## Polar version of the Cauchy Riemann equations? [closed]

Not every function $f(x,y) = u(x,y) + iv(x,y)$ is holomorphic. We need $\frac{\partial f}{\partial \overline{z}} = 0$. One might check $\frac{\partial f}{\partial \overline{z}} = \frac{\partial }{\partial x} - i \frac{\partial }{\partial y}$. Then use the product rule

$$\left( \frac{\partial }{\partial x} - i \frac{\partial }{\partial y}\right)( u+ iv) = \left( \frac{\partial u }{\partial x} + \frac{\partial v}{\partial y} \right) + i\left( \frac{\partial v}{\partial x} - \frac{\partial u }{\partial y} \right) = 0$$

If we had used a polar representation $f(x,y) = R(x,y) e^{i\theta(x,y)}$, how would the Cauchy-Riemann equations relate $R , \theta$ ?

If I had the real part of $f$ on the real line $u(x,0)$, to what extent is it possible to reconstruct $u(x,y),v(x,y)$? Actually, I am hoping to reconstruct $R(x,y)$ from $\theta(x,y)$.

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$f = r e^{i \theta}$ is analytic if and only if $\log(f) = \log r + i \theta$ is analytic. Just use the Cauchy-Riemman equations on the latter. – David Speyer May 23 2012 at 12:17
Firstly, your statement of the Cauchy Riemann equation is wrong: $\partial/\partial \bar{z} = \frac12(\partial_x + i \partial_y)$ and so you have the wrong signs. Secondly, for any harmonic function $u$ that agrees with the given $u(x,0)$ on the real line, $u+y$ is again harmonic and agrees. And similarly the harmonic conjugate $v$ is determined only up to constant. So you are definitely not going to get unique solutions. – Willie Wong May 23 2012 at 12:26