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?

With only partial differentiation and algebra. From what we already know of complex functions:
$ƒ(z) = u(r,\theta) + iv(r,\theta)$
$z = re^{i\theta} = r\cos\theta +ir\sin\theta = x + iy$
$x(r,\theta)= r\cos\theta,\quad y(r,\theta) = r\sin\theta $
Apply partial derivative and chain rule:
$\frac{∂f}{∂r} = \frac{df}{dz}\frac{∂z}{∂r} = \frac{df}{dz}e^{i\theta} = \frac{df}{dz}\frac1rz$
$\frac{∂f}{∂\theta} = \frac{df}{dz}\frac{∂z}{∂\theta} = \frac{df}{dz}ire^{i\theta} = \frac{df}{dz}iz$
which are also equal to:
$\frac{∂f}{∂r} = \frac{∂u}{∂r} + i\frac{∂v}{∂r}$
$\frac{∂f}{∂\theta} = \frac{∂u}{∂\theta} + i\frac{∂v}{∂\theta}$
then set the equality of $r\frac{∂f}{∂r} = \frac1i\frac{∂f}{∂\theta}$:
$r(\frac{∂u}{∂r} + i\frac{∂v}{∂r}) = -\frac1i(\frac{∂u}{∂\theta} + i\frac{∂v}{∂\theta}) $
and separate the real and imaginary:
$\frac{∂u}{∂r} = \frac1r\frac{∂v}{∂\theta}, \quad \frac{∂v}{∂r} = -\frac1r\frac{∂u}{∂\theta}$
yey :3
