What are the Second Order Cauchy-Riemann equations in terms of polar co-ordinates?

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For a complex function $f(x + i y) = u(x,y) + i v(x,y)$, in Cartesian co-ordinates the Cauchy-Riemann equations are

$u_x = v_y$ and $u_y = -v_x$,

which can be used to find the second order Cauchy-Riemann equations:

$u_{xx} = -u_{yy}$ and $v_{xx} = -v_{yy}$.

In polar co-ordinates, the Cauchy-Riemann equations are

$u_r = \frac{1}{r}v_{\theta}$ and $v_r = -\frac{1}{r}u_{\theta}$.

In an attempt to find the second order Cauchy-Riemann equations for polar co-ordinates, I came out with two separate answers depending on whether I differentiated the polar Cauchy-Riemann equations with respect to $r$ or $\theta$. With respect to $\theta$, I attained:

$u_{rr} = -\frac{1}{r^2}u_{\theta \theta}$ and $v_{rr} = -\frac{1}{r^2}v_{\theta \theta}$,

but when I differentiate with respect to $r$, I get

$u_{rr} = -\frac{1}{r^2}v_{\theta} - \frac{1}{r^2}u_{\theta \theta}$ and $v_{rr} = \frac{1}{r^2}u_{\theta} -\frac{1}{r^2}v_{\theta \theta}$,

due to using the product rule.

Are either of these versions correct, and why does one, if not both, of them not give the correct answer?

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Differentiating $$ u_r = v_{\theta}/r $$ with respect to $r$ and $\theta$ gives $$ u_{rr} = -\frac{1}{r^2}v_{\theta} + \frac{1}{r}v_{\theta r}, \qquad u_{r\theta} =\frac{1}{r}v_{\theta\theta}, $$ and doing the same to $v_r = -u_{\theta}/r$ gives $$ v_{rr} = \frac{1}{r^2}u_{\theta}-\frac{1}{r}u_{\theta r}, \qquad v_{r\theta} = -\frac{1}{r}u_{\theta\theta}. $$ You have to use all four of these, since you have eight second-order derivatives possible: $u_{rr},u_{r\theta},u_{\theta r},u_{\theta\theta}$ and the same for $v$, and you have two other equations: provided all the second derivatives are continuous, $u_{r\theta}=u_{\theta r}$ and $v_{r\theta} = v_{\theta r}$. Combining, we find $$ u_{rr} = \frac{1}{r^2}(-v_{\theta}-u_{\theta\theta}), \qquad v_{rr} = \frac{1}{r^2}(u_{\theta}-v_{\theta\theta}). $$

We can't disentangle the first derivative terms, essentially because what you are dealing with here is the vector Laplacian, which has to mix components in curvilinear coordinates such as polar coordinates, since the basis vectors on which the derivatives are based change from point to point.