What am I doing wrong?
Let $\omega =f(z)=R(r,\theta)e^{i \,\Theta(r,\theta)} $
I want to find Cauchy Riemann conditions by differentiating in respect to $\theta $ and r and then equalizing the results, i.e.: $$ \frac{\partial}{\partial r}(Re^{i \,\Theta})=\frac{\partial}{\partial (i\theta)}(Re^{i \,\Theta})$$
where $R \,and \,\Theta$ are functions of $r \,and \,\theta$
From the left side of the equation I get (applying the product and chain rule):
$\frac{\partial}{\partial r}(Re^{i \,\Theta})= R ie^{i\Theta}\frac{\partial \Theta}{\partial r}+e^{i\Theta}\frac{\partial R}{\partial r}$
$\frac{\partial}{\partial (i\theta)}(Re^{i \,\Theta})= (-i \cdot i) R e^{i\Theta}\frac{\partial \Theta}{\partial \theta}-i\cdot e^{i\Theta}\frac{\partial R}{\partial \theta} = R e^{i\Theta}\frac{\partial \Theta}{\partial \theta}- i\cdot e^{i\Theta}\frac{\partial R}{\partial \theta}$
If we equalize the real an imaginary parts then we have: $$ \frac{\partial R}{\partial r} = R \frac{\partial}{\partial \theta}$$ and $$ R\frac{\partial \Theta}{\partial r} = -\frac{\partial R}{\partial \theta}$$
However, in my book (Arfken) the Cauchy Riemann relation in polar coordinates are: $$ \frac{\partial R}{\partial r} = \frac{R}{r} \frac{\partial}{\partial \theta}$$ and $$-R\frac{\partial \Theta}{\partial r} = \frac{1}{r}\frac{\partial R}{\partial \theta}$$
Why am I missing a factor of $\frac{1}{r}$?
The derivative in the direction of $\theta$ is actually wrong
The derivative in that direction should be: $$ \frac{\partial R e^{i \Theta}}{\partial (ri d\theta)} = \frac{\partial R e^{i \Theta}}{ri \,\partial \theta}$$
This because the length differential isn't only $\partial \theta$ but $r\partial \theta$
This because the arc is longer when we move the same $d \theta$ in a larger distance (radius) from the origin
So the derivatives that should be equalized are:
$$\frac{\partial R e^{i \Theta}}{ \,\partial r} = \frac{1}{r}\frac{\partial Re^{i\Theta}}{i\partial \theta}$$
This explains the factor $\frac{1}{r}$