This is a question about representing smooth functions in polar coordinates. I'll write $\mathbb{C}$ for the complex plane with coordinates denoted interchangeably by $z=x+iy=(x,y)$. Meanwhile, I'll write $\mathbb{R}^2$ for the $(r,\theta)$-plane. I'll give a bit of a lengthy preamble, but you are welcome to scroll down for the question. I titled it in bold.
OK, let's suppose $f:\mathbb{C}\to\mathbb{C}$ is a $C^\infty$ smooth function. Correspondingly, we get a $C^\infty$ smooth function $g:\mathbb{R}^2 \to \mathbb{C}$ defined by $g(r,\theta)=f(re^{i\theta})$. Let's ask ourselves the question "what kind of smooth functions $g$ can arise this way?". Clearly it's not all of them. For one thing, $g(0,\theta)=f(0)$ for all $\theta$. Another thing: $g$ is $2\pi$-periodic in the $\theta$ variable [Correction: actually it must satisfy the stronger symmetry condition $g(r,\theta)=g(-r,\theta+\pi)$]. But is that the whole story? No it is not. We also have to account for infinitesimal behaviour. For example, it is quite easy to check that $$ g_r(0,\theta) = f_x(0) \cos \theta + f_y(0) \sin \theta,$$ $$g_{rr}(0,\theta) = f_{xx}(0)\cos^2 \theta + 2f_{xy}(0)\cos\theta \sin\theta +f_{yy}(0) \sin^2 \theta $$ and so on, using subscript shorthand for partial derivatives.
OK, so it seems that the $r$ partial derivatives of $g$ are forced to have very particular forms along the $\theta$-axis. What is really going on here? How can we formalize this and understand better what exactly the restriction on $g$ is?
Here is my attempt to make sense of things. We have a usual Taylor expansion for $f$ at the origin: $$ f(x,y) \sim \sum_{m,n \geq 0} a_{m,n} x^m y^n.$$ That is not too helpful for understanding what's going on with $g$. However, I believe it should also make sense (and be equivalent) to talk about the Taylor expansion of $f$ in the variables $z=x+iy$ and $\overline z = x-iy$: $$ f(x,y) \sim \sum_{m,n \geq 0} b_{m,n} z^m \overline z^n.$$ I guess that, into this expansion, we can directly substitute $z=re^{i\theta}$, $\overline z=re^{-i\theta}$ to obtain: $$g(r,\theta) \sim \sum_{m,n \geq 0} b_{m,n} r^{m+n} e^{i(m-n)\theta}$$ where this should be thought of as a kind of asymptotic expansion along the $\theta$-axis, as $r \to 0$. We could also rewrite this as: $$g(r,\theta) \sim \sum_{n \geq 0} p_n(e^{i\theta}) r^n$$ where $p_n$ is a polynomial of the form $p_n(u) = \sum_{m=0}^n c_{m,n} u^{n-2m}$.
The above is all to sketch a proof of the following claim.
Claim: If $f(x+iy)$ smooth function $\mathbb{C} \to \mathbb{C}$, then the function $g(r,\theta)=f(re^{i\theta})$ has the property $\frac{\partial^n}{\partial r^n} g(0,\theta) = p_n(e^{i\theta})$ where $p_n(u) = \sum_{m=0}^n c_{m,n} u^{n-2m}$ where all the $c_{m,n}\in\mathbb{C}$.
OK, so now we have a pretty good idea of what the restriction should be and the question is:
Question: Does the converse hold? If $g(r,\theta)$ is a smooth function with the property in the claim above that is also $2\pi$-periodic in $\theta$ (and constant on the $\theta$-axis), does it follow that $g(r,\theta)=f(re^{i\theta})$ for some smooth function $f:\mathbb{C}\to\mathbb{C}$?
Assuming the answer is "yes", I would be happy to be pointed to a reference. I'm not overly concerned with seeing a proof, just wanted some reinforcement that this is correct. I can somewhat imagine how this might go even. Given such a function $g$, I suppose we can subtract off a function $h(re^{i\theta})$ with the same expansion as $g$ along the $\theta$ axis using Borel's theorem. Then we have a $g$ with an identically vanishing expansion and the problem becomes to show under that assumption that we can push it forward to a smooth function while keeping it smooth which sounds pretty believable to me.
Also welcome would be a nice rephrasing of what I've written in better language! I feel like I should be trying to package this more in terms of harmonic analysis or representation theory, but I'll leave it in the above form for now.
Thanks for reading this long rambling question!
Added 13/Aug/2023
Two things to add:
Firstly, it should be pointed out that I missed a necessary condition above. Given $f(x,y)$ smooth, the corresponding $g(r,\theta) =f(re^{i \theta})$ is clearly not simply $2\pi$-periodic in $\theta$, but actually satisfies the stronger symmetry condition $g(-r,\theta+\pi) = g(r,\theta)$ of which $2\pi$-periodicity in $\theta$ is a consequence.
Secondly, here is a bit more information about pushing forward a given smooth function $g(r,\theta)$ that satisfies the symmetry condition $g(-r,\theta+\pi) = g(r,\theta)$ and vanishes to infinite order along $r=0$. Let me write $T(r,\theta)=re^{i\theta}$ for the polar coordinates transformation, which is a local diffeomorphism away from $r=0$. Note that $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial_y}$ pull back under $T$ to mercifully nice operators vector fields. Specifically: \begin{align*} T^*(\frac{\partial}{\partial x}) = \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} && T^*(\frac{\partial}{\partial y}) = \sin \theta \frac{\partial}{\partial r} + \frac{\cos \theta}{r} \frac{\partial}{\partial \theta} \end{align*} From the symmetry condition alone, it follows that there is a unique function $f$ such that $g=f \circ T$. The problem is smoothness of $f$. Firstly, clearly $f$ is $C^\infty$ away the origin (where $T$ is a local diffeomorphism). It is also clear that $f$ vanishes to infinite order at the origin, in the sense that it vanishes faster than any power of $r$. This already implies that $f$ is differentiable at $0$ with vanishing Jacobian. On the other hand, from the above pullback formulas, it can be seen that the first order partials of $f$ are the pushforwards of functions in the same class as $g$ (vanishing to infinite order on $r=0$ and with the same symmetry condition). So, by induction, all the partial derivatives of $f$ exist and vanish at the origin.