Gaussian curvature of the graphs of the real and imaginary parts of an analytic function.

Consider an analytic function $f$ of complex variable $z$: $ f(x + i y) = f(z) $.

At any point $z$ in the domain, are the Gaussian curvatures of the surfaces $ (x,y, \operatorname{Re} f(z) )$ and $(x,y, \operatorname{Im} f(z))$ the same?

The motivation for asking this question is to recognize intrinsic character of $f$ dependence.

For $f(x+i\, y) = u(x,y) + i\, v(x,y)$ the graphs of both functions $u$,$v$ have the same curvature at the points over $(x,y)$

$$ k(x,y) = -\frac{u_{xy}^2 + v_{xy}^2}{ (1 + u_x^2 + u_y^2)^2}$$

Use Cauchy-Riemann relations and the formula for the curvature of the graph of a function $\phi(x,y)$

$$k(x,y) = \frac{ \left| \begin{array}{cc} \phi_{xx} & \phi_{xy}\\ \phi_{yx} &\phi_{yy} \end{array} \right|}{(1 + \phi_x^2 + \phi_y^2)^2}$$

Note that the metric on the surface $\{(x,y,u(x,y)\}$ is

$$ds^2 = (1+ u_x^2)\,dx^2 + 2\,u_x\cdot u_y\, dx\, dy + (1+u_y^2)\, dy^2$$

so the map $(x,y,u(x,y)) \mapsto (x,y,v(x,y))$ is $\it{not}$ an isometry in general (since $\ v_x v_y = - u_x u_y$).