Riemann sphere, metric derivation-Completed I have been calculated Riemann sphere, but i got stuck with calculating its metric. 
Consider complex plane $\mathbf{C}$ and its point $\zeta=\xi+i\eta$. And consider a point in $S^2 / (0,0,1)$ which $(x_1, x_2, x_3)\neq (0,0,1)$.
Now consider a line which connect two points, North pole $(0,0,1)$ and point $(x_1, x_2, x_3)$, 
\begin{align}
t(x_1, x_2, x_3) + (1-t)(0,0,1) = (t x_1, t x_2, t(x_3-1)+1)
\end{align}
and the intersection with plane $\mathbf{C}$ to be $(\xi, \eta, 0)$. At the intersection point $t=\frac{1}{1-x_3}$ we have
\begin{align}
\xi=\frac{x_1}{1-x_3}, \quad \eta =\frac{x_2}{1-x_3}
\end{align}
Note from $x_1^2 +x_2^2 +x_3^2=1$, I found that
\begin{align}
x_1 =\frac{2\xi}{||\zeta||^2+1}, \quad x_2 =\frac{2\eta}{||\zeta||^2+1}, \quad x_3 =\frac{||\zeta||^2-1}{||\zeta||^2+1}
\end{align}

Here are the problems. 
I try to obtain following expression in textbook, but my calculation does not fit with this.
Also i got trouble for $ds^2$, Plug $dx_1$ to $dx_3$ below, i could not obtain the last equation. Is the textbook wrong? 
\begin{align}
&dx_1 = \frac{2(1-\xi^2+\eta^2)d\xi -4\xi\eta d\eta}{(1+||\zeta||^2)^2} \\
&dx_2 = \frac{2(1+\xi^2-\eta^2)d\eta -4\xi\eta d\xi}{(1+||\zeta||^2)^2} \\
& dx_3 = \frac{4\xi d\xi +4\eta d\eta}{(1+||\zeta||^2)^2}
\end{align}
\begin{align}
ds^2 = dx_1^2+dx_2^2 +dx_3^2 =\frac{4(d\xi^2 + d\eta^2)}{(1+\xi^2+\eta^2)^2}
\end{align}

It was trivial calculations... 
Just using complex variable defined above $\zeta$, properly, i have been finished my calculations.  I will post it next time. 
 A: Followings are my solutions
\begin{align}
x_1 =\frac{\zeta-\bar{\zeta}}{||\zeta||^2+1}, \quad x_2 =\frac{-i(\zeta-\bar{\zeta})}{||\zeta||^2+1}, \quad x_3 =\frac{||\zeta||^2-1}{||\zeta||^2+1}
\end{align}
Then Note $||\zeta||^2 =\zeta \bar{\zeta}$ and $\zeta=\xi+i\eta$.
\begin{align}
dx_1  &= \frac{(d\zeta +d\bar{\zeta})(\zeta\bar{\zeta}+1)-(\zeta+\bar{\zeta})(\bar{\zeta}d\zeta + \zeta d\bar{\zeta})}{(1+||\zeta||^2)^2}\\
&=\frac{d\zeta +d\bar{\zeta} -\bar{\zeta}\bar{\zeta}d\zeta -\zeta\zeta d\bar{\zeta}}{(1+||\zeta||^2)^2}=\frac{2(1-\xi^2+\eta^2)d\xi -4\xi\eta d\eta}{(1+||\zeta||^2)^2}
\end{align}
\begin{align}
dx_2& = \frac{-i(d\zeta-d\bar{\zeta})(\zeta\bar{\zeta}+1)+i(\bar{\zeta}d\zeta + \zeta d\bar{\zeta})(\zeta -\bar{\zeta})}{(1+||\zeta||^2)^2}\\
&=\frac{-id\bar{\zeta}+id\bar{\zeta}-i\bar{\zeta}\bar{\zeta}d\zeta + i \zeta\zeta d\bar{\zeta}}{(1+||\zeta||^2)^2}= \frac{2(1+\xi^2-\eta^2)d\eta -4\xi\eta d\xi}{(1+||\zeta||^2)^2}
\end{align}
\begin{align}
dx_3 = \frac{(\zeta d\bar{\zeta} +\bar{\zeta} d\zeta)(\zeta\bar{\zeta}+1)-(\zeta d\bar{\zeta} +\bar{\zeta} d\zeta)(\zeta\bar{\zeta}-1)}{(1+||\zeta||^2)^2}=\frac{2(\zeta d\bar{\zeta}+\bar{\zeta}d\zeta)}{(1+||\zeta||^2)^2}= \frac{4\xi d\xi +4\eta d\eta}{(1+||\zeta||^2)^2}
\end{align}
Summarize we have
\begin{align}
&dx_1 = \frac{2(1-\xi^2+\eta^2)d\xi -4\xi\eta d\eta}{(1+||\zeta||^2)^2} \\
&
dx_2 = \frac{2(1+\xi^2-\eta^2)d\eta -4\xi\eta d\xi}{(1+||\zeta||^2)^2} \\
& dx_3 = \frac{4\xi d\xi +4\eta d\eta}{(1+||\zeta||^2)^2}
\end{align}
So one computes striaghtforward calculations as
\begin{align}
ds^2 = dx_1^2+dx_2^2 +dx_3^2 =\frac{4(1+\xi^2 +\eta^2)^2(d\xi^2 + d\eta^2)}{(1+||\zeta||^2)^4}=\frac{4(d\xi^2 + d\eta^2)}{(1+\xi^2+\eta^2)^2}
\end{align}
