Could Residue theorem be seen as a special case of Stokes' theorem?

Question

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?

Answer 1

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 $f(x+iy) = u(x+iy) + i v(x+iy)$. 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 $u_y+v_x=0$. $\square$

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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.

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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.

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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.