Stereographic projection is conformal --- from the line element I'm looking over some fairly basic stuff on complex methods and the book I'm using takes the formula for the stereographic projection: 
$$z = \cot(\beta/2)e^{i\phi} $$
as well as the line element on the sphere:
$$ ds^2 = d\beta^2 + \sin^2 \beta d\phi^2 $$
(where $\beta$ is the polar angle and $\phi$ the azimuthal angle) and uses this to derive
$$ \frac{4 dz d \bar{z}}{(1 + z\bar{z})^2} = d\beta^2 + \sin^2 \beta d\phi^2 \,.$$
It then says "from this we deduce that the stereographic projection is conformal". I don't understand this step. At the moment the only mention made of conformal maps is that holomorphic maps from C to C are conformal.
 A: A map is conformal if it induces a rescaling of the metric, ie. for $ f: (M,g_M) \rightarrow (N,g_N) $, $ f^* g_N = \phi \; g_M $ for some $ \phi \in C^\infty(M) $.
If above is not helpful, then you can picture it in the following way: pick a point on the sphere $ (\beta, \phi ) $ on the sphere, and take two tiny vectors tangent to the surface at this point, say $ v_1 = (d \beta_1, d \phi_1 )$ and $ v_2 ( d \beta_2, d \phi_2 ) $. The line element $ ds^2 = d\beta^2 + \sin^2 \beta d\phi^2 $ tells you how to take the dot product of the vectors, eg. $ v_1 \cdot v_2 = (d \beta_1 d \beta_2 + \sin(\beta)^2 d \phi_1 d \phi_2) $, and also the angle between $ v_1 $ and $ v_2 $, eg. $ \frac{v_1 \cdot v_2 }{ |v_1| |v_2| } $.
The stereographic projection map takes the point $ (\beta, \phi) $ to the point $ (z, \bar{z}) = (\cot(\beta/2)e^{i\phi}, \cot(\beta/2)e^{-i\phi}) $, and sends the tiny tangent vectors $ v_1 $ and $ v_2 $ to tiny tangent vectors starting at $ (z, \bar{z}) $, say $ v_1' = (dz_1, d\bar{z}_1) $ and $ v_2' $. In the complex plane, the line element $ dz \; d \bar{z} $ again tells you how to take inner product, eg $ v_1' \cdot v_2' =  dz_1 d\bar{z}_2+dz_2 d\bar{z}_1 $. (You can check that if you write $ z = x + i y $, then it gives you $ x_1 x_2 + y_1 y_2 $, or times some factor of 2).
The calculation he does shows that the the dot product $ v_1' \cdot v_2'  = (1 + z \bar{z})^2 \; v_1 \cdot v_2 $. So in particular, the angle between $ v_1 $ and $ v_2 $, is the same as $ v_1' $ and $ v_2' $.
