So I know that the regular Cauchy Riemann equations are:
$\begin{equation} \frac{\partial u}{\partial x} =\frac{\partial v}{\partial y} \ \ \ \ \ \text{and} \ \ \ \ \frac{\partial u}{\partial y} =-\frac{\partial v}{\partial x} \ \ \ \ \ \ \ \end{equation}$
And to get to the polar form of these equations we can just use the fact that $z=x+iy=r(\cos\theta + i \sin\theta)$
Where x is a function of r and $\theta$ , the same for y, and the equations can be obtained easily in the form:
$\left( \frac{\partial u}{\partial r}\right) = \frac{1}{r} \left( \frac{\partial v}{\partial \theta}\right) \ \ \ \ \ \text{and} \ \ \ \ \left(\frac{\partial v}{\partial r} \right) = \frac{-1}{r} \left( \frac{\partial u}{\partial \theta}\right)$
But my question is, how to obtain these polar form equations by starting with the "complex" form of Cauchy Riemann's equations which are:
$\frac{\partial }{\partial \bar z}f(z) = 0 \ \ \ \ \ \text{and} \ \ \ \ \frac{\partial }{\partial z}f(z) = f'(z)$
Where z and $\bar z$ are complex conjugates
Your first equation in the "complex form" says that $$ \frac{\partial}{\partial \overline{z}} f = \frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)(u+iv) = \frac{1}{2}\left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) + \frac{i}{2}\left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right) = 0, $$ which, after equating the real and imaginary parts with zero, gives you the two Cauchy-Riemann equations in the "regular form" that you know how to work with.