Show that satisfaction of Cauchy-Riemann Equations in polar coordinates implies analyticity Suppose that $U(r,\theta),V(r, \theta)$ are continuously differentiable functions on some polar rectangle $R = \{(r, \theta) \colon r \in (a,b), \theta \in (\theta_1, \theta_2) \} \subseteq \mathbb{R}^2.$ Furthermore, assume that $U$ and $V$ satisfy the polar Cauchy-Riemann equations in $R$:
$$rU_r = V_\theta, U_\theta = -rV_r.$$
If we now view $R$ as a subset of $\mathbb{C}$ rather than $\mathbb{R}^2$, we can define the function $f : R \to \mathbb{C}$ by $f(re^{i\theta}) = U(r, \theta) + iV(r,\theta).$ Prove that $f$ is analytic on $R$.
I am linking this problem to a previous post: Proof of Cauchy Riemann Equations in Polar Coordinates. I believe I am asking a similar question. However, to my best knowledge, the answers to the linked post actually establish the converse of my statement above. That is, they show that analyticity of $f$ implies that these polar Cauchy-Riemann equations are satisfied.      
Here's what I have so far: I do know that a function $f(x + iy) = U(x,y) + iV(x,y)$ is analytic when its real and imaginary parts are continuously differentiable and satisfy the rectangular Cauchy-Riemann equations $U_x = V_y, U_y = -V_x$. The proof I have seen of this fact comes from Stein, and the key to the argument is to expand $U$ and $V$ via Taylor's formula for $C^1$ functions. That is, for a point $(x_0, y_0) \in \mathbb{R}^2$, we can write:
$$U(x,y) = U(x_0,y_0) + U_x(x_0,y_0)(x - x_0) + U_y(x_0, y_0)(y - y_0) + R(x,y),$$ 
and a similar formula for $V(x,y)$. Here, $R(x,y)$ is a remainder term with $\frac{R(x,y)}{|(x,y) - (x_0,y_0)|} \to 0$ as $(x,y) \to (x_0,y_0)$. I'm wondering if there is some way I can adapt this proof from Stein to the polar case?
Hints are solutions are greatly appreciated.
 A: Here's the Taylor expansion approach. First, we need a linear algebra fact: 

Any $\mathbb R$-linear map $T:\mathbb C\to\mathbb C$ can be written as $Tz=\alpha z+\beta\bar z$ for some (unique) $\alpha,\beta\in \mathbb C$. 

Uniqueness should be clear, and existence follows by first writing $T(x+iy)=\gamma x+\delta y$ with complex $\gamma,\delta$ and then replacing $x=(z+\bar z)/2$, $y=(z-\bar z)/(2i)$. 
This linear algebra fact is worth remembering, as it simplifies various computations in complex analysis. 
Back to the problem. The first-order real Taylor expansion of $f$ at a point $z\in\mathbb C$ takes the form 
$$
f(z+h)=f(z)+\alpha h+\beta \bar h+o(|h|) 
$$
and we want to show that $\beta=0$ here. The equations you are given say that $r f_r + i f_\theta = 0$.
Write $h=r e^{i\theta}$, and take the derivatives of $\alpha h+\beta \bar h$: 
$$
\begin{align}
f_r &= \alpha e^{i\theta} + \beta e^{-i\theta}   \\
f_\theta & = \alpha  ir e^{i\theta} - \beta i r e^{-i\theta} \\
rf_r+f_\theta & =  2\beta r e^{-i\theta} 
\end{align}
$$ 
Thus $\beta=0$ as desired. 
