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I am an undergraduate mathematical physics student doing a summer project. I am very familiar with "mathematicicans complex analysis", but am having difficulty with how fast and loose with the rules the physicists seem to be with their notation.

Let $\gamma(a \to z)$ be some path in $\mathbb{C}$ from $a$ to $z$, and let $f(z,\overline{z})$ denote some (most certainly not holomorphic function). In my case, stuff like $1/z, 1/\overline{z}, 1/(1-z), 1/(1-\overline{z})$.

I keep coming across expressions like: $$\int_{\gamma(a \to z)} f(z, \overline{z}) d\overline{z}.$$

What does this mean? I can certainly evaluate $\int_{\gamma(a \to z)} f(z, \overline{z}) d{z} = \int_0^1 f(\gamma(t),\overline{\gamma(t)})\gamma'(t) dt$, but I do not know how to handle the first equation. What does this mean?

Specific context:

The specific context I am encountering is for evaluating the path ordered exponential (from $0$ to $z$) of $$\begin{pmatrix}0&\frac{dz}z+\frac{d\bar{z}}{\bar{z}}&\frac{dz}{1-z}+\frac{d\bar{z}}{1-\bar{z}}&0\\0&0&0&-\frac{dz}{1-z}+\frac{d\bar{z}}{1-\bar{z}}\\0&0&0&\frac{dz}z-\frac{d\bar{z}}{\bar{z}}\\0&0&0&0\end{pmatrix}.$$ I'll admit, I am already a little bit uncomfortable with this whole treating the $z$ and $\overline{z}$ as independent variables. But I am willing to work with the understanding that we can simply make a variable change from $x,y$ to $x+iy,x-iy$ and go from there. Thus a path $\gamma(t)$ is just a path in $x(t)$ and $y(t)$. I do not however, understand how this allows us to "integrate around the origin in $z$" and "integrate around the origin in $\\overline{z}$". Regardless, this is all secondary, but I include it for some context.

Jack
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    The interpretation is almost exactly the same, the only difference is the last term in the integrand is $\overline{\gamma}'(t)$ instead of $\gamma'(t)$. This no different than if you had an integral $$\int_\gamma f(z,\overline{z})|dz|$$ where the last term in the integrand would now be $|\gamma'(t)|$. Simply substitute $z=\gamma$ and every other relation follows. – Ninad Munshi Jun 12 '24 at 12:02
  • This newer question is closely related, but it doesn't deal with $f(z,\overline z)$. – Mark S. Feb 13 '25 at 14:36

1 Answers1

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It has nothing to do with complex analysis per se. Let's take a step back and compare $\Bbb{C}$ with $\Bbb{R}^2$ for the moment, in order to work in a more familiar context. A path $\gamma$ in $\Bbb{R}^2$ can be parametrized in general in the cartesian plane as the map $\gamma : [0,1] \to \Bbb{R}^2, t \mapsto (x(t),y(t))$, with $t$ the curvilinear coordinate.

Yet, you may choose $x = t$, in such a way that $\mathbf{\gamma} = (x,y)$, where $y = f(x)$. This setup corresponds for example to the standard parametrization of a path representing the graph of a real function. Nevertheless, you may choose another parametrization instead, such as $y = t$, whence $\gamma = (f^{-1}(y),y)$. This is actually a mere change of variable with respect to the previous parametrization (letting aside considerations about invertibility/bijectivity).

In consequence, if you intended to integrate a vector field $\mathbf{F}(x,y)$ along the path $\gamma(t) \in \Bbb{R}^2$, it would have to be done with respect to the curvilinear coordinate initially, i.e. $\int_\gamma \mathbf{F}(x,y) \cdot \mathrm{d}\mathbf{s} = \int_0^1 \mathbf{F}(x(t),y(t)) \cdot \dot{\mathbf{\gamma}} \,\mathrm{d}t$, but you are allowed to make a change the variable in order to carry the integration with respect to $x$ or $y$ instead.

Mutatis mutandis, these considerations are also valid in the case of a path in the complex plane, i.e. $\gamma(t) = x(t) + iy(t)$. However, it is often practical to consider the following unitary transformation (up to normalization) : $$ \begin{pmatrix} z \\ \bar{z} \end{pmatrix} := \begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x+iy \\ x-iy \end{pmatrix} \tag{$*$} $$ The pair $(z,\bar{z})$ is kind of "complex-rotated" version of the usual cartesian coordinates $(x,y)$. Since this transformation is unitary, $z$ and $\bar{z}$ form an orthogonal basis of $\Bbb{C}$ too. Here, it is important to highlight that $\bar{z}$ is not independent of $z$ in general, but it is linearly independent of it $-$ because the complex conjugation is not a $\Bbb{C}$-linear operation $-$, that is why $\bar{z}$ is allowed to play the role of a complementary basis vector.

Finally, the parametrization $\gamma(t) = x(t) + iy(t) = z(t)$ is the analog of the choice $x = t$ in $\Bbb{R}^2$, in the sense that one of the two (cartesian) coordinates is identified to the curvilinear coordinate. However, as discussed previously, it is possible to choose another parametrization by setting $\bar{z} = t$, which implies $\gamma = z = \phi(\bar{z})$ (with $\phi$ the complex conjugation, obviously). Now, given that the curvilinear coordinate is $\bar{z}$, integrating a function along the path $\gamma$ will be done with respect to $\bar{z}$.

Conclusion. You can make a change of variable, in such a way that the curvilinear coordinate parametrizing the path $\gamma$ is identified with one of the cartesian coordinates, and a fortiori with $\bar{z}$ after the transformation $(*)$ above.

Abezhiko
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