Consider the following theorem.
Theorem: Suppose $u(r, \theta)$ and $v(r, \theta)$ (which we shall denote by $u$ and $v$ here afterwards) are real valued functions on $(0, \infty) \times (-\pi, \pi]$. Suppose $u$ and $v$. Suppose all the partial derivatives of $u$ and $v$ (with respect to $r$ and $\theta$ exist and are continuous at $z_0$. Also suppose that following hold at $z_0= r_0e^{i\theta_0}\neq 0$: $$ r \frac{\partial u}{\partial r} = \frac{\partial v}{\partial \theta} \quad \text{and} \quad r\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial \theta}.$$ Then $f = u+iv$ is analytic at $z_0$.
$\textbf{Question I had got:}$ Is the above theorem correct? Or do we need any additional condition or modification?
$\textbf{My attempt to answer the above question:}$ I feel that an additional condition that $\theta_0 \neq \pi$ is required as $\frac{\partial \theta}{\partial y}$ does not exist for any point on negative real axis. So assuming that $\theta_0\neq \pi$, if I write $$f(z) = p(x,y)+iv(x,y),$$ where $z=x+iy$, then the usual chain rule implies the following: $$\frac{\partial p}{\partial x} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial y}$$
$$\frac{\partial p}{\partial y} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial y}$$
$$\frac{\partial q}{\partial x} = \frac{\partial v}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial v}{\partial \theta}\frac{\partial \theta}{\partial x}$$
$$\frac{\partial q}{\partial y} = \frac{\partial v}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial v}{\partial \theta}\frac{\partial \theta}{\partial y}$$
By our assumption that partial derivatives of $u$ and $v$ with respect to $r$ and $\theta$ are continuous, we conclude that each of $\frac{\partial p}{\partial x}, \frac{\partial p}{\partial y}, \frac{\partial q}{\partial x}$ and $\frac{\partial q}{\partial y}$ is continuous [We have used the facts that partial derivatives of $r$ and $\theta$ with respect to both $x$ and $y$ are also continuous whenever $\theta\neq \pi$]. Further, we can manipulate the above mentioned equations to arrive at
$$\frac{\partial p}{\partial x} = \frac{\partial q}{\partial y} \quad \text{and} \quad \frac{\partial p}{\partial y} = -\frac{\partial q}{\partial x}.$$
Thus $f$ must be differentiable at $z_0 = r_0e^{i\theta_0}$ with $\theta_0\neq \pi$.
$\textbf{Clarification required.} $
Am I correct that $\theta_0\neq \pi$ for the Theorem to be true? Or is that condition not needed at all?
Am I correct when I say that $\frac{\partial \theta}{\partial y}$ does not exist for a point on negative real axis (whenever my domain of $\theta$ is $(-\pi, \pi]$ interval)?
I was unable to search for an example wherein $u(r,\theta)$ and $v(r, \theta)$ satisfy CR equations in polar form and the partial derivatives of $u$ and $v$ are all continuous at a point $z_0$ on negative real axis, yet the function $f(z) = u + iv$ is not differentiable at $z_0$. Does such thing exist?