Computing some differential forms using complex coordinates I was computing some things in the Poincaré disk $\mathbb{H}^2$ in complex coordinates and then I tried to show that $\sigma_r(z) = \frac{r^2}{z}$ is an isometry. However $d\sigma_r = \frac{-r^2}{z^2}dz$ and $g = \frac{4 dz\otimes d\overline{z}}{ (1 - z\overline{z})^2} $, then $d\sigma_r(\partial z) = \frac{-r^2}{z^2}$, therefore $g(d\sigma_r(\partial z), d\sigma_r(\partial z)) = ?$ I do not know how to compute it since there is no $\partial z$ or $\partial \overline{z}$ in $d\sigma_r(\partial z)$. In real coordinates it appears $\partial x$ and $\partial y$ however when I try using complex coordinates, $C^{\infty}(\mathbb{H}^2)$ changes.
Thanks in advance.

$d\sigma_r = \frac{-r^2}{z^2}dz$ and $g = \frac{4 dz\otimes d\overline{z}}{ (1 - z\overline{z})^2} $, then $d\sigma_r(\partial z) = \frac{-r^2}{z^2} \partial z$, and not $d\sigma_r(\partial z) = \frac{-r^2}{z^2}$, therefore $g(d\sigma_r(\partial z), d\sigma_r(\partial z)) = g (\partial z, \partial z)$. The problem is that $\partial z$ is NOT dual to $dz$ in the sense that $dz (\partial z) \neq 1$, since $dz = 1$ the differential of the identity and $\partial z$ is just an abstract stuff living the cotangent bundle (as a section). 
