Find the points on the sphere xi^2+eta^2+zeta^2=1 whose…

Find the points on the sphere $\xi^2+\eta^2+\zeta^2=1$ whose stereographic projections (onto the plane $\zeta=-1$) are: $\ 1^\circ\ z=0;\quad2^\circ\ z=i;\quad3^\circ\ z=-1;\quad4^\circ\ z=e^{-\pi i/4}$.

Step 1 · Derive the inverse formulas

From $z=\dfrac{2(\xi+i\eta)}{1-\zeta}$: $\ |z|^2=\dfrac{4(\xi^2+\eta^2)}{(1-\zeta)^2}=\dfrac{4(1-\zeta^2)}{(1-\zeta)^2}=\dfrac{4(1+\zeta)}{1-\zeta}$. Solving for $\zeta$ (write $R=|z|^2$):

$$ R(1-\zeta)=4(1+\zeta) \ \Longrightarrow\ \zeta=\frac{R-4}{R+4} $$

and then $1-\zeta=\dfrac8{R+4}$, so from $\xi+i\eta=\dfrac{z(1-\zeta)}2$:

$$ \boxed{\xi=\frac{4\,\text{Re}(z)}{|z|^2+4}, \qquad \eta=\frac{4\,\text{Im}(z)}{|z|^2+4}, \qquad \zeta=\frac{|z|^2-4}{|z|^2+4}} $$

Case $1^\circ$: $z=0$

$R=0$: $\ \xi=0,\ \eta=0,\ \zeta=\dfrac{-4}4=-1$. $$ \boxed{(0,0,-1)} \quad\text{(south pole)} $$

Case $2^\circ$: $z=i$

$\text{Re}(z)=0,\ \text{Im}(z)=1,\ R=1$: $\ \xi=0,\ \eta=\dfrac45,\ \zeta=\dfrac{1-4}5=-\dfrac35$.

$$ \boxed{\left(0,\tfrac45,-\tfrac35\right)} $$

Check: $0+\frac{16}{25}+\frac9{25}=1$ ✓

Case $3^\circ$: $z=-1$

$\text{Re}(z)=-1,\ \text{Im}(z)=0,\ R=1$: $\ \xi=-\dfrac45,\ \eta=0,\ \zeta=-\dfrac35$.

$$ \boxed{\left(-\tfrac45,0,-\tfrac35\right)} $$

Check: $\frac{16}{25}+0+\frac9{25}=1$ ✓

Case $4^\circ$: $z=e^{-\pi i/4}=\dfrac{\sqrt2}2-i\dfrac{\sqrt2}2$

$\text{Re}(z)=\dfrac{\sqrt2}2,\ \text{Im}(z)=-\dfrac{\sqrt2}2,\ R=1$: $\ \xi=\dfrac{4\cdot\frac{\sqrt2}2}5=\dfrac{2\sqrt2}5,\ \eta=-\dfrac{2\sqrt2}5,\ \zeta=-\dfrac35$.

$$ \boxed{\left(\tfrac{2\sqrt2}5,-\tfrac{2\sqrt2}5,-\tfrac35\right)} $$

Check: $\frac8{25}+\frac8{25}+\frac9{25}=1$ ✓

Verification & observation

All three points with $|z|=1$ (cases $2^\circ$–$4^\circ$) share the same $\zeta=-\frac35$, consistent with Problem 44's result that $|z|=r$ maps to the parallel at height $\zeta=\dfrac{r^2-4}{r^2+4}$ — here $r=1$ gives $\zeta=\dfrac{1-4}5=-\dfrac35$ exactly.