3

In Gameline's Complex Analysis Chapter 8, the notation $d(\log f(z))$ is used:

enter image description here enter image description here

Here are my questions:

In the real case, suppose for any $x\in\mathbb{R}$, one has $f(x)\neq 0$ and $f$ is differentiable. Then one has $$ d(\log(f(x))=\frac{f'(x)}{f(x)}dx $$ by the chain rule.

  • But in the complex case, if the $\log$ function is not differentiable on the curve $\gamma$, how should one makes sense of $d\log f(z)$?

  • Similarly, how should one understand $d\arg(z)$?

  • $fdx+gdy$ is exact if $dh=fdx+gdy$ for some $h$ according to Gameline's book. But what does "$d\arg(z)$ is not exact" mean?

  • when you write $\int_\gamma d \log f(z)$ you suppose that you chose $\log$ to be "unwrapped", hence that locally (on every point of the contour $z \in \gamma$) it is true that $(\log f(z))' = \frac{f'(z)}{f(z)}$. this doesn't make sense since you get that $\log(f(z_0))$ depends on the path from which you arrived to $z= z_0$, but what makes sense is considering for a (smooth) curve of the complex plane $\gamma(t), t \in [0,1]$ some function $L(t)$ such that $L(t) = \log(f(\gamma(t)))+ 2i k(t) \pi$ where $k(t)$ is a function with integer values, and $L'(t)$ exists for every $t\in [0,1]$. – reuns Mar 29 '16 at 02:21
  • it is what is explained there https://en.wikipedia.org/wiki/Instantaneous_phase : the key idea is that for some given contour $\gamma$, you can consider the unwrapped $\log,arg$ on that contour, but it will depend on the contour – reuns Mar 29 '16 at 02:23
  • SEE THIS ANSWER for a detailed development of $\int_{\gamma} \frac{1}{z},dz$ where $\gamma$ is rectifiable. Apply the same procedure to $\frac{f'(z)}{f(z)}$. – Mark Viola Mar 29 '16 at 03:42

2 Answers2

1

If $f:\>\Omega\to{\mathbb R}$ or $f:\>\Omega\to{\mathbb C}$ is a "bona fide" function defined on some domain $\Omega\subset{\mathbb C}$ its differential $df$ is a "closed one-form", and can be expanded as $$df=\quad adx+bdy,\quad g dz+h d\bar z,\quad gdz\ ,$$ depending on the context. Closed one-form means that the integral $\int_\gamma df$ has the same value $f(q)-f(p)$ for all curves beginning at $p$ and ending at $q$, and is $=0$ for all closed curves.

Now already in Calculus 102 we write $d\phi$ when working with polar coordinates, even though the variable $\phi$ is not a bona fide function in the punctured plane $\dot{\mathbb R}^2$ or $\dot{\mathbb C}$. Usually one glosses over this point and "integrates from $\phi=0$ to $\phi=2\pi$". It is however true that each point $z_0\in \dot{\mathbb C}$ has a pretty large neighborhood in which one can select a well defined branch of the variable $\phi$ that serves all desired purposes. This implies that the one-form denoted by $d\phi$ is exact. By the way, it expands as $$d\phi={-y\over x^2+y^2}dx+{x\over x^2+y^2}dy\ ,\tag{1}$$ whereby the right hand side is uniquely defined in all of $\dot{\mathbb R}^2$.

It is in this "abuse de language" sense that you have to understand the $d$ used in typographical assemblies like $d\log$ or $d{\rm arg}$. Corresponding to $(1)$ one has $$d\log(z)={dz\over z}\ ,$$ whereby the right hand side is well defined on all of $\dot{\mathbb C}$, and $d{\rm arg}$ is nothing else but the $d\phi$ alluded to above.

Christian Blatter
  • 226,825
0

[Partial answer]

  • Consider the case when $f(z)=z$ and $\gamma$ is the unit circle $\{z:|z|=1\}$. Note that the function $\frac{1}{z}$ does not have an antiderivative on $\mathbb{C}\backslash\{0\}$, or even on a neighbourhood of the unit circle (or any other curve going around the origin). Therefore, one cannot interpret the formula $ \frac{1}{z}\ dz=d\log(z) $ as $$ \frac{d\log(z)}{dz}=\frac{1}{z}, \quad z\in\gamma. $$

  • Therefore, one should really think (1.1) as written in a "formal" way. (Conway pointed out in his Functions of One Complex Variable that one cannot define $\log f(x)$ on $\gamma$ since otherwise one would have $\int_\gamma f'/f=0$, which is nonsense.) On the other hand, $\int_\gamma d\ln|z|$ is defined in Chapter III of Gameline's book as aline integral: $$ \int_\gamma d\ln|z|=\int_\gamma d\big(\frac{1}{2}\log(x^2+y^2)\big) =\int_\gamma \frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy $$

  • Another way to look at the integral $\int_\gamma\frac{f'(z)}{f(z)}\ dz$ without using the annoying discussion of differentials is via change of variables: $$ \frac{1}{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}\ dz=\frac{1}{2\pi i}\int_{f\circ\gamma}\frac{1}{z}\ dz\tag{*} $$ the right hand side of which is defined to be the winding number of the (closed) curve $f\circ\gamma$ around $z=0$. If ${f\circ\gamma}$ is homotopic in ${{\mathbb C} \backslash \{0\}}$ (as a closed curve, up to reparameterisation) to a curve of the form ${t \mapsto z_0 + r e^{imt}}$, ${t \in [0,2\pi]}$, then $(*)$ can be calculated by Cauchy's theorem.