I'm trying to proof the following statement:
$\frac{df}{dz}=(\frac{cos\theta}{1} -i\frac{sin\theta}{1})\frac{df}{dr}$
and
$\frac{df}{dz}=-(\frac{sin\theta}{r}+i\frac{cos\theta}{r}) \frac{df}{d\theta}$
for a complex analytical function $f(z)$.
My attempt: taking the Polar form for the Cauchy-Riemann relation, supposing that f has the form $f(z)=u+iv$. Anyways, that got me to nothing, so I tried to verify what implications the statement has:
$r\frac{dr}{dz} = (rcos\theta - i rsin\theta) \frac{dz}{dr}$ and then:
$\to r\frac{dr}{dz}=(Re(z)-iIm(z))\frac{dz}{dr}$
but I'm not sure if this is right, and is not the most elegant way to construct a proof (then going backwards).