Laurent
```bash
uv run python -m runtime.harness scripts/sympy_compute.py series "exp(1/z)" --var z --at 0
```

### Z3_Pole_Inside
```bash
uv run python -m runtime.harness scripts/z3_solve.py prove "abs(z0) < R"
```

## Key Techniques

*From indexed textbooks:*

- [Complex analysis  an introduction to... (Z-Library)] The fact that the calculus of residues yields complex rather than real integrals is no dis¬ (49) with g(z) — z, we obtain <»» i>(”)=25 / f^w) = 2vi / /'(*) /(z) - w z dz. If (49) is applied with g(z) = zm, equation (50) is replaced by 2iri I |z-zo| = /'(*) f(z) - w zm dz. The right-hand member represents an analytic function of w for \w — ir0| < 8.
- [Complex analysis  an introduction to... (Z-Library)] What are the possible values of r dz J \/l — z2 over a closed curve in the region? THE CALCULUS OF RESIDUES The results of the preceding section have shown that the determination of line integrals of analytic functions over closed curves can be reduced to the determination of periods. Under certain circumstances it turns out that the periods can be found without or with very little computation.
- [Complex analysis  an introduction to... (Z-Library)] Hint: Sketch the image of the imaginary axis and apply the argument principle to a large half disk. Evaluation of Definite Integrals. The calculus of residues pro¬ vides a very efficient tool for the evaluation of definite integrals.
- [Complex analysis  an introduction to... (Z-Library)] The particular function 1 /(z — ay) has a vanishing period. The constant Rj which produces this result is called the residue of f(z) at the point ay. We repeat the definition in the following form: It is helpful to use such self-explanatory notations as R = Res!
- [Complex Analysis (Elias M. Stein, Ram... (Z-Library)] Cauchy, 1826 There is a general principle in the theory, already implicit in Riemann’s work, which states that analytic functions are in an essential way charac- terized by their singularities. That is to say, globally analytic functions are “eectively” determined by their zeros, and meromorphic functions by their zeros and poles. While these assertions cannot be formulated as precise general theorems, there are nevertheless signicant instances where this principle applies.

## Cognitive Tools Reference

See `.claude/skills/math-mode/SKILL.md` for full tool documentation.

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