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This is a practice exercise from a geometry textbook by P. Wilson. Suppose we have a Riemannian metric of the form $|dz|^2/h(r)^2$ on an open disc of radius $\delta>0$ centered on the origin in $\mathbb C$, where $h(r)>0 $ for all $r<\delta$. I want to show that the gaussian curvature is K=hh'/r+hh"-(h')^2. Could someone please kindly help me out? I am thinking of using $K=-\sqrt{G}_{uu}/\sqrt{G}$ for a first fundamental form being $du^2+G(u,v)dv^2$. I can change variables to get the first fundamental form into the form $da^2+db^2$ but it isn't helpful at all here... Please help!

Perhaps a more fundamental question: What is $|dz|^2$? What is it in terms of $r$, say?

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    I've never looked at the textbook you are using, but near the definition of Gaussian curvature there should be a formula for it in terms of $E$, $F$, $G$ and their first and second derivatives. In your case you have $F = 0$ and $E = G$ so the formula should greatly simplify. I wouldn't worry about trying to change to a different coordinate system.2012-03-12

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The given metric is a so-called conformal metric. This means it is of the form $ds^2\ = g(z)\ |dz|^2$ where $|dz|^2=dx^2+dy^2$ and $g(\cdot)$ is a scalar function of the complex variable $z=x+iy$. The Riemannian structure so defined on the $z$-plane is conformally equivalent to the usual euclidean metric: The angles between tangent vectors are the same, but lengths are different. The Gaussian curvature corresponding to such a $g$ can be computed by going through the theorema egregium (Christoffel symbols, etc.). Since our parametrization is orthogonal a considerable simplification results, see here.