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If I have the function $\phi (x)= \frac{-z+i} {-iz+1} z \in \mathbb{C} $
from D to H where $ D = { z \in \mathbb{C} | |z| <1} $ and H is the upper half plane
It's not that hard to see that $\phi^{-1} (x)= \frac{z-i} {iz-1} $
g is given as the standard metric of the upper half plane
how can i show that: $(\phi^* g)[id](z) = \frac 4 {(1-|z|^2)^2} id_{[\mathbb{R}^2]} $
The metric on the upper half-plane is given by
$$g_{|w} = \frac{\left|dw\right|^2}{\mathrm{Im}(w)^2}$$
To find the expression of the the pull-back of this metric by $\phi$ at some point $z$, just do $w = \phi(z)$ in the previous expression. You find $$(\phi^*g)_{|z} = \frac{\left|d\phi(z)\right|^2}{\mathrm{Im}(\phi(z))^2}$$ with $$d\phi(z) = \phi'(z) dz = {2 dz \over (z+i)^2}$$ and $$\mathrm{Im}(\phi(z)) = {1 - \left|z\right|^2 \over \left|z+i\right|^2}$$ you indeed get $$(\phi^*g)_{|z} = \frac{4 \left|dz\right|^2}{( 1 - \left|z\right|^2)^2}$$
