In a set of lecture notes (not available online) that I'm currently working through, one is given the following set-up:

Consider rotational vector fields on the plane
$\mathbb{R}^{2}\cong\mathbb{C}$ with coordinates $(x,y)\cong
z=x+\imath y$. Let $X$ be the vector field given by $X(z)=\imath
z=-y+\imath x\in\mathbb{C}\cong T_{z}\mathbb{C}$, i.e.
$X(x,y)=-y\partial_{x}+x\partial_{y}$.

This is all fine by me, but there is one thing that I don't understand: How do I express $X(z)$ in terms of the standard notation for tangent vectors (i.e. I'm looking for the *complex* equivalent of $X(x,y)=-y\partial_x+x\partial_y$). There are two approaches that seem sensible, but they give irreconcilable results---at least so it seems:

Use $z=x+\imath y$ and $\partial_z=\partial_x-\imath\partial_y$ and directly "translate" the expression in real coordinates. This yields
$$X(z)=-\frac{z-\bar z}{2}(\partial_z+\partial_{\bar z})
+\frac{z+\bar z}{2}(\partial_{\bar z}-\partial_z)
=\bar z\partial_{\bar z}-z\partial_z $$

Assume that the isomorphism $\mathbb C\cong T_z\mathbb C $ identifies $X(z)\in \mathbb{C}$ with $X(z)\partial_z+0\cdot\partial_{\bar z}$ so that we obtain
$$ X(z)=\imath z \partial_z $$
But when "translating back" to real variables one then obtains
$$ X(z)=\frac{1}{2}\imath(x+\imath y)(\partial_x-\imath \partial_y)
=\frac{1}{2}\Big((\imath x-y)\partial_x+(x+\imath y)\partial_y\Big)$$

At first sight, it seems reasonable to simply say that the second method is based on a wrong assumption, and to simply dismiss it (even though I don't know *why* it's wrong). However, continuing in the lecture notes I find

Its trajectories are given
by $c_{z}(t)=e^{\imath t}z$. Indeed, $\dot{c}_{z}(t)=\imath
e^{\imath t}z=\imath c_{z}(t)$.

Assuming the first approach is correct we indeed obtain something different: It seems like the "translation" to the tangent space notation was not only useless (since the solution simply seems to solve the equation $c'(t)=\imath c$ with initial condition $c(0)=z$) but leads to the wrong solution! Of course, the *second* approach here does give the correct result... My question is therefore twofold:

What does the vector field $X$ correspond to in terms of the notation for derivations of smooth functions on $\mathbb C$, and is this at all relevant for finding its integral curves (trajectories)? If the answer to the latter is "no", please elaborate!

This post imported from StackExchange Mathematics at 2015-09-13 11:44 (UTC), posted by SE-user Danu