Proof of Cauchy Riemann Equations in Polar Coordinates How would one go about showing the polar version of the Cauchy Riemann Equations are sufficient to get differentiability of a complex valued function which has continuous partial derivatives? 
I haven't found any proof of this online.
One of my ideas was writing out $r$ and $\theta$ in terms of $x$ and $y$, then taking the partial derivatives with respect to $x$ and $y$ and showing the Cauchy Riemann equations in the Cartesian coordinate system are satisfied. A problem with this approach is that derivatives get messy.
What are some other ways to do it?

We can derive using purely polar coordinates. Start with
\begin{align}
z(r,\theta) &=r\,\mathrm{e}^{\mathrm{i}\theta} \\
f(z) &= u(r,\theta) + \mathrm{i} v(r,\theta)
\end{align}
We define $f'(z)$ using the limit
$$ f'(z) = \lim_{z\to 0} \frac{\Delta f}{\Delta z} $$
where
\begin{align}
\Delta f &= \Delta u + \mathrm{i} \Delta v \\
\Delta z &= z(r+\Delta r,\theta + \Delta\theta) - z(r,\theta) \\
&= (r+\Delta r)\,\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)} - r \,\mathrm{e}^{\mathrm{i}\theta}
\end{align}
Next, we try to first approach from $\Delta\theta\to 0$,
$$ \Delta z = (r+\Delta r)\,\mathrm{e}^{\mathrm{i}\theta} - r \,\mathrm{e}^{\mathrm{i}\theta} = \Delta r\,\mathrm{e}^{\mathrm{i}\theta} $$
Therefore when we take the limit,
$$ f'(z) = \lim_{\Delta r\to 0} \frac{\Delta u + \mathrm{i} \Delta v}{\Delta r \,\mathrm{e}^{i\theta}} = \frac{1}{\mathrm{e}^{\mathrm{i}\theta}}(u_r + \mathrm{i} v_r) \label{a}\tag{1}$$
On the other hand, approaching from $\Delta r\to 0$ first yields
$$\Delta z = r\,\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)} - r \,\mathrm{e}^{\mathrm{i}\theta}$$
Using the derivative
$$\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)}-\mathrm{e}^{\mathrm{i}\theta} = \frac{\mathrm{d}\mathrm{e}^{\mathrm{i}\theta}}{\mathrm{d}\theta}\Delta\theta = \mathrm{i}\mathrm{e}^{\mathrm{i}\theta}\Delta\theta$$
we get
$$\Delta z = \mathrm{i}r\,\mathrm{e}^{\mathrm{i}\theta}\Delta\theta$$
Therefore
$$f'(z) = \lim_{\Delta\theta\to 0} \frac{\Delta u + \mathrm{i} \Delta v}{\mathrm{i}r\,\mathrm{e}^{\mathrm{i}\theta}\Delta\theta} = \frac{1}{\mathrm{i}r\,\mathrm{e}^{\mathrm{i}\theta}}(u_\theta + \mathrm{i}v_\theta) \label{b}\tag{2}$$
Finally, comparing the real and imaginary parts of ($\ref{a}$) and ($\ref{b}$) gives us what we want:
$$u_r=\frac{1}{r}v_\theta \quad,\quad v_r = -\frac{1}{r}u_\theta$$

Reference: Kwong-tin Tang, Mathematical Methods for Engineers and Scientists 1 (Springer, 2007)
