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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).

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  • $\begingroup$ This is an interesting exercise in partial derivatives and cauchy riemann - one way to do it is to use the matrix for the partial derivatives of $z,\bar z$ with respect to $r,\theta$ to get by inversion the partial derivatives of $r,\theta$ w.r. $z, \bar z$. Then apply the chain rule for both $f_z, f_{\bar z}$ and use that the latter is zero for analytic $f$ to get the polar cauchy riemann and plug that back in $f_z$ to get your result; try and do it as it's a really useful computation; note that for example $z_r=e^{i\theta}, \bar z_{r}=e^{-i\theta}$ etc $\endgroup$ – Conrad Oct 14 '19 at 0:31
  • $\begingroup$ Thanks you very much @Conrad, I'll try to compute it using that method! $\endgroup$ – holahola Oct 14 '19 at 0:35

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