# Associated family of isometric surfaces of a helicoid (or a catenoid)

This question is to understand the local isometry between a helicoid and a catenoid using complex analysis. Given the usual parametrizations, $$x(u,v)=(-\cosh{u}\sin{v},\cosh{u}\cos{v},u)$$ $$y(u,v)=(\sinh{u}\cos{v},\sinh{u}\sin{v},v)$$ let $z=u+iv$ and $Z(u,v)=x(u,v)+iy(u,v)$. I have showed that $Z=(\sin{iz},\cos{iz},z)$. The next step would be to describe how to construct from here a one-parameter family of isometric surfaces between the above helicoid and catenoid.

The expression of $Z$ reminds me of the Cauchy-Riemann equations or harmonic functions and I suppose we should be using that to prove something about the family of surfaces sharing the same gauss curvature (so that they are locally isometric). The idea of interpolating between $x(u,v)$ and $y(u,v)$ makes me think perhaps I could have something like $Z=\cos(t)x(u,v)+\sin(t)y(u,v)$. But that does not incorporate the complex variables. Also, $Z=\cos(t)x(u,v)+i\sin(t)y(u,v)$ does not seem to yield anything. Any suggestions on how to proceed from here?

• math.colostate.edu/~clayton/courses/660/660_4.pdf ... Uses Codazzi Mainardi compatibility directly without complex notation, – Narasimham Feb 24 '16 at 6:41
• Can you elaborate? I feel that the point of the exercise is to use complex variables to express the associated family of surfaces. – user269715 Feb 24 '16 at 6:49

$\newcommand{\Cpx}{\mathbf{C}}$Your instincts are good. :)
You've constructed a map $Z:\Cpx \to \Cpx^{3}$ whose real part is $x$ and whose imaginary part is $y$. A natural way to interpolate is to rotate $\Cpx^{3}$ and take the real part, i.e., to define \begin{align*} Z_{t}(u, v) &= \operatorname{Re} \bigl[e^{-it}Z(u, v)\bigr] \\ &= x(u, v) \cos t + y(u, v) \sin t. \end{align*} It's easy to show directly that the first fundamental form of $Z_{t}$ is independent of $t$, i.e., that the interpolation occurs through local isomatries.