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I am reading some lecture notes where it applies green's theorem on the holomorphic function $f(z)=u+iv$. The conclusion is that $\oint_{dD}f(z)dz=2i\iint_D \frac{df}{d\overline{z}}dxdy$.

The main step is to treat $dz=dx+idy$ and do the algebra as if $dx$ and $dy$ are variables, then separate real part and imaginary part, then apply the green's theorem. For me at least in calculus class, $dx$ and $dy$ are just a "code" that we integrate with respect to that variable. How do I understand this case where I treat this as a variable? I learned somewhere about how to calculate wedge products, but I didn't really understand the underlying connection with calculus.

What specific concepts do I need to learn in order to understand this one? My guess is the later chapter in baby Rudin's analysis book on the differential form. Could someone give me some insight?

BCLC
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jk001
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    maybe it's the same kind of mistake as in this question: https://math.stackexchange.com/questions/2866072/discussing-fracdd-thetaei-theta-aka-cis-before-complex-derivatives-and now following to see how this unfolds... – BCLC Sep 21 '21 at 15:12
  • I think my question is unclear and this might be irrelevant to wedge product/differential forms, even though I don't know well about them. What confuses me in this case is how I am able to say $dz=dx+idy$ as if they are variables. In a similar vain, I see someone write $\mid dz\mid$ instead of $dz$. I think this is relevant to "orientation." but I'm not sure. I hope someone please gives me some direction – jk001 Sep 21 '21 at 15:14
  • The short answer is that $\text dz,\text d x$ and $\text dy$ are differential forms (which are the natural object to integrate on a 1D path. – SolubleFish Sep 21 '21 at 15:22
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    One way to think about this is to parameterize the curve $dD$ as $z(t) = x(t)+iy(t)$. Then $dx+idy$ can be thought of as $dt(x'(t)+iy'(t))$, in which case you get a standard real one-dimensional integral (with a complex integrand). – march Sep 21 '21 at 15:30
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    Do not read Rudin's chapter if you want an understanding of multivariable analysis and differential forms. You might try my YouTube lectures on differential forms, etc. (linked in my profile). $|dz|$ represents the element of arclength, usually written $ds$ in multivariable calculus. Here it's better to work with $dz$ and $d\bar z$ as a basis (over $\Bbb C$) for the $1$-forms, rather than $dx$ and $dy$. Green's Theorem is just $\int_{\partial D}\omega = \int_D d\omega$, and you can compute $d$ in the $z,\bar z$ world. – Ted Shifrin Sep 21 '21 at 15:47
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    If you don't want to go the differential forms route, it's generally understood that there are situations where you can treat the differentials as variables as shorthand for a more thorough proof. In particular, a complex valued integral is defined as a sum of real valued integrals, one scaled by $i$, so this is one of those "legal" situations. – While I Am Sep 21 '21 at 15:47

2 Answers2

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I interpret this question as asking "What is the minimum I need to understand why I can manipulate $dx$ and $dy$ algebraically?"

At the simplest level (without going into differential forms and such), we can think of $dz=dx+idy$ in terms of a parameterization of the curve $\partial D$. We start by breaking up the curve into $N$ small chunks bounded by the points $z_{i=0,\dots,N}$, with $z_0=z_N$, and approximate each part of the curve as $\Delta z_i=z_{i+1}-z_i$. Then, we can sum up the function along the segmented curve as $$ \sum_{i=0}^{N-1}f(z_i)\Delta z_i. $$ We can break apart $\Delta z_i$ into real and imaginary parts as $\Delta z_i=\Delta x_i +i \Delta y_i$. In the limit, these little differences $\Delta x_i$ and $\Delta y_i$ correspond to the $dx$ and $dy$ in the integral. To make that correspondence more rigorous, we parameterize the curve as $$ z(t) = x(t) +i y(t), $$ with $t\in[0,1]$ and identify $z(t_i)=z_i$. Then, the sum above becomes $$ \sum_{i=0}^{N-1}f(z(t_i))\frac{\Delta z_i}{\Delta t}\Delta t, $$ where $\Delta t = t_{i+1}-t_i$ and $$ \Delta z_i= z(t_{i+1})-z(t_i) =\left(x(t_{i+1})-x(t_i)\right) + i\left(y(t_{i+1})-y(t_i)\right) =\left(\Delta x_i\right) + i\left(\Delta y_i\right). $$ Taking the limit of more and more, smaller and smaller chunks, the sum becomes $$ \int_0^1f(z(t))z'(t)dt =\int_0^1f(z(t))(x'(t)+iy'(t))dt. $$ After expanding $f$ into real and imaginary parts, this integral can be separated into two integrals with real integrand which are the real and imaginary parts of the complex integrand.

Finally, then, we can interpret (in a sense) that $dx=x'(t) dt$ and $dy = y'(t) dt$, and the algebraic manipulations therefore follow from normal rules of algebra (because really, we're manipulating the integrand at that point and not differentials).

Notes

  • First of all, of course, we have to assume convergence issues have been dealt with.
  • Second, all of this also requires definitions and proofs involving of derivatives and integration of complex functions that show that both differentiation and integration are linear with respect to complex numbers, by which I mean, for instance, that $z'(t) = x'(t) + i y'(t)$ if $z(t) = x(t) +i y(t)$, with $t$ real.
march
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The main concepts needed here are :

  • complex differential forms (most introductory books only deal with real valued differential forms, but it is not much more complicated to deal with complex valued ones).
  • their integration
  • Stoke's theorem
SolubleFish
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  • is the issue in this question perhaps similar to the issue in that question https://math.stackexchange.com/questions/2866072/discussing-fracdd-thetaei-theta-aka-cis-before-complex-derivatives-and… i.e. the author/s might've forgotten to define something? or not really? – BCLC Sep 21 '21 at 15:52