Bundle connection with 'Strain' I am working on exercise 15.14 of Roger Penrose's "A Road to Reality" (Section 15.8 Bundle curvature).
We will be working on a bundle $\mathcal{B}^{\mathbb{C}}$ where the base space is the complex plane $\mathbb{C}$ and the fibers are copies of $\mathbb{R}$. Informally speaking - and to make the bundle interesting - we want to introduce some form of stretch (formally, the transformation $v \rightarrow 2v$) as we go around the unit circle on the base space $\mathbb{C}$.
Let's start with a REAL function $\Phi = \Phi(z, \bar{z})$ on the complex plane $\mathbb{C}$ to represent our cross-section.
We can assign the following operator (acting on $\Phi$): $\nabla = \frac{\partial}{\partial z} -ik\bar{z}$ where $k$ is a real constant and $ik\bar{z}$ is to be thought of as the "correction term" due to the 'strain' of the bundle (scalar multiplication). In other words, $\nabla \Phi = 0$ would correspond to a "horizontal" cross-section.

Question: What is the value of the scalar $k$ so that the "stretching factor" (over the unit circle $S^1$) is equal to exactly $2$?

Naively, what (I would think) that would entail would be to solve the following equation for $k$, (assuming $\Phi = 1$):
$$\int_{S^1} \nabla{\Phi} dz  = 2$$
which for $\Phi = 1$, it would be equivalent to
$$\int_{S^1} -ik\bar{z} dz  = 2$$
Or
$$2\pi k =2$$
Which yields $k = \frac{1}{\pi}$
The issue I am having is that I am not using the fact that $\Phi$ is a function of both $z$ and $\bar{z}$, so I feel like I am missing a $d\bar{z}$ integral term. Yet, I could not justify it, since the incremental change $\nabla{\Phi} dz$ seems to include all there is, if we think of the unit circle as parametrized by $z=e^{it}$ where $t$ ranges from $0$ to $2 \pi$.
Ultimately, I suspect the issue boils down to my incomplete understanding of the insistence that $\Phi = \Phi(z, \bar{z})$ is to be thought as a function of two $2$-dimensional but "related" inputs (namely, as a function from $\mathbb{C}^2 \rightarrow \mathbb{R}$), as opposed to a function of two $1$-dimensional inputs (namely, $\mathbb{R}^2 \rightarrow \mathbb{R}$) which is its intended visual representation.
What is the missing link here? Any insights would be highly appreciated (the exact answer to the problem is immaterial)
 A: I think the subtlety is in the "dual" character of the definition of the operator. On the one hand, it seems as if $\nabla = \frac{\partial}{\partial z} -ik\bar{z}$ is the "whole" operator. That would be true, if $\Phi$ was just a function of $z$
In actuality, since $\Phi = \Phi(z, \bar{z})$, the above operator is only the first component of what would be a two dimensional covariant operator.
In other words, the above is better written as: $\nabla_z = \frac{\partial}{\partial z} -ik\bar{z}$
The subscript is necessary because we also have another component in the operator, one that corresponds to the $\bar{z}$ variable. (Similar to how in the real case one can write the first component of the gradient as: $\nabla_x f = \partial_x f$). This is of course the reflection (see: conjugate - the two are after all related!) of what we did before, which means that:
$$\nabla_{\bar{z}} = \bar{\nabla_z}=\frac{\partial}{\partial \bar{z}} +ikz$$
From there, we can easily calculate the differential $1$-form via:
$$d\Phi = \nabla_z \Phi dz + \nabla_{\bar{z}}\Phi d \bar{z} = -ik\bar{z} dz + ikz d\bar{z}$$
Passing this onto the integral we get:
$$\int_{S^1} d\Phi = \int_{S^1} -ik\bar{z} dz + ikz d\bar{z} = 2k\pi + 2k\pi := 2$$
which means that $k= \frac{1}{2\pi}$
Credits: Huge thanks to @EricTowers for all the help and references! All typos and errors are my responsibility.
