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This cross-ratio convention is the value of $z_3$ under the Möbius transformation sending $z_1,z_2,z_4$ to $1,0,\infty$. It is one minus the convention $(z_1-z_3)(z_2-z_4)/[(z_1-z_2)(z_3-z_4)]$. Exchanging the first two points replaces $r$ by $1-r$. For four points in cyclic order on a circle, this convention gives $r<0$ and $1-r>1$.

Every Möbius transformation preserves the cross-ratio. The identity $M(u)-M(v)=(ad-bc)(u-v)/[(cu+d)(cv+d)]$ makes all extra factors cancel in the ratio; limiting conventions handle points at infinity.

Cross-ratio conventions differ by permutations. Use the following ordering, which gives the signs requested in part (c):Values involving infinity are defined by limits. This is the cross-ratio with second point mapped to zero: the Möbius transformationsends $z_1,z_2,z_4$ to $1,0,\infty$ and sends $z_3$ to $r$.

For $M(z)=(az+b)/(cz+d)$, $ad-bc\ne 0$, the identityshows that all factors cancel in the cross-ratio. This proves Möbius invariance of the cross-ratio, with the cases involving infinity following by limits. Algebraic subtraction also givesThe latter expression is another common ordering for the cross-ratio; keeping the convention explicit prevents an incorrect reciprocal or sign.

Write $z_j=c+R{e}^{i{\theta }_j}$ with ${\theta }_1<{\theta }_2<{\theta }_3<{\theta }_4<{\theta }_1+2\pi$. In the identitythe phase factors cancel in the cross-ratio. In our convention the two numerator sine factors are negative, while the denominator factors have opposite signs. Thus $r<0$, and the exchanged cross-ratio $1-r$ is greater than one. Consequently $|r|+1=1-r=|1-r|$.

Taking absolute values of the two formulas in part (a) givesMultiplying $|r|+1=|1-r|$ by the common positive denominator proves Ptolemy theorem:The product of the diagonals equals the sum of the products of opposite sides.

Use the Riemann sphere convention for evaluating a Möbius transformation at a pole and at infinity. When the three prescribed points are finite, takeIts zero is $w_2$, its pole is $w_1$, and substitution at $w_3$ gives one. The remaining three cases areAll coefficient determinants are nonzero because the points are distinct. There is only one such Möbius transformation: the composite of any two candidate maps, one inverted, fixes $\infty ,0,1$; fixing infinity makes it affine, and fixing zero and one makes it the identity.

For this question use the cross-ratio normalizationIn the finite case this is $(w_3-w_1)(w_4-w_2)/[(w_3-w_2)(w_4-w_1)]$, with the corresponding limits if a point is infinite. It is finite and distinct from zero and one, since $w_4$ is distinct from the other three points and $f$ is a bijection. Ordering conventions for the cross-ratio differ; the definition here is fixed by the specified images, rather than by importing another ordering formula.

A generalized circle means either a circle or a straight line completed by infinity. Let $K$ be the unique generalized circle through the first three points. Its image under $f$ is the unique generalized circle through $\infty ,0,1$, namely $\mathbb{R}\cup \{\infty \}$. Therefore $w_4\in K$ exactly when $f(w_4)$ is real. Conversely a real $f(w_4)$ lies on that extended real line, whose inverse image is $K$. This proves the real cross-ratio criterion for a generalized circle in both directions.

A Möbius transformation of the Riemann sphere is a mapwith its natural values at the pole and at infinity.

For three distinct points $z_1,z_2,z_3$, defineThe usual limiting conventions cover an infinite $z_i$. This Möbius transformation sends $(z_1,z_2,z_3)$ to $(1,0,\infty )$. Defining $T_w$ similarly, the mapsends $z_i$ to $w_i$. If two Möbius transformations do so, their quotient fixes $0,1,\infty$. A Möbius transformation fixing infinity is affine, and fixing zero and one then makes it the identity. This proves uniqueness.

With this convention, the cross-ratio isSubstitution shows that translations, nonzero scalings, and inversion preserve it; since these generate the Möbius group, every Möbius transformation preserves cross-ratios.

Conversely, suppose a bijection $f$ of the Riemann sphere preserves every cross-ratio. Let $m$ be the unique Möbius transformation agreeing with $f$ at three chosen points $z_1,z_2,z_3$. For any other $z$, preservation by $f$ and $m$ givesThe last coordinate in a cross-ratio with three fixed distinct entries is injective, so $f(z)=m(z)$. The equality already holds at the three base points, hence $f=m$ everywhere and $f$ is Möbius.

Finally, the map $z\mapsto a\overline{z}+b$ is constant when $a=0$ and therefore is not Möbius. If $a\ne 0$, it is bijective and fixes infinity. Were it Möbius, it would have the affine form $\alpha z+\beta$. Equality on real $z$ forces $\alpha =a$ and $\beta =b$, whereas equality at $z=i$ would require $ai=-ai$, contradicting $a\ne 0$. Thus

Use an orientation-preserving isometry to send $P$ to $i$ and $\ell$ to the imaginary axis, whose boundary points are $0,\infty$. A geodesic $m$ through $i$ meeting it at angle $\theta$ has boundary pointsIndeed $uv=-1$, so the semicircle with endpoints $u,v$ passes through $i$, and its tangent there makes angle $\theta$ with the vertical.

The cross-ratio is invariant under the normalizing Möbius transformation. With the suitable ordering $(u,v,\infty ,0)$ and the convention used here,

Normalize three distinct points of the Riemann sphere by a Möbius transformation $f$ sending them to $\infty ,0,1$. A fourth point lies on their generalized circle exactly when $f$ takes a real value there: the image of that generalized circle is the extended real line. Other standard cross-ratio ordering conventions differ by real fractional-linear transformations and preserve this reality criterion.