Giải tích phức và thặng dư
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| residue/ˈresɪdjuː/ | Lư thừa |
| pole/pəʊl/ | An isolated singularity with only finitely many negative powers in its Laurent series. |
| essential singularity/ɪˈsenʃl ˌsɪŋɡjʊˈlærɪti/ | An isolated singularity with infinitely many negative powers in its Laurent series. |
Smooth, yet not complex-differentiable
- $f(z) = \bar z = x - iy$ has perfectly good real partial derivatives everywhere, yet it is complex-differentiable nowhere.
- GRE complex analysis tests the Cauchy–Riemann equations, singularity types, residues 留数 and the residue theorem.
Cauchy–Riemann
- Write $f = u + iv$. Complex differentiability requires $u_x = v_y$ and $u_y = -v_x$. With continuous partials on an open set, these give analyticity there.
- For $\bar z$: $u = x$, $v = -y$, so $u_x = 1 \neq -1 = v_y$. Fails everywhere.
- $|z|^2 = x^2 + y^2$ satisfies the equations only at $0$ — differentiable at one point, analytic nowhere.
Which function is analytic on all of $\mathbb C$?
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Polynomials in z satisfy Cauchy–Riemann everywhere.
Singularities and Laurent series
- Removable: bounded near the point ($\frac{\sin z}{z}$ at 0). Pole 极点 of order $m$: finitely many negative powers, the lowest $(z-a)^{-m}$. Essential singularity 本性奇点: infinitely many ($e^{1/z}$ at 0).
- The residue is the coefficient of $(z - a)^{-1}$ — not the leading negative power.
Match each function with its singularity at $z = 0$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Count negative powers in the Laurent series.
Computing residues at poles
- If $f = \dfrac{g(z)}{(z - a)^m}$ with $g$ analytic and $g(a) \neq 0$: $\operatorname{Res}_a f = \dfrac{g^{(m-1)}(a)}{(m-1)!}$.
$\dfrac{3}{z-1}$: residue 3. $\dfrac{e^z}{(z-1)^2}$: $g' = e^z$, residue $e$. $\dfrac{e^{2z}}{(z-1)^3}$: $\tfrac{1}{2}g''(1) = \tfrac{1}{2} \cdot 4e^2 = 2e^2$. A constant over a pure double pole, $\dfrac{5}{(z-1)^2}$, has residue 0.
What is the residue at z = 1?
Compute residues from pole order and numerator.
The residue of $\dfrac{e^{3z}}{(z-2)^2}$ at $z = 2$ is
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$g = e^{3z}$, $g'(2) = 3e^6$.
The residue theorem
- For a positively oriented simple closed curve with $f$ analytic on it and inside except at isolated points: $\oint f\,dz = 2\pi i \sum \operatorname{Res}$ (enclosed only).
- $\oint_{|z|=2} \dfrac{dz}{z(z-3)}$: only $z = 0$ is inside, residue $-\tfrac13$, so the integral is $-\tfrac{2\pi i}{3}$. Clockwise flips the sign.
A singularity on the contour blocks the theorem; one outside contributes nothing.
$\oint_{|z|=1} \dfrac{z^2 + 1}{z}\,dz = 2\pi i \cdot k$. Find $k$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
(z² + 1)/z = z + 1/z, so the residue at the only pole z = 0 (inside |z| = 1) is 1. The integral is 2πi · 1, giving k = 1.
A pole outside the contour adds its residue to the integral.
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Only enclosed singularities count.
Recap
- Test Cauchy–Riemann on an open set, not one point.
- Residue = coefficient of $(z-a)^{-1}$; for order-$m$ poles use $g^{(m-1)}(a)/(m-1)!$.
- Sum only enclosed residues; mind orientation.
- pole: An isolated singularity with only finitely many negative powers in its Laurent series.
- essential singularity: An isolated singularity with infinitely many negative powers in its Laurent series.