Cauchy-Riemann equations for $z=x+iy$ and $f(z)=R(x,y)e^{i\theta(x,y)}$

Question

I know the following forms of the Cauchy-Riemann equations (please correct me if I made a mistake):

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$, $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$, where $z=x+iy$, $f(z)=u(x,y)+iv(x,y)$,

$r\frac{\partial u}{\partial r} = \frac{\partial v}{\partial \theta}$, $\frac{\partial u}{\partial \theta} = -r\frac{\partial v}{\partial r}$, where $z=re^{i\theta}$, $f(z)=u(r,\theta)+iv(r,\theta)$,

$\frac{\partial \ln R}{\partial \ln r} = \frac{\partial \varphi}{\partial \theta}$, $\frac{\partial \ln R}{\partial \theta} = -\frac{\partial \varphi}{\partial \ln r}$, where $z=re^{i\theta}$, $f(z)=R(r,\theta)e^{i\varphi(r,\theta)}$.

Can someone show me the fourth form, i.e. with $z=x+iy$ and $f(z)=R(x,y)e^{i\varphi(x,y)}$, preferably with a proof if it is not too bothersome?

EDIT:

I have managed to find the fourth form, but not from a very reliable source.

$\frac{\partial R}{\partial x} = R\frac{\partial \theta}{\partial y}, \frac{\partial R}{\partial y} = -R\frac{\partial \theta}{\partial x}$, where $z=x+iy$, $f(z)=R(x,y)e^{i\theta(x,y)}$

Can someone please verify that this is correct?

Answer 1

It is actually a good exercise to deduce all the variants from the classical CR equations:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$, $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$.

So for example $u = Rcos(\theta)$, so $\frac{\partial u}{\partial x} = \frac{\partial R}{\partial x}cos(\theta) - Rsin(\theta)\frac{\partial \theta}{\partial x}$ and all the others, so using the classical CR and bringing terms together you get:

$(\frac{\partial R}{\partial x} - R\frac{\partial \theta}{\partial y})cos(\theta)$ = $(\frac{\partial R}{\partial y} + R\frac{\partial \theta}{\partial x})sin(\theta)$, $(\frac{\partial R}{\partial x} - R\frac{\partial \theta}{\partial y})sin(\theta)$ = -$(\frac{\partial R}{\partial y} + R\frac{\partial \theta}{\partial x})cos(\theta)$,

so you conclude with indeed the relations $\frac{\partial R}{\partial x} = R\frac{\partial \theta}{\partial y}, \frac{\partial R}{\partial y} = -R\frac{\partial \theta}{\partial x}$