\[\pdd{\bar{z}} \pqty{\frac{1}{z}} = \pi\,\delta(\vec{z})\]For English version, scroll to end or click here.
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我最早是在 Qualls 的 CFT 讲义 arxiv:1511.04074 计算自由 boson 场处看到的这个公式，没想到在著名的“黄书” Di Francesco 中也出现了（在引入二维共性对称不久），还给出了仔细的证明，可见比较重要。下面我们也尝试证一证，并以此复习一下早已忘却的复变函数 
首先，式中 $\pdd{z}, \pdd{\bar{z}}$ 常简记为 $\partial, \bar{\partial}$, 是将 $(z,\bar{z})$ 看成两个独立变量 的求导，而 $\delta(\vec{z})$ 是 $\mbb{C} \simeq \mbb{R}^2$ 二维平面上的 Dirac $\delta$ 函数。若你知道二维点电荷的势函数，可以迅速而巧妙地证明该式（参见 Qualls）；太长不看版：
\[\newcommand{\pd}{\partial} \newcommand{\pdbar}{\bar{\partial}} \pdbar \pqty{\frac{1}{z}} = \pdbar \pd \ln \abs{z}^2 = \frac{1}{4} \laplacian \ln \abs{z}^2 = \pi\,\delta(\vec{z})\]下面我们仔细说明一下这个巧妙证明，以及介绍一下直接证明该式子的常规方法。Note: 过于巧妙的证明有一大问题，即对于像我这样比较愚钝的人而言难以学会；相反，常规方法往往具有很高的普适性！
一些背景 + 巧妙证明
这里，我们首先给出 $(z,\bar{z})$ 变量的微分法则——数学上这东西应该是可以严格化的（大概要用复几何？），而我们暂且投机取巧，就把他们当成 $(x,y)$ 的坐标变换来处理——有：
\[x = \frac{z + \bar{z}}{2},\ \ y = \frac{z - \bar{z}}{2i} \quad\Longrightarrow\quad \pdd{z, \bar{z}} = \frac{1}{2} \pqty{\pdd{x} \mp i\,\pdd{y}}\]这里的 $\frac{1}{2}$ 万万不可忘记，它来自坐标变换的 Jacobian 行列式。
关于 $(z,\bar{z})$ 的补充说明： 另一种看待 $(z,\bar{z})$ 的视角则是将原来的 $(x,y) \in \mbb{R}^2$ 视为复变量 $(x,y) \in \mbb{C}^2$, 那么 $(x,y) \mapsto (z,\bar{z})$ 不过是 $\mbb{C}^2 \to \mbb{C}^2$ 的坐标变换，显然此时 $\bar{z}$ 不一定是 $z$ 的共轭，这便解决了自由度方面的困惑。参见 Di Francesco. 当然，最终的物理结果还是在 $(x,y) \in \mbb{R}^2$ 上考察，这有点像物理中在壳 / 离壳（on / off shell）的关系。
此外，对这个问题而言，我们首先应当知道如下事实：
\[\pdbar\pd = \frac{1}{4}\,\pqty{\pdd{x} + i\,\pdd{y}}\pqty{\pdd{x} - i\,\pdd{y}} = \frac{1}{4}\laplacian\]巧妙证明之所以巧妙，即在于你一看就懂，但死活想不到；知道 $\pdbar\pd$ 正是 Laplace 算符，接下来就好办了。众所周知，据 Gauss 定理 / 定律，二维空间中单位点源的场强为 $\frac{1}{2\pi r}$, 相应地势函数为其积分，也就是说我们可以直接写出 $\laplacian$ 的 Green 函数：
\[\delta(\vec{z}) = \laplacian \pqty{\frac{1}{2\pi}\,\ln \abs{z}} = \frac{1}{4\pi} \laplacian \ln z\bar{z}\]胜利在望，接下来我们只需代入 $\pdbar\pd$, 并注意到 $\pd \ln z\bar{z} = \frac{\bar{z}}{z\bar{z}} = \frac{1}{z}$, 即完成了原式的证明。细节留待读者自行补充 
常规证明
常规证明由 Di Francesco 和这个问答给出，基于确定 $\delta$ 系数的一般方法：在奇点附近积分。一个有意思的问题是，这两个参考来源给出的答案相差一个符号；个人以为 Di Francesco 的结果是正确的，因为和前面介绍的“巧妙方法”结果一致。符号的问题很可能来源于定向的混淆；具体而言，采用外微分记号，我们有：
\[\dd{z} \wedge \dd{\bar{z}} = \pqty{\dd{x} + i\dd{y}} \pqty{\dd{x} - i\dd{y}} = -2i \dd{x} \wedge \dd{y}\]可见此时 Jacobian 行列式为 $-2i$, 是个虚数！大家知道，坐标变换 Jacobian 的正负号体现变换是否改变流形的定向，不过此前我们遇到的 Jacobian 都是实的，这里却是个虚数，应当如何理解呢？我也不懂，如有谁知道请告诉我…
不过，即便如此，我们也并非束手无策；事实上，只要始终保证积分变量为实变量 $(x,y)$ 或 $(r,\theta)$, 视 $z = z(x,y; i)$, 其中虚单位 $i$ 视为积分的一个参数，则总可以自洽地完成 $\mbb{C} \simeq \mbb{R}^2$ 上的积分 $\int \dd{x} \dd{y}$. 如有需要复平面上的围道积分，再于适当的时候利用：
\[\oint \dd{z} = \int_0^{2\pi} R\,e^{i\theta} i \dd{\theta}\]进行代换（这里取了圆形围道 $r = R$）。
前期准备
为证明某函数是 $\delta$ 分布，较为严格的做法是引入“测试函数” $f(\vec{z})$ 然后与之积分，看看是否得到 $f(\vec{z_0})$. 不过，这里我们再次投机取巧，直接假定结果就是 $\propto \delta$ 函数，只是通过积分求出待定的系数；这是基于如下几点观察：
- 首先，$\pdbar \pqty{\frac{1}{z}}$ 在 $z = 0$ 以外处处为 0, 这是显然的；
- 对于 $z = 0$ 处的奇点，有如下对称性：
-
旋转不变： 考虑 $z \mapsto e^{i\phi} z$, 则 $(z,\bar{z}) \mapsto (e^{i\phi} z,\,e^{-i\phi} \bar{z})$,
\[\pdbar \pqty{\frac{1}{z}} \longmapsto \pdv{}{e^{-i\phi}} \frac{\pd}{\pd\bar{z}} \pqty{\frac{1}{e^{i\phi} z}} = \pdbar \pqty{\frac{1}{z}}\] -
共轭不变： 考虑 $z \mapsto \bar{z}$,
\[\pdbar \pqty{\frac{1}{z}} \longmapsto \pd \pqty{\frac{1}{\bar{z}}}\]左右两边其实是一样的！为什么呢？这其实再次回到了前面的“巧妙证明”，不过这次我们有明确的目的——将 $\pdbar$ 转化为 $\pd$, 因此这一步是相对容易想到的：
\[\pdbar \pqty{\frac{1}{z}} = \pdbar \pd \ln z\bar{z} = \pd \pdbar \ln z\bar{z} = \pd \pqty{\frac{1}{\bar{z}}}\]
-
有意思的是，考察这个奇点的对称性时，我们的思路和分析黑洞对称性的过程是 异曲同工 的：只有旋转对称时，并不能区分 Schwarzschild （静态）和 Kerr （旋转）黑洞，还需要判定是否反演不变。
综合上述观察，我们发现 $\pdbar \pqty{\frac{1}{z}} = \pd \pqty{\frac{1}{\bar{z}}}$ 给出 $z = 0$ 处的一个旋转对称、镜面对称的奇点，这大概只可能是 $\delta$ 函数了。当然，这并不是一个严谨的证明，但对物理学家而言大概是足够好了吧！ 
此外，反演对称性表明结果意味着 $\delta$ 分布 不像 是复平面上的解析函数；同时，解析函数的一个重要特征是不存在局部极值，而 $\delta$ 分布（或更严格地，形如 $\delta$ 分布的函数）极大地违背了这一要求。因此我们最好用矢量记号标记其变量，即写作 $\delta(\vec{z})$.
积分定系数
综上，我们有：
\[\pdbar \pqty{\frac{1}{z}} = p\,\delta(\vec{z})\]下面确定系数 $p$, 只需计算面积分：
\[p = \int \dd{x}\dd{y}\, \pdbar \pqty{\frac{1}{z}}\]为了避开奇点、干净地化简这货，我们需要“正规化” $\frac{1}{z} \mapsto \frac{1}{\sqrt{z^2 + a^2}}$. 或者，利用 Gauss 散度公式：
\[\int_{\mcal{M}} \mrm{d}^{n+1}{x} \ \pdd{\mu} J^{\cdots\,\mu\,\cdots} = \int_{\pd\mcal{M}} \mrm{d}^n{x} \ n_\mu J^{\cdots\,\mu\,\cdots}\]这里 $n^\mu$ 是边界 $\pd\mcal{M}$ 上垂直面元 $\mrm{d}^n{x}$ 向外的法矢量。此时，$\mcal{M}$ 可取为覆盖原点的一小圆盘，原式化为：
\[p = \oint \dd{\ell}\ n_\mu\,J^\mu\]先别急着代入计算！由于我们处于美妙的二维，此时可将退化的“面”积分化为沿曲线的围道积分；按照通常约定，取积分沿逆时针方向，则：
\[n_\mu \dd{\ell} \sim (\dd{y}, -\dd{x}) \sim \epsilon_{\mu\nu} \dd{x^\nu}\]这里 $\epsilon_{\mu\nu}$ 是 Levi-Civita 反对称记号；最终得到：
\[p = \oint \epsilon_{\mu\nu}\,J^\mu \dd{x^\nu}\]取 $\mu = x,y,\ J^\mu\sim (J^x, J^y) = \frac{1}{2z}\pqty{1, i}$, 参数化 $x = \cos\theta,\,y = \sin\theta$, 而 $z = x + iy$, 我们得到：
\[p = \int_0^{2\pi} \frac{1}{2z}\, \pqty{-i, 1}\cdot\pqty{-\sin\theta, \cos\theta} \,\dd{\theta} = \pi\]搞定！这一过程甚至没有用到复变围道积分。
不过，在一个旋转对称的系统中用直角坐标进行计算似乎有些丑陋；下面我们试着用极坐标再算一遍。须注意，引入坐标变换前，我们得先将积分式写为协变形式，即需要将 Levi-Civita 记号补上度规系数使之成为张量：
\[\epsilon_{\mu\nu} \longmapsto \sqrt{\det g}\,\epsilon_{\mu\nu}\]变换后再令 $r \equiv \mrm{const}$, 即沿圆形围道积分，这导致：
\[p = \int_0^{2\pi} rJ^r \dd{\theta} = \oint \frac{J^r}{i\,e^{i\theta}}\,\dd{z}\]我们需要知道 $J^r$. 利用 $z = r\,e^{i\theta}$, 可得 $ \pdbar \mapsto (\pdbar{r})\,\pdd{r} = \frac{1}{2} \frac{z}{r} \pdd{r} = \pdd{r} \circ \frac{1}{2} \frac{z}{r}$, 这里利用了 $ \pdd{r} \frac{z}{r} = \pdd{r} e^{i\theta} = 0$. 此外，我们无需 $J^\theta$, 故直接丢掉了 $\pdd{\theta}$ 的贡献。由此得到：
\[J^r = \pqty{\frac{1}{2} \frac{z}{r}} \frac{1}{z} = \frac{1}{2r} = \frac{e^{i\theta}}{2z}\]下一步可以选择对 $\theta$ 积分或对 $z$ 进行围道积分，均给出 $p = \pi$.
总结人生经验
是正是负？这是一个问题
上述证明过程中我们谨小慎微，直到最后才引入 $\oint \dd{z}$, 就是为了避免 $(x,y)\to (z,\bar{z})$ 时复系数造成的符号问题。前已提及，如果直接计算面积分 $\int \dd{z}\wedge\dd{\bar{z}}$, 可能难以判断定向，从而不能确定结果的正负号；有趣的是，将面积分转化为线积分 $\oint \dd{x^\mu}$ 后，这一问题依然可能复现！仍应倍加小心。
具体而言，我们从前述：
\[p = \oint \epsilon_{\mu\nu}\,J^\mu \dd{x^\nu}\]出发，若此时直接用坐标 $(z,\bar{z})$ 进行计算，同样我们需要将 Levi-Civita 张量 协变化，即乘上 $\sqrt{\det g}$. 对 $\mu = z,\bar{z}$, 计算给出：
\[g_{\mu\nu} = \delta_{ij} \pdv{x^i}{z^\mu} \pdv{x^j}{z^\nu} = \vec{x}_\mu \cdot \vec{x}_\nu \sim \pqty{ \begin{smallmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{smallmatrix} },\\ \det g = - \frac{1}{4}\]很不幸，我们再次碰上了符号问题：这里的 $\sqrt{\det g} = \frac{1}{2} \sqrt{-1}$, 其中 $\sqrt{-1}$ 应当取为 $+i$ 还是 $-i$ 呢？似乎并不清楚！暂且令 $\mcal{I} \equiv \sqrt{-1}$, 我们发现，有：
\[p = \oint \pqty{ - \frac{\mcal{I}}{2} } J^{\bar{z}} \dd{z} = \pqty{ - \frac{\mcal{I}}{2} } \oint \frac{1}{z} \dd{z} = - \mcal{I}\cdot \pi i\]比较前面给出的答案，可见应当取 $\mcal{I} = +i$, 方能给出 $p = +\pi$ 的一致结果。这也与 Di Francesco 给出的 Levi-Civita 张量吻合：
\[\sqrt{\det g}\,\epsilon_{\mu\nu} \sim \frac{i}{2} \pqty{ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} }\]关于共轭变量的导数
前文涉及“巧妙证明”和论证反演不变时，我们用到了一个关键的式子：
\[\pd\,\ln \abs{z}^2 = \pd\,\ln z\bar{z} = \frac{1}{z}\]这个式子仔细一看是有点不对劲的，按道理来说是 $\pd\,\ln z = \frac{1}{z}$, 其中多值函数 $\ln z = \ln \abs{z} + i\arg z$, 而 $\ln\abs{z}^2 = 2\ln \abs{z}$, 如果两者差一常数则没有问题，可是看上去却是相差两倍有余，这是不是一个 bug 呢？
事实上，在 $\pd$ 看来，两者确实相差一个“常数”：注意有：
\[\begin{align} \ln\abs{z}^2 - \ln z &= 2\ln \abs{z} - \pqty{\ln \abs{z} + i\arg z} \\ &= \ln \abs{z} - i\arg z \\ &= \ln \bar{z} \end{align}\]由于引入了 $\pdbar$, 在 $\pd$ 看来，$\ln \bar{z}$ 正像是一个常数，并没有产生矛盾。
上式实际上正是对数的“化乘为加”性质：$\ln z\bar{z} = \ln z + \ln \bar{z}$. 自然，对于多值函数 $\ln z$, 上式只有在恰当选取单值分支时才严格成立，否则可能相差若干个常数 $2\pi i$. 当然这并不影响我们的结论。
应当注意，在引入 $\pdbar$ 之前，由于 $\ln \abs{z}^2$ 并不是解析函数（或者，按数学家的用词——全纯 / 亚纯函数），$\pd\,\ln \abs{z}^2$ 根本就是没有定义的！引入 $\pdbar$ 后，解析函数的 $\pd$ 导数与原本一致，同时我们还可以额外处理这一类非解析的函数，故有上述表观“矛盾”，实则没有问题。
English Version
\[\pdd{\bar{z}} \pqty{\frac{1}{z}} = \pi\,\delta(\vec{z})\]
Notes / 学习笔记is a tag dedicated for my unfinished notes. The content of these notes is mostly extracted from classic textbooks. Normally, these references are stated at the very beginning of an article, or given within the post itself, therefore do not strictly follow the format of bibliography. For this I would like to apologize in advance.On the other hand, if I am lucky, there just might be some original ideas from these notes. Sharing of this post is fully welcomed; however, please share in accordance with the
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I first noticed this peculiar formula from Qualls’ lecture notes about CFT arxiv:1511.04074. It is related to the free boson propagator in 2D. What I didn’t expect is that it also appears in the famous “Yellow Book” Di Francesco (somewhere around the introduction of two-dimensional conformal symmetry). A detailed proof is also given there, which hints at its importance.
Here we will try to prove it in multiple ways, and by doing so try to refresh our memories of complex analysis, which have long been forgotten (at least by me).
At first, let’s clear up some of the conventions. Operators $\pdd{z}, \pdd{\bar{z}}$ are usually abbreviated as $\partial, \bar{\partial}$. They denote differentiations while taking $(z,\bar{z})$ as two independent variables. $\delta(\vec{z})$ is the Dirac delta function on the two dimensional plane $\mbb{C} \simeq \mbb{R}^2$.
If you happen to know the potential of a two dimensional point charge, you can prove this formula in a most elegant and concise way (see Qualls). TLDR:
\[\newcommand{\pd}{\partial} \newcommand{\pdbar}{\bar{\partial}} \pdbar \pqty{\frac{1}{z}} = \pdbar \pd \ln \abs{z}^2 = \frac{1}{4} \laplacian \ln \abs{z}^2 = \pi\,\delta(\vec{z})\]Let’s discuss this elegant proof in detail, and further introduce the usual method of proving this kind of formulas. Note: The problem with an elegant proof is that it is hard to master for general population (like me), while the usual method is usually easy to generalize.
Backgrounds + Elegant Proof
Here we first lay out the differentiation rules with respect to $(z,\bar{z})$. I believe they can be given in a mathematically rigorous way (possibly using complex geometry?), but for now let’s be lazy, and just regard them as the result of coordinates transformation from $(x,y)$. We have:
\[x = \frac{z + \bar{z}}{2},\ \ y = \frac{z - \bar{z}}{2i} \quad\Longrightarrow\quad \pdd{z, \bar{z}} = \frac{1}{2} \pqty{\pdd{x} \mp i\,\pdd{y}}\]Here we should not forget the $\frac{1}{2}$ factor, which comes from the Jacobian of coordinates transformation.
Additional notes regarding $(z,\bar{z})$: Another way of seeing these $(z,\bar{z})$ variables is to take the original coordinates $(x,y) \in \mbb{R}^2$ as complex variables $(x,y) \in \mbb{C}^2$, then $(x,y) \mapsto (z,\bar{z})$ is truly some transformed coordinates from $\mbb{C}^2 \to \mbb{C}^2$. Note that now $\bar{z}$ is not necessary the complex conjugate of $z$, which resolves the issue about degrees of freedom (see Di Francesco).
Of course, the final physical result should be obtained by restricting $(x,y) \in \mbb{R}^2$. This is kind of similar to the relation between on / off shell conditions.
Besides, for this particular problem, we should know the following fact before proceeding:
\[\pdbar\pd = \frac{1}{4}\,\pqty{\pdd{x} + i\,\pdd{y}}\pqty{\pdd{x} - i\,\pdd{y}} = \frac{1}{4}\laplacian\]The elegance of the proof lies in the fact that you know it the minute you see it, but it is almost impossible to figure it out by yourself, before knowing the trick. Once we know that $\pdbar\pd$ is precisely the Laplacian, the rest is natural. According to the well-known Gauss’s theorem / law, the field strength of a 2D point source is $\frac{1}{2\pi r}$, which when integrated gives potential; we can therefore write down the Green function of $\laplacian$：
\[\delta(\vec{z}) = \laplacian \pqty{\frac{1}{2\pi}\,\ln \abs{z}} = \frac{1}{4\pi} \laplacian \ln z\bar{z}\]Next we need only substitute in $\pdbar\pd$, note that $\pd \ln z\bar{z} = \frac{\bar{z}}{z\bar{z}} = \frac{1}{z}$, and the proof is done. Details are left as an exercise for the readers
Usual (conventional) Proof
A more conventional proof is given by Di Francesco and this answer, based on the usual procedure of fixing a $\delta$ distribution — by integrating over the singularity.
An interesting note: these two references actually produce different results, off by a sign; I believe that the result in Di Francesco should be correct, since it agrees with our previous result from the elegant proof. The sign issue might be related to some confusion about the orientation; in differential forms, we have :
\[\dd{z} \wedge \dd{\bar{z}} = \pqty{\dd{x} + i\dd{y}} \pqty{\dd{x} - i\dd{y}} = -2i \dd{x} \wedge \dd{y}\]We see that the Jacobian is given by a imaginary number $-2i$. As we know, the sign of the Jacobian labels the change orientation of a coordinate transformation; however, we have only encountered real Jacobians before. How should we make of this imaginary Jacobian? This remains a mystery to me…
Fortunately, despite this minor inconvenience, we can still carry out the calculations, just with some extra care: we keep the integration variables “real”, as e.g. $(x,y)$ or $(r,\theta)$, while taking $z = z(x,y; i)$ as a function of the real variables, where the imaginary unit $i$ is only a parameter of the integration. With this trick, we shall see that it is possible to complete the integral $\int \dd{x} \dd{y}$ on $\mbb{C} \simeq \mbb{R}^2$. If we do need to perform a complex contour integral, we can then substitute:
\[\oint \dd{z} = \int_0^{2\pi} R\,e^{i\theta} i \dd{\theta}\]Here we have taken a circular contour with $r = R$.
Preliminary work
A more rigorous procedure to proof that some function is the $\delta$ distribution is by integrating with some “test function” $f(\vec{z})$, and checking if we have $f(\vec{z_0})$. But again we decide to cheat a little, by assuming the result is already $\propto \delta$ function, and only use the integral to fix the proportionality constant. This is based on the following observations:
- Firstly, $\pdbar \pqty{\frac{1}{z}}$ vanishes unless $z = 0$; this is obvious；
- For the singularity at $z = 0$, we note the following symmetries:
-
Invariance under rotation: Consider $z \mapsto e^{i\phi} z$, then $(z,\bar{z}) \mapsto (e^{i\phi} z,\,e^{-i\phi} \bar{z})$,
\[\pdbar \pqty{\frac{1}{z}} \longmapsto \pdv{}{e^{-i\phi}} \frac{\pd}{\pd\bar{z}} \pqty{\frac{1}{e^{i\phi} z}} = \pdbar \pqty{\frac{1}{z}}\] -
Invariance under conjugation： Consider $z \mapsto \bar{z}$,
\[\pdbar \pqty{\frac{1}{z}} \longmapsto \pd \pqty{\frac{1}{\bar{z}}}\]We see that the left and right hand side is, in fact, identical. The reason goes back to our elegant proof before, but this time we have a clear objective: we need to transform $\pdbar$ into $\pd$, hence this step is easy to come by:
\[\pdbar \pqty{\frac{1}{z}} = \pdbar \pd \ln z\bar{z} = \pd \pdbar \ln z\bar{z} = \pd \pqty{\frac{1}{\bar{z}}}\]
-
An interesting note: our thought process is similar to the analysis of the symmetry of a blackhole: with only rotational symmetry, we cannot distinguish Schwarzschild (static) and Kerr (rotating) blackholes; we need to check for invariance under time-reversal, which is realized by conjugation in math.
With the above observations, we see that $\pdbar \pqty{\frac{1}{z}} = \pd \pqty{\frac{1}{\bar{z}}}$ gives a singularity at $z = 0$ that is invariant under rotation and reflection; we could see no other possibilities other than it being a $\delta$ function. Obviously, this is not a rigorous proof, but I believe it is adequate for most physicists
Besides, we note that the invariance under conjugation means that the $\delta$ distribution is not an analytic (holomorphic or meromorphic) function. Recall that a central property of an analytic function is that it has no local extremum, which the $\delta$ distribution (or any $\delta$-like function) violates to the maximal extent. Therefore, it is more appropriate to treat the arguments of this $\delta$ distribution as a vector, written as $\delta(\vec{z})$.
Coefficient from integration
In summary, we have：
\[\pdbar \pqty{\frac{1}{z}} = p\,\delta(\vec{z})\]Next we fix the coefficient $p$; we need only compute the following surface integral:
\[p = \int \dd{x}\dd{y}\, \pdbar \pqty{\frac{1}{z}}\]To simplify this, we need to first regularize the singularity: $\frac{1}{z} \mapsto \frac{1}{\sqrt{z^2 + a^2}}$. Alternatively, we can use Gauss’s divergence theorem：
\[\int_{\mcal{M}} \mrm{d}^{n+1}{x} \ \pdd{\mu} J^{\cdots\,\mu\,\cdots} = \int_{\pd\mcal{M}} \mrm{d}^n{x} \ n_\mu J^{\cdots\,\mu\,\cdots}\]Here $n^\mu$ is the outward normal vector on the boundary $\pd\mcal{M}$, perpendicular to the surface element $\mrm{d}^n{x}$. For $\mcal{M}$ a small disk covering $z = \bar{z} = 0$, we have:
\[p = \oint \dd{\ell}\ n_\mu\,J^\mu\]We don’t need to compute this directly — since we are in the wonderful 2D, the ``surface’’ integral is actually a contour integral along the circle; with the usual convention of a counter-clockwise contour, we get:
\[n_\mu \dd{\ell} \sim (\dd{y}, -\dd{x}) \sim \epsilon_{\mu\nu} \dd{x^\nu}\]Here $\epsilon_{\mu\nu}$ the Levi-Civita anti-symmetric symbol; in the end, we have:
\[p = \oint \epsilon_{\mu\nu}\,J^\mu \dd{x^\nu}\]With $\mu = x,y,\ J^\mu\sim (J^x, J^y) = \frac{1}{2z}\pqty{1, i}$, and parametrization $x = \cos\theta,\,y = \sin\theta$, $z = x + iy$, we get：
\[p = \int_0^{2\pi} \frac{1}{2z}\, \pqty{-i, 1}\cdot\pqty{-\sin\theta, \cos\theta} \,\dd{\theta} = \pi\]Done! We didn’t even use the complex contour integral.
However, on second thought, using orthogonal coordinates in a rotational symmetric system is indeed less than elegant, so let’s try to do it again in polar coordinates. Before we introduce a coordinate transformation, we should first restore the integral to the covariant form, i.e. we need to add the appropriate metric factor to the Levi-Civita symbol, so that it becomes a tensor:
\[\epsilon_{\mu\nu} \longmapsto \sqrt{\det g}\,\epsilon_{\mu\nu}\]Setting $r \equiv \mrm{const}$ after coordinate transformation, we get:
\[p = \int_0^{2\pi} rJ^r \dd{\theta} = \oint \frac{J^r}{i\,e^{i\theta}}\,\dd{z}\]Now we need to know $J^r$. With $z = r\,e^{i\theta}$, we have $ \pdbar \mapsto (\pdbar{r})\,\pdd{r} = \frac{1}{2} \frac{z}{r} \pdd{r} = \pdd{r} \circ \frac{1}{2} \frac{z}{r} $; here we’ve used the fact that $ \pdd{r} \frac{z}{r} = \pdd{r} e^{i\theta} = 0 $. Besides, we don’t need $J^\theta$, so we’ve thrown away the $\pdd{\theta}$ contributions. In the end, we have：
\[J^r = \pqty{\frac{1}{2} \frac{z}{r}} \frac{1}{z} = \frac{1}{2r} = \frac{e^{i\theta}}{2z}\]Then we can integrate over $\theta$ or contour integrate over $z$; either way, we get $p = \pi$.
The Lessons
Plus or minus? This is a problem
We have been extremely careful in our derivations above, and we’ve steered clear of $\oint \dd{z}$ until the very last momentum, just to avoid the sign complication of the transformation $(x,y)\to (z,\bar{z})$. As is mentioned before, if we try to compute the surface integral $\int \dd{z}\wedge\dd{\bar{z}}$, it would be difficult to determine the orientation, thus confuse the sign in the result. Interestingly, even after we reduce the surface integral in to line integral $\oint \dd{x^\mu}$, we could still encounter a similar issue.
More specifically, we start from:
\[p = \oint \epsilon_{\mu\nu}\,J^\mu \dd{x^\nu}\]If we decide to go to the $(z,\bar{z})$ coordinates now, we still have to make the Levi-Civita symbol covariant by adding a $\sqrt{\det g}$ factor. For $\mu = z,\bar{z}$, direct calculations reveal that:
\[g_{\mu\nu} = \delta_{ij} \pdv{x^i}{z^\mu} \pdv{x^j}{z^\nu} = \vec{x}_\mu \cdot \vec{x}_\nu \sim \pqty{ \begin{smallmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{smallmatrix} },\\ \det g = - \frac{1}{4}\]Unfortunately, we run into the sign issue again; here $\sqrt{\det g} = \frac{1}{2} \sqrt{-1}$, should we take $\sqrt{-1}=+i$ or otherwise? It seems unclear. For now let’s set $\mcal{I} \equiv \sqrt{-1}$, we find that：
\[p = \oint \pqty{ - \frac{\mcal{I}}{2} } J^{\bar{z}} \dd{z} = \pqty{ - \frac{\mcal{I}}{2} } \oint \frac{1}{z} \dd{z} = - \mcal{I}\cdot \pi i\]When compared with the result before, we see that we should take $\mcal{I} = +i$ for a consistent result: $p = +\pi$. This also matches the Levi-Civita tensor given by Di Francesco:
\[\sqrt{\det g}\,\epsilon_{\mu\nu} \sim \frac{i}{2} \pqty{ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} }\]About the derivative of the conjugate variable
When we discuss the “elegant proof” and the invariance under conjugation, we utilize the following key result:
\[\pd\,\ln \abs{z}^2 = \pd\,\ln z\bar{z} = \frac{1}{z}\]This equation seems fishy after some close inspection; normally we have $\pd\,\ln z = \frac{1}{z}$, where the multi-valued function $\ln z = \ln \abs{z} + i\arg z$. On the other hand, $\ln\abs{z}^2 = 2\ln \abs{z}$; if they differ by some constant then everything is fine, but they seem to differ by more than a factor of two! This sounds like a serious bug.
In fact, they do differ by a “constant”, at least from the perspective of $\pd$. Note that:
\[\begin{align} \ln\abs{z}^2 - \ln z &= 2\ln \abs{z} - \pqty{\ln \abs{z} + i\arg z} \\ &= \ln \abs{z} - i\arg z \\ &= \ln \bar{z} \end{align}\]Since we’ve introduced $\pdbar$, $\ln \bar{z}$ behaves as a constant under $\pd$ action. There is no contradiction.
This is in fact a consequence of the fundamental property of logarithm: $\ln z\bar{z} = \ln z + \ln \bar{z}$. Naturally, for the multi-valued $\ln z$, this result only holds when we choose some compatible branches, otherwise it may differ by some $2\pi i$’s; but this do not affect our result.
We should note that before the introduction of $\pdbar$, since $\ln \abs{z}^2$ is not analytic (holomorphic / meromorphic), $\pd\,\ln \abs{z}^2$ is not well-defined at all! After the introduction of $\pdbar$, the $\pd$ derivative of a analytic function agrees with the original result, meanwhile we can also deal with these non-analytic functions, which lead to the “apparent” contradiction.