The standard norm also called Euclidean norm of $ v=(x,y,z)\in\mathbb R^3$ is $$\|v\|_E=\sqrt{x^2+y^2+z^2}$$
Then a stereographic projection is given by
$$\begin{align} x=\xi_0^2-\xi_1^2,\\ y=i(\xi_0^2+\xi_1^2), \\ z=-2\xi_0\xi_1, \end{align}$$ where $\xi_0,\xi_1\in\mathbb C$. One can easily see that
$$x^2=\xi_0^4+\xi_1^4-2\xi_0^2\xi_1^2, \\ y^2=-\xi_0^4-\xi_1^4-2\xi_0^2\xi_1^2, \\ z^2=4\xi_0^2\xi_1^2$$
and hence $\|v\|_E=x^2+y^2+z^2\equiv0$ (!)
What's more, by regarding $(x,y,z)\in\mathbb R^3\subset \mathbb C^3$ and applying Hermitian norm
$$\|v\|_H=xx^\ast+yy^\ast+zz^\ast$$
we have
$$ xx^\ast=|\xi_0|^4+|\xi_1|^4-(\xi_0^2\xi_1^{*2}+\xi_1^2\xi_0^{*2}), yy^\ast=|\xi_0|^4+|\xi_1|^4+(\xi_0^2\xi_1^{*2}+\xi_1^2\xi_0^{*2}), zz^\ast=4|\xi_0|^2|\xi_1|^2 $$
$$\|v\|_H=(|\xi_0|^2+|\xi_1|^2)^2$$
which may take any real value including zero.
However, since we have let $x,y,z\in\mathbb R$ and there must be $x=x^\ast$ etc. A contradiction appeared when we applying Hermitian and Euclidean norm on that transformed by stereographic projection.
Why such a difference appears? Is there any mathematical concepts and tools can explain this contradiction?
Supplement about the relation between $(\xi_0,\xi_1)$ and $(x,y,z)$ mentioned in comment.
Let $\xi_0=\alpha+i\beta,\xi_1=\gamma+i\delta,\alpha,\beta,\gamma,\delta\in\mathbb R$. Substitute them into $x,y,z\in\mathbb R$ and uses the condition that a real number have zero imaginary part, which made us derive 3 equations, remaining one degree of freedom. $ \alpha\beta=\gamma\delta,\alpha\gamma=\beta\delta, \alpha^2+\gamma^2=\beta^2+\delta^2$ Moreover, we have $$\alpha=\delta,\beta=\gamma$$ or $$\alpha=-\delta,\beta=-\gamma$$
So, from this could we conclude that any $(x,y,z)\in\mathbb R^3$ corresponds to two $(\xi_0,\xi_1)\in \mathbb C^2$??