7
$\begingroup$

Is there any single variable complex analysis textbook which uses $\textrm{d}\bar{z}$?

Every single variable text I have found defines what a complex line integral with respect to $\textrm{d}z$ means, and also $\textrm{d}s$, but none of them defines an integral with respect to $\textrm{d}\bar{z}$, or $\textrm{d}\bar{z} \wedge \textrm{d} z$.

I want to be able to write computations like

$$ \begin{align*} \int_{d\Omega} \frac{f(z)}{z-\zeta} \textrm{d}z &= \iint_\Omega d\left( \frac{f(z)}{z-\zeta}\textrm{d}z \right)\\ &= \iint_\Omega \frac{\partial f}{\partial \bar{z}} \frac{1}{z-\zeta} \textrm{d}\bar{z} \wedge\textrm{d}z \\ \end{align*} $$

without needing to convert everything back into $\textrm{d}x$ and $\textrm{d}y$ terms. No textbook I have uses these symbols at all, much less develops how to do this kind of calculus.

Something I find to be completely crazy is that I have never seen the following computation in print anywhere that I can remember, even though it is surely one of the most fundamental computations in all of complex analysis:

$$ \begin{align*} \int_{S^1} \frac{1}{z} \textrm{d}z &= \int_{S^1} \bar{z} \textrm{d}z \textrm{, since $\frac{1}{z} = \bar{z}$ on $S^1$}\\ &= \iint_D \textrm{d}\bar{z} \wedge \textrm{d}z\\ &= \iint_D 2i\textrm{d}x \wedge \textrm{d}y \textrm{, since $\textrm{d}\bar{z} \wedge \textrm{d}z = 2i\textrm{d}x \wedge \textrm{d}y$}\\ &=2\pi i \textrm{, since the area of the unit circle is $\pi$} \end{align*} $$

I think this is also because the textbooks essentially ignore non-holomorphic functions and their exterior derivatives entirely.

I never felt like I really understood single variable complex analysis until I learned how to do these kinds of computations while learning several complex variables.

I am looking for such a book because I cannot really imagine teaching the subject without these tools.

Improve this question
$\endgroup$
10
  • $\begingroup$ Cool. I'm not sure I've seen what you do here before. I think part of what you want is in Gamelin, Maybe I'm remembering something from Freitag and Busam’s text or possibly Remmert. Try to say what I find tomorrow. $\endgroup$
    – James S. Cook
    Dec 2, 2020 at 0:45
  • $\begingroup$ @JamesS.Cook Ya, this stuff is all in several complex variables and complex differential geometry books, but it doesn't seem like even the basics have filtered down to introductory textbooks. $\endgroup$
    – Steven Gubkin
    Dec 2, 2020 at 0:48
  • 2
    $\begingroup$ Maybe the feeling is that in one-variable complex variables courses (which service engineering students as well as math students), that level of abstraction would be too cumbersome. $\endgroup$
    – Gerald Edgar
    Dec 2, 2020 at 11:54
  • $\begingroup$ @GeraldEdgar I think teaching the rules of the calculus is actually simpler than the hoops you have to jump through to do these kinds of calculations without the calculus available. Students also get this impression that holomorphic functions are the only kinds of functions around, and are often at a loss for which computations apply only to analytic functions, and which would apply to any smooth function. Many complex analysis students would try to apply Cauchy's theorem to the integral of $\bar{z} $ for instance. $\endgroup$
    – Steven Gubkin
    Dec 2, 2020 at 14:39
  • 1
    $\begingroup$ Gamelin has discussion of $\partial/ \partial \bar{z}$ and presents Theorem 8.5 as "Pompeiu's Formula". It is close to what you have here... but no wedges in sight. Maybe you need to write a book. $\endgroup$
    – James S. Cook
    Dec 4, 2020 at 4:56

2 Answers 2

Reset to default
5
$\begingroup$

The book you are looking for is:

Complex Variables An Introduction

Authors: Berenstein, Carlos A., Gay, Roger

In my opinion, there is no way this book is an introduction except for truly ambitious students. I would classify it as an advanced graduate course. The book is very well-written and I think this is the only book in which the Weierstrass factorization theorem is proved using the solution to the $\overline{\partial}$-equation.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks, I did not know about this book. It is refreshing to see a book organized this way: it is certainly how I think about complex analysis. I agree that it is extremely hard for an introduction. I will wait another week or two before accepting an answer on the hope that a more elementary text might be organized the same way. $\endgroup$
    – Steven Gubkin
    Dec 3, 2020 at 16:11
  • $\begingroup$ I will mention that Krantz's several complex variables book also proves Weierstrass Factorization this way. Joseph Taylor's book on SCV and Algebraic Geometry also does this in the first chapter (I think) as a motivating example of a $\bar{\partial}$ problem. $\endgroup$
    – Steven Gubkin
    Dec 3, 2020 at 16:13
  • $\begingroup$ Bought this book yesterday. Thanks for bringing this to my attention! $\endgroup$
    – James S. Cook
    Dec 4, 2020 at 16:30
2
$\begingroup$

I do appreciate the idea of using $\dfrac{\partial}{\partial\bar{z}}$ so that one can bring non-holomorphic functions into the course, and applications to physics. Your question brings the Poisson kernel to my mind.

But let's assume we are talking about a first course on functions of a single complex variable $w=f(z)$. A philosophy of the course is to proceed as one does in a course such as Calculus I, where the focus is a function of a single real variable $y=f(x)$. One learns the definition of $f'(z)$, about $\int_Cf(z)\, dz$, Taylor series $\Sigma c_n z^n$, etc. One comes to understand the similarities and surprising differences between the two "categories" $y=f(x)$ and $w=f(z)$.

So from this perspective, we should hesitate to focus too much on $\dfrac{\partial f}{\partial\bar{z}}$ because this suggests we are working in the framework of $w=f(z,\bar z)$, not the framework of a function of a single complex variable $w=f(z)$.

Of course we can think of the function $g(z)=\bar z$ in the framework of $w=f(z)$. But we quickly move past this example after discovering that $g'(z)$ does not exist by the limit definition of a function of a single variable $g'(z)=\lim_{h\to 0}\dfrac{g(z+h)-g(z)}{h}$. As instructor of such an introductory course, I would prefer to not be forced to dance around the difference (or abuse of notation?) between $f(z)$ and $f(z,\bar z)$.

When we get to a course in several complex variables with creatures like $f(z,w)$, which is akin to multivariable Calculus III with creatures like $f(x,y)$, we are forced to confront partial derivatives $\dfrac{\partial}{\partial{z}},\dfrac{\partial}{\partial{w}}$ from the very beginning. In this course, we are ready to open the flood gates to $\dfrac{\partial}{\partial\bar{z}}$ and $ \dfrac{\partial}{\partial\bar{w}}$

Improve this answer
$\endgroup$
10
  • 1
    $\begingroup$ Notation like $w = f(z,\bar{z})$ really does not make sense, and is to be avoided in my opinion. $f$ is a function taking a single complex input and returning a single complex output, whether it is holomorphic or not. It is exactly this sort of confusion that I hope to avoid by learning how to do calculus with the full suite of smooth functions $\mathbb{C} \to \mathbb{C}$. $\endgroup$
    – Steven Gubkin
    Dec 4, 2020 at 18:26
  • $\begingroup$ @StevenGubkin-I agree, hence my reference to "abuse of notation." But if $f(z,\bar{z})$ does not make sense, then what variable does $ \frac{\partial}{\partial\bar{z}}$ reference? I suppose we can go down the path of formally defining it as a linear combination of $\frac{\partial}{\partial{x}}$ and $ \frac{\partial}{\partial y}$. $\endgroup$
    – user52817
    Dec 4, 2020 at 18:39
  • $\begingroup$ I would expect students taking this course to have taken linear algebra. I would make sure they understand the derivative of a multivariable function $\mathbb{R}^n \to \mathbb{R}^m$ at a point as an $\mathbb{R}$-linear map. Then the derivative of a function $\mathbb{C} \to \mathbb{C}$ is also an $\mathbb{R}$-linear map, but it can be split into a complex linear part and a complex conjugate-linear part. $\frac{\partial f}{\partial z} \textrm{d}z$ is the complex linear part, and $\frac{\partial f}{\partial \bar{z}} \textrm{d}\bar{z}$ is the other. $\endgroup$
    – Steven Gubkin
    Dec 4, 2020 at 19:24
  • $\begingroup$ This makes the is clear why $\frac{\partial f}{\partial \bar{z}} = 0$ is an appealing condition: it is saying that the derivative map is $\mathbb{C}$ -linear, which is to say that it is just multiplication by a complex number (which we denote $f'(z)$ in this case). $\endgroup$
    – Steven Gubkin
    Dec 4, 2020 at 19:25
  • 1
    $\begingroup$ @StevenGubkin, your comment that $f(x,y)=f(z,\bar{z})$ are not the same function is reminiscent of the OSU Vector Calculus Bridge Project "joke." Suppose the temperature on a rectangular slab of metal is given by $T(x,y)=k(x^2+y^2)$. What is $T(R,\theta)$? Physicists say $T(R,\theta)=k R^2$ and mathematicians say $T(R,\theta)=k(R^2+\theta^2)$. $\endgroup$
    – user52817
    Dec 4, 2020 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.