Transforming the measure in $CP^1$ mapping from Riemann sphere to $\mathbb{C}^2$-plane

Transforming the measure in $CP^1$ mapping from Riemann sphere to
$\mathbb{C}^2$-plane

I would like to know how the measure changes in $CP^1$ mapping from
Riemann sphere (2-sphere) to $\mathbb{C}^2$-plane.
Let a point on the 2-sphere is given by the vector
$\hat{n}(\mathbf{x})=(n^x(\mathbf{x}),n^y(\mathbf{x}),n^z(\mathbf{x}))$
with constraint $|\hat{n}|^2=1$. Now the $CP^1$ mapping is defined as
follows \begin{equation} n^\alpha(\mathbf{x}) = \mathbf{z}^\dagger
(\mathbf{x})\sigma^\alpha \mathbf{z}(\mathbf{x}), \quad \alpha=x,y,z
\end{equation} where
$\mathbf{z}(\mathbf{x})=[z_1(\mathbf{x}),z_2(\mathbf{x})]$ is a two
component spinor, $z_1, z_2\in \mathbb{C}$ are two complex numbers, and
$\sigma$ are Pauli matrices. The $CP^1$ mapping leads \begin{eqnarray} n^x
&=& z_1^\ast z_2 + z_2^\ast z_1= 2\Re \left( z_1^\ast z_2 \right)\\ n^y
&=& -iz_1^\ast z_2 + i z_2^\ast z_1=2\Im \left( z_1^\ast z_2 \right) \\
n^z &=&z_1^\ast z_1 - z_2^\ast z_2 \end{eqnarray} and \begin{equation}
|\hat{n}|= |\mathbf{z}|^2 = z_1^2 + z_2^2=1. \end{equation}
Now my question is how to find measure $dn^xdn^ydn^z$ in terms of complex
numbers $z_1$ and $z_2$ with appropriate normalization constant.