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Residue theorem in complex analysis is seems like Stokes' theorem in real calculus, so a question arose that could Residue theorem be seen as a special case of Stokes' theorem?

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Yes; but then again most people prefer to prove it within the framework of complex analysis. –  Fabian Sep 14 '12 at 9:02
    
I should check, but I remember that the approach à la Stokes requires more regularity than it is needed. –  Siminore Sep 14 '12 at 9:24
    
This is mentioned in Wikipedia. –  lhf Sep 14 '12 at 12:40
    
Stokes' theorem works on smooth real manifolds, so I consider that if $\mathbb{C}$ can be seen as $\mathbb{R}^2$ but with metric matrix $\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$ $\ldots$ –  Popopo Sep 14 '12 at 13:06
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Since I left this answer mathoverflow.net/questions/23478/… , I feel obliged to write up how this works. –  David Speyer Sep 14 '12 at 14:42
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up vote 7 down vote accepted

As a first version of the residue theorem, consider the following statement:

Residue theorem without poles: Let $D$ be a bounded region of $\mathbb{C}$ with boundary $\gamma$. Let $f: D \to \mathbb{C}$ be an analytic function. Then $\oint_{\gamma} f(z) dz=0$.

Proof by Stokes: Write $f(x+iy) = u(x+iy) + v(x+iy)$. So $$\oint_{\gamma} f(z) dz = \oint_{\gamma} {\Large (} u(x+iy) dx - v(x+iy) dy {\Large )} + i \oint_{\gamma} {\Large (} u(x+iy) dy + v(x+iy) dx {\Large )}$$ where the right hand side is a path integer in the sense of real calculus. I'll show that the first integral vanishes; the second is similar.

Let $w$ be the vector field $w(x,y) = (u(x+iy), - v(x+iy))$. So we want to show $\oint_{\gamma} w=0$. By Stokes theorem, $\oint_{\gamma} w = \int_D \nabla \times w$. But $\nabla \times w = \frac{\partial u(x+iy)}{\partial y} - \frac{\partial (-v(x+iy))}{\partial x}=u_y+v_x$. By the Cauchy-Riemann equations, we have $u_y+v_x=0$. $\square$


OK, now what if $f$ has poles? Let $\delta$ be a union of circles, one small circle around each pole of $f$ within $\gamma$. So there are no poles of $f$ in the region between $\gamma$ and $\delta$ and the above theorem shows that $\oint_{\gamma} f(z) dz - \oint_{\delta} f(z) dz=0$. (The minus sign because $\delta$ is oriented backwards.) So $\oint_{\gamma} f(z) dz = \oint_{\delta} f(z) dz$. We just see that we can reduce the computation of $\oint_{\gamma} f(z) dz$ to integrals around small loops $\delta$ enclosing the poles of $f$.

At this point, Stokes theorem does not directly apply because Stokes theorem wants $f$ to be defined everywhere inside the curve, not to have a pole. I think the easiest thing to do is look at how your textbook does this case and stop trying to use Stokes.


But, if you insist on using Stokes... For the specific case of $f(z) = 1/z$, I can do it. Let $\delta$ be the circle of radius $1$ around $0$. We want to compute $\oint_{\delta} z^{-1} dz$. On the circle $\delta$, $(x+iy)^{-1} = x-iy$. So we want to compute $$\oint_{\delta} (x-iy) (dx+idy) = \oint_{\delta} {\Large (} x dx + y dy {\Large )} + i \oint_{\delta} {\Large (} x dy - i dx {\Large )}.$$ I'll compute the second integral. Fill the interior of $\delta$ with the vector field $w=(y,-x)$. Notice that this is NOT the imaginary part of $z^{-1} dz$: that would be $\frac{1}{x^2+y^2} (y,-x)$. It's some other vector field, which doesn't have a pole, and matches the integral which we want to compute on the circle $\delta$. So $$\oint_{\delta} w = \oint_D \nabla \times w.$$ We have $\nabla \times w = \frac{\partial y}{\partial y} - \frac{\partial (-x)}{\partial x} = 1 - (-1) = 2$. So our integral is $\int_D 2 dA=2 \mathrm{Area}(D) = 2 \pi$. We see that the imaginary part of $\oint_{\delta} z^{-1} dz$ is $2 \pi$. Working a bit harder, one can show the real part vanishes so $\oint_{\delta} z^{-1} dz = 2 \pi i$.

Conceptually, $(y,-x)$ has curl $2$ everywhere inside the unit disc. The vector field $\frac{1}{x^2+y^2} (y,-x)$ has curl $0$ everywhere but at the origin, where you should think of it has having curl $2 \pi$ times a Dirac delta function.


Can we repeat this trick for other analytic functions without poles; replace them with other functions which have the same value on the unit circle but extend smoothly inside it? The only way that I know how to do it is to use material which is usually proved USING the Cauchy residue formula. Namely, write $f(z)$ as an absolutely convergent sum $\sum_{n=-\infty}^{\infty} a_n z^n$. Set $g(z) = \sum_{n = -\infty}^{-1} a_n \overline{z}^n + \sum_{n=0}^{\infty} a_n z^n$. On the unit circle, $\overline{z} = z^{-1}$ and, inside the unit circle, $|\overline{z}| < |z|^{-1}$, so the sum will still converge. It is tractable to compute curl of the real and imaginary parts of $g(z) dz$ and recover the residue theorem. But the fact that you can write your $f$ as such as sum is usually proved using Cauchy residue. So I think that the best approach is to use Stokes to get down to doing integrals around small circles, and then switch to another method.

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