Complex analysis and residues · Análisis complejo y residuos
| English | Español |
|---|---|
| residue/ˈresɪdjuː/ | residuo |
| analytic/ˌænəˈlɪtɪk/ | analítico |
A decision before an answer
- A function smooth in two real coordinates can still fail complex differentiability.
- Your goal: Test complex differentiability.
Read the relationship
- Write f(z)=u(x,y)+iv(x,y). Complex differentiability imposes u_x=v_y and u_y=−v_x; with continuous first partials locally, the Cauchy–Riemann equations establish analyticity there. They differ from real differentiability of a two-coordinate map. The conjugate function x−iy fails these equations on every open neighbourhood, although real partial derivatives exist. State the region being checked, not only a convenient point.
- Use contour integrals and residues.
The residue of 3/(z−1) at z=1 is:
The coefficient of the simple-pole term is 3.
Use the defining rule
- An analytic function is complex differentiable throughout a neighbourhood and has a local convergent power series. A removable singularity can be filled analytically when the function is bounded near the missing point. A pole has a finite principal part in its Laurent series; an essential singularity has infinitely many negative-power terms. The residue is the coefficient of (z−a)⁻¹, not necessarily the leading or largest negative-power term.
- Evaluate contour integrals with simple and higher-order pole residues.
A positively oriented circle enclosing only that pole gives integral:
Multiply residue 3 by 2πi.
Check the conditions
- For an isolated pole of order m, write f(z)=g(z)/(z−a)^m with g analytic at a. The residue is g^(m−1)(a)/(m−1)!. A simple pole uses g(a); a double pole uses g′(a). This follows by expanding g into its Taylor series. Thus e^z/(z−a)² has residue e^a, while a constant numerator over a pure double pole has zero residue. The pole order alone does not determine the contour integral.
- Evaluate contour integrals with simple and higher-order pole residues.
For f(z)=conjugate(z), u=x and v=−y. Then u_x=1 but v_y=−1, so f is not complex differentiable. For 1/(z−2), a positively oriented circle |z−2|=1 encloses a simple pole of residue 1, hence the integral is 2πi.
The residue of 5/z at zero is ____.
The Laurent coefficient of z^-1 is 5.
Apply the task format
- For a positively oriented contour enclosing isolated singularities, the residue theorem gives ∮f(z)dz=2πi times the sum of enclosed residues, provided the function is analytic on the contour and elsewhere in the required interior. Reversing orientation changes the sign. Cauchy’s derivative formula is ∮g(z)/(z−a)^(m+1)dz=2πi g^(m)(a)/m! under its analytic-domain hypotheses. A singularity on the contour prevents direct application; a singularity outside contributes nothing to this contour.
- Evaluate contour integrals with simple and higher-order pole residues.
Continuity or real differentiability alone does not imply complex analyticity.
Which answer fits this case? · ¿Qué respuesta se ajusta a este caso?
Test complex differentiability · Probar diferenciabilidad compleja
Every continuous complex function is analytic.
Complex differentiability imposes Cauchy–Riemann conditions.
What is the residue of e^(2z)/(z−1)³ at z=1?
For a triple pole use g″(1)/2! with g(z)=e^(2z). Its second derivative is 4e^(2z), so the residue is 2e².
Keep the distinctions
- analytic 解析的 — Complex differentiable throughout a neighbourhood.
- residue 留数 — Coefficient of (z−a)^−1 in a Laurent series.
- Test complex differentiability.
- Use contour integrals and residues.
- Evaluate contour integrals with simple and higher-order pole residues.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.