The inverse has the same nonzero determinant. Each formula extends over its pole and infinity, and the compositions are the identity there as well.

This decomposition uses translations, a nonzero complex multiplication, and reciprocal inversion.

Solve for \(z=(w-i)/(w+i)\). The unit-circle equation becomes \(|w-i|=|w+i|\), the perpendicular bisector of \(i\) and \(-i\): the real axis. The pole \(z=1\) supplies infinity.

Set \(d=1\). Then \(b=i\) and \(a=-c\). The remaining condition gives \(-c+i=4c+4\), so \(c=(i-4)/5\). Multiplying coefficients by five yields:

Its determinant is \(25+15i\ne0\). Direct substitution checks the three prescribed images.

Compose the first with the inverse of the second. The resulting map fixes three distinct points. For a nonidentity fractional linear map, the fixed-point equation is a nonzero quadratic, with infinity counted when appropriate, and has at most two solutions. Thus the composite is identity.

As the last point tends to infinity, the ratio of the two factors involving it tends to one.

Normalize the last three source points to one, zero, infinity. Normalizing their images and then applying the original map does exactly the same thing to those three points. Uniqueness makes these two composites identical, so their values at the remaining point agree. Those values are the two cross ratios.

If \(u=z-a\), the reflected displacement is \(R^2/\bar u\). Conjugate that displacement and apply the rule again to obtain \(R^2/(R^2/u)=u\). Points of the circle are fixed because \(u\bar u=R^2\).

This is an Apollonius circle, one member of the second Steiner family.

The fixed-point equation gives \(2z^2-2z=0\), so the fixed points are zero and one. For \(S(z)=z/(z-1)\):

The multiplier is a positive real number different from one, so the transformation is hyperbolic.

The starting coordinate is \(S(2)=2\), so \(S(T^n2)=2/3^n\). Solving \(w=z/(z-1)\) gives \(z=w/(w-1)\), hence \(T^n2=2/(2-3^n)\). For integer n≥0 the denominator is never zero. The sequence tends to zero, the attracting fixed point. The other fixed point one stays at one. A large normalized coordinate and a large original coordinate are not the same statement.

Infinity maps to one, so it is not fixed. The sole fixed point is zero.

Approach.

Use the denominator \(c(cz+d)\).

Put \(\Delta=ad-bc\ne0\). Bringing the proposed expression to a common denominator gives \[\frac ac-\frac{\Delta}{c^2}\frac1{z+d/c}=\frac{a(cz+d)-(ad-bc)}{c(cz+d)}=\frac{az+b}{cz+d}.\] The equality extends to the pole and infinity as an equality of maps of the Riemann sphere.

Approach.

Start with denominator \(1\) for the first three maps.

The coefficient quadruples \((a,b,c,d)\) may be \((e^{i\theta},0,0,1)\) for rotation; \((1,v,0,1)\) for translation; \((k,0,0,1)\) for homothety, \(k>0\); and \((0,1,1,0)\) for the book's inversion \(1/z\). Their determinants are respectively \(e^{i\theta},1,k,-1\), all nonzero.

Approach.

Check the numerator and denominator separately at each point.

At \(z_1\) the two nonzero ratios cancel to \(1\). At \(z_2\) the numerator is zero and denominator nonzero. At \(z_3\) the denominator is zero but the numerator is nonzero, so the sphere value is \(\infty\). If an input is infinity, take the corresponding limit.

Approach.

Choose an ordering of each set.

Order the points as \((z_1,z_2,z_3)\) and \((w_1,w_2,w_3)\). The fundamental theorem provides a Möbius map sending each \(z_j\) to \(w_j\). It therefore maps the first three-point set onto the second. Different orderings can give different maps, but existence is all congruence requires.

Approach.

A translation preserves a displacement vector.

Any chosen point \(z\) can be sent to any chosen point \(w\) by the unique translation \(T(u)=u+w-z\). Two ordered pairs are translation-congruent precisely when \(z_2-z_1=w_2-w_1\). Thus arbitrary one-point figures, but not arbitrary two-point figures, are congruent.

Approach.

Recognize each term as a number minus its conjugate.

The remaining equation is \(\alpha z-\bar\alpha\bar z+\beta-\bar\beta=0\). Since \(u-\bar u=2i\operatorname{Im}u\), this is exactly \(2i\operatorname{Im}(\alpha z+\beta)=0\). Also \(\alpha\ne0\). If \(c=0\), the determinant condition gives \(a,d\ne0\), hence \(\alpha=a\bar d\ne0\). If \(c\ne0\), the leading-coefficient condition says \(a/c=k\) is real. Were \(\alpha=0\), then \(k\bar d=\bar b\), so \(b=kd\) and \(ad-bc=0\), a contradiction. Thus this really is a line.

Approach.

Use three points and preservation of clines.

Yes. Choose three distinct points on each cline. The unique Möbius map matching those triples maps the first cline to a cline through the second triple. A triple of distinct sphere points determines a unique cline, so its image is the second cline.

Approach.

Write the cross ratio in linear-fractional form.

For finite distinct reference points set \(K=(z_1-z_3)/(z_1-z_2)\ne0\). Then \(N(z)=(Kz-Kz_2)/(z-z_3)\) has determinant \(K(z_2-z_3)\ne0\). Substitution gives \(N(z_1)=1\), \(N(z_2)=0\) and \(N(z_3)=\infty\). Limits handle a reference point at infinity and still give a nonzero-determinant map.

Approach.

The ratio \((u-a)/(u-b)\) tends to \(1\) as \(u\to\infty\).

Divide the factors involving the variable tending to infinity by that variable. The limits are, respectively, \[\frac{z_0-z_2}{z_0-z_3},\qquad\frac{z_1-z_3}{z_0-z_3},\qquad\frac{z_0-z_2}{z_1-z_2}.\] These formulas assume the other three points are finite and distinct; the cross ratio is interpreted on the sphere at a remaining pole.

Approach.

Normalize one coefficient, then impose the three prescribed values.

(a) Set \(d=1\), so \(b=i\) and \(a=-c\). The condition at \(1\) gives \(-c+i=4c+4\), hence \(c=(i-4)/5\). Thus \(T(z)=((4-i)z+5i)/((i-4)z+5)\).

(b) Set \(b=0,d=1\). The two remaining conditions give \(ai=ci+1\) and \(-ai=2(-ci+1)\); hence \(c=-3i,a=-4i\). Thus \(T(z)=-4iz/(1-3iz)\).

(c) With \(d=1\), equations \(a+b=2c+2\), \(2a+b=6c+3\), \(3a+b=3c+1\) give \(c=-3/7,a=-5/7,b=13/7\). Thus \(T(z)=(13-5z)/(7-3z)\). Each displayed map has nonzero determinant, and direct substitution checks all three values. Uniqueness follows from the fundamental theorem.

Approach.

First map the circle to the real axis, then rotate and translate.

The Cayley map \(u=i(1+z)/(1-z)\) takes the unit circle to the extended real axis: for \(|z|=1\), conjugating \(u\) leaves it unchanged. Then \(w=1+(1-i)u\) has real part \(1+u\) and imaginary part \(-u\) when \(u\) is real, so \(x+y=1\). Consequently \[T(z)=1+(1+i)\frac{1+z}{1-z}\] works, with \(z=1\) mapped to the line's point at infinity.

Approach.

Rotate \(z-1\), not \(z\).

Translate the center to zero, multiply by \(i\), and translate back: \(T(z)=1+i(z-1)=iz+1-i\). It fixes \(1\), preserves distances and multiplies every displacement from \(1\) by \(i\).

Approach.

Use \(cz^2+(d-a)z-b=0\), then check infinity.

Each denominator has nonzero coefficient of \(z\), so infinity is not fixed. Solve after cross multiplication and check that no root is a pole.

(a) \(3z^2-3z=0\), so \(z=0,1\). (b) \(2z^2-4z+2=0\), so the sole fixed point is \(z=1\). (c) \(z^2+z+2=0\), giving \(z=(-1\pm i\sqrt7)/2\). (d) \((1-i)z^2+(i-1)z=0\), so \(z=0,1\).

Approach.

The determinant condition prevents numerator and denominator from vanishing together.

For \(c\ne0\), divide numerator and denominator by \(z\): \(T(\infty)=a/c\). At \(z=-d/c\), the numerator equals \((bc-ad)/c\ne0\), so \(T(-d/c)=\infty\). If \(c=0\), then \(a,d\ne0\), and \(T(z)=(a/d)z+b/d\) has no finite pole and fixes infinity.

Approach.

Move the fixed points to \(0,\infty\), or move the single fixed point to infinity.

(a) Conjugate by \(S(z)=(z-1)/(z+1)\). A map fixing \(0,\infty\) is \(w\mapsto\lambda w\), where \(\lambda\ne0\). Solving \(S(Tz)=\lambda S(z)\) gives \[T(z)=\frac{(1+\lambda)z+1-\lambda}{(1-\lambda)z+1+\lambda},\quad\lambda\ne0,1.\] Exclude \(\lambda=1\) because the identity has every point fixed.

(b) Conjugate by \(S(z)=1/(z+1)\). A nonidentity map with sole fixed point infinity is \(w\mapsto w+\beta\), \(\beta\ne0\). Hence \(T(z)=-1+(z+1)/(1+\beta(z+1))\). These and only these maps have the requested fixed-point sets.

Approach.

Use the matrix of the composition.

Represent \(T,S\) by matrices \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\) and \(B=\begin{pmatrix}e&f\\g&h\end{pmatrix}\). Substitution represents \(T\circ S\) by \(AB\). Expansion gives \(\det(AB)=(ad-bc)(eh-fg)=\det A\det B\). This statement concerns the chosen coefficient matrices: a Möbius map itself has no uniquely specified determinant, since multiplying all coefficients by \(k\ne0\) multiplies the determinant by \(k^2\).

Approach.

Linearly dependent coefficient rows produce a constant ratio.

If the denominator row \((c,d)\) is nonzero and \(ad-bc=0\), then \((a,b)=k(c,d)\) for some \(k\), so the quotient equals \(k\) wherever defined. Any common zero is a removable defect, not a bijective sphere map. If \((c,d)=(0,0)\) the quotient is not a valid finite-valued formula at all; projectively a nonzero numerator would give the constant infinity map away from its zero. A constant map on the sphere is not injective and has no inverse, so cannot be a transformation.

Approach.

An affine map with slope other than \(1\) has a finite fixed point.

Fixing infinity forces \(c=0\), so \(T(z)=\alpha z+\beta\), \(\alpha\ne0\). If \(\alpha\ne1\), then \(\beta/(1-\alpha)\) is a finite fixed point, contradiction. Thus \(\alpha=1\). Also \(\beta\ne0\), since otherwise every point is fixed. Hence it is a nonzero translation.

Approach.

Scaling all coefficients does not change the map.

Choose a square root \(s\) of \(\Delta=ad-bc\ne0\). Divide all four coefficients by \(s\). The map is unchanged and the new determinant is \(\Delta/s^2=1\). The two choices \(s,-s\) give opposite matrices. Any two matrices for the same Möbius map differ by a nonzero scalar; determinant one forces that scalar's square to be \(1\). Therefore there are exactly two such matrices, \(A\) and \(-A\).

Approach.

At a pole, differentiate the reciprocal target coordinate.

At any finite nonpole, \(T'(z)=(ad-bc)/(cz+d)^2\ne0\). Its differential is multiplication by a nonzero complex number, which rotates and scales all tangent vectors equally and preserves oriented angles. At a pole use the target coordinate \(v=1/T(z)\); its derivative there is nonzero. At infinity use the source coordinate \(u=1/z\), and also the reciprocal target coordinate if needed. The same nonzero-determinant calculation proves a nonzero derivative in these charts. Thus conformality holds at every sphere point.

Approach.

The quotient of the displacement vectors must have modulus \(1\).

Two ordered pairs of distinct points \((p,q)\) and \((r,s)\) can be matched by exactly one orientation-preserving Euclidean motion if and only if \(|q-p|=|s-r|\). Necessity follows from distance invariance. For sufficiency set \(\alpha=(s-r)/(q-p)\); equality of distances gives \(|\alpha|=1\). The map \(T(z)=r+\alpha(z-p)\) sends the pairs correctly. Any such motion must have this multiplier and translation, proving uniqueness. Arbitrary pairs with unequal separation cannot be matched.

Approach.

A translation is determined by its displacement; a rotation preserves radius.

Translations act uniquely transitively on individual points: \(T(z)=z+w-p\) is the unique translation sending \(p\) to \(w\). For rotations about the origin, two nonzero points \(p,w\) can be matched if and only if \(|p|=|w|\); then the unique rotation is \(T(z)=(w/p)z\). Zero can only map to zero, and every rotation does that, so uniqueness fails there.

Approach.

Use the three-point theorem, not just a sketch.

Choose three distinct points on each cline. A Möbius map carries the first triple to the second. It maps clines to clines, and a cline through a prescribed triple of distinct sphere points is unique. Thus the entire first cline maps to the second.

Approach.

Send a common tangency point to infinity and count preserved intersections.

The indicated incidence classes are \((a,f)\), \((b,e)\) and \((c,d)\). In (a), send the unique symmetric limiting pair of the two disjoint clines to \(0,\infty\). They become concentric circles, while their common perpendicular becomes a line through zero, giving (f). In (b), send the common tangency to infinity: all three clines become parallel lines, giving (e). In (c), send the tangency of the outer circle and one inner circle to infinity. Those two become parallel lines and the third becomes a circle tangent to both; the same construction on (d), using a circle–line tangency, gives this canonical configuration.

The three classes differ by preserved incidence: the first has transverse intersections; the second has one common tangency of all three clines; the third has three distinct pairwise tangencies. A qualification is important: arbitrary drawings within a class can carry numerical invariants. For example three parallel lines have a spacing ratio (up to permutation) preserved by affine maps fixing infinity. The scan is a schematic congruence puzzle, not metric data sufficient to certify every arbitrary realization.

Approach.

Divide each point by its squared modulus.

Use \(z^*=1/\bar z=z/|z|^2\). In the stated order the answers are \(1,2,i,2i,(1+i)/2,1+i\). Each pair lies on the same ray and its moduli multiply to \(1\); points on the unit circle are fixed.

Approach.

Rationalize \(4/(x+i)\).

For real \(x\), the formula becomes \(z^*=i+4/(x+i)=4x/(x^2+1)+i(1-4/(x^2+1))\). Substituting gives, in order, \(-3i\), \(2-i\), \(8/5+i/5\), \(\sqrt3\), and \(6/5+3i/5\). The fixed value \(\sqrt3\) lies on the circle since \(|\sqrt3-i|=2\).

Approach.

Write the two tangent equations after placing the center at zero.

Rotate and translate so the center is \(0\), the point is \(x\in(0,R)\) and the chord is vertical. Then \(B=(x,\sqrt{R^2-x^2})\) and \(D=(x,-\sqrt{R^2-x^2})\). The tangent equations are \(xX\pm\sqrt{R^2-x^2}Y=R^2\). Their intersection is \((R^2/x,0)\), so the radial distances multiply to \(R^2\), exactly the symmetry formula. Reversing the construction proves the exterior case. At the center the tangents are parallel and meet at infinity; a point on the circle is its own symmetric point, handled as a limit.

Approach.

Invert the normalizing Möbius map.

Choose three distinct reference points on the circle and let \(N\) be their cross-ratio normalizer. The defining equation is \(N(z^*)=\overline{N(z)}\). Since \(N\) is a bijection of the sphere, exactly one solution exists: \(z^*=N^{-1}(\overline{N(z)})\). This includes the center and infinity. For a circle of center \(a\) and radius \(R\), the explicit formula is \(a+R^2/(\bar z-\bar a)\), with \(a\leftrightarrow\infty\).

Approach.

First treat the real axis.

For the real axis take real reference points; their normalizer has real coefficients, so \(N(\bar z)=\overline{N(z)}\). Uniqueness therefore gives \(z^*=\bar z\). For a general line choose a Euclidean motion \(U\) mapping it to the real axis. Symmetry is Möbius-invariant, hence its reflection is \(U^{-1}(\overline{U(z)})\), exactly the Euclidean mirror construction.

Approach.

Evaluate both maps at \(i\).

Unit-circle symmetry is \(J(z)=1/\bar z\); the book's inversion is \(I(z)=1/z\). They differ, for example \(J(i)=i\) while \(I(i)=-i\). They are related by \(I=K\circ J=J\circ K\), where \(K(z)=\bar z\) is reflection in the real axis. \(I\) preserves orientation and \(J\) reverses it.

Approach.

Equidistance from conjugate points forces the center onto the real axis.

Move \(C\) to the real axis. Write \(z=x+iy\) with \(y\ne0\), so \(z^*=x-iy\). A circle orthogonal to the real axis has center on that axis: at an intersection its radius must be horizontal, since its tangent is vertical. It is then invariant under conjugation and contains both points. Conversely, if its center is \(u+iv\), equality of its squared distances from \(x+iy\) and \(x-iy\) gives \(4vy=0\), hence \(v=0\). Its radius exceeds \(|v|=0\), so it crosses the real axis orthogonally. Undo the motion.

Approach.

Normalize the reflecting cline and include the straight-line case.

Choose a Möbius map taking \(C\) to the extended real axis and avoiding infinity at the two points. Symmetry, incidences and angles are preserved. For a circle the claim is Exercise 24. A line through conjugate nonreal points is vertical and thus perpendicular to the real axis; conversely a perpendicular line through one conjugate contains the other. These cover all image clines. Transforming back proves both directions, including configurations whose original points involve infinity.

Approach.

Move \(p,q\) to \(0,\infty\).

Every first-kind cline passes through both distinguished points \(p,q\), so at either of these there are infinitely many choices. At every other point \(z\), the triple \(p,q,z\) determines exactly one cline. For the second kind, neither \(p\) nor \(q\) belongs to any member: in the coordinate \(w=(z-p)/(z-q)\) those points are \(0,\infty\), whereas the members are \(|w|=k\) with \(0<k<\infty\). Every other point lies on exactly one such level set, namely \(k=|w|\).

Approach.

Combine radial dilation and angular rotation in the normalized plane.

In normalized coordinates write \(w(t)=e^{t(\log k+i\theta)}w_0\), where \(k>0\), \(k\ne1\), and \(\theta\not\equiv0\pmod{2\pi}\). Radius changes by \(k^t\) while argument changes by \(t\theta\), so this is a logarithmic spiral. Apply \(S^{-1}\), where \(S(z)=(z-p)/(z-q)\), to obtain \(z(t)=(q w(t)-p)/(w(t)-1)\). Draw first-kind and second-kind Steiner clines as the coordinate grid and the orbit crossing both families. For \(k<1\) forward motion tends to \(p\); for \(k>1\) it tends to \(q\). The discrete orbit consists of the integer-time points.

Approach.

Solve the normal-form equation for \(\lambda\).

For \(z\ne p,q\), divide by \((z-p)/(z-q)\) to obtain \[\lambda=\frac{Tz-p}{Tz-q}\frac{z-q}{z-p}=[Tz,z,p,q].\] Sphere limits handle an infinite entry. The fixed points themselves must be excluded from this evaluation: the displayed quotient otherwise contains an indeterminate cancellation. The constant multiplier is determined by any nonfixed point.

Approach.

Evaluate at the pole and at infinity.

Substitute \(z=z_0\): the left side is zero, so \(\beta=-1/(z_0-p)\). Substitute \(z=\infty\): the term \(1/(z-p)\) is zero, so \(\beta=1/(T(\infty)-p)\). Here \(\beta\ne0\) because the map is not the identity; its pole is finite and distinct from \(p\).

Approach.

Solve \(Tz=z\) before choosing the normalizing map.

(a) The fixed equation is \(2z(z-1)=0\). With \(p=0,q=1\), \(S(z)=z/(z-1)\) gives \(S(Tz)=-S(z)\), so \(\lambda=-1\): an elliptic half-turn. Orbits lie on second-kind Steiner clines.

(b) The fixed equation is \((z-2)^2=0\). Direct simplification gives \(1/(Tz-2)=1/(z-2)+1\). Thus \(p=2\), \(\beta=1\), a parabolic map. In the \(1/(z-2)\) plane motion is to the right along horizontal lines; pull these back to circles tangent at \(2\).

(c) The fixed points are \(0,1\). For \(S(z)=z/(z-1)\), \(S(Tz)=\tfrac12S(z)\). This is hyperbolic; forward nonfixed orbits tend to \(0\) along first-kind clines.

(d) The fixed equation gives \(0,i\). For \(S(z)=z/(z-i)\), \(S(Tz)=-iS(z)\). This is an elliptic quarter-turn, with orbits on the second-kind family for \(0,i\). These normal forms provide the coordinate grids for the sketches.

Approach.

Separate rotations from translations before solving for the fixed point.

Put \(\alpha=e^{i\theta}\). If \(\alpha\ne1\), the finite fixed point is \(p=b/(1-\alpha)\), the other is infinity, and \(Tz-p=\alpha(z-p)\) is its normal form. If \(\alpha=1,b\ne0\), the map is already the parabolic normal form \(Tz=z+b\) with fixed point infinity. If \(\alpha=1,b=0\), it is the identity and has no distinguished fixed-point normal form.

Approach.

Use the normal forms from Exercise 4.

If \(e^{i\theta}\ne1\), its multiplier has modulus \(1\) and it is elliptic. If \(e^{i\theta}=1\) and \(b\ne0\), it is parabolic. If both \(e^{i\theta}=1\) and \(b=0\), it is the identity. A Euclidean motion of this orientation-preserving form is never nonidentity hyperbolic or loxodromic: those types require a multiplier with modulus different from \(1\).

Approach.

Solve \(z^2=1\) and normalize those points.

Its fixed points are \(1,-1\). For \(S(z)=(z-1)/(z+1)\), direct substitution gives \(S(1/z)=-S(z)\). Thus \(\lambda=-1\), so inversion is elliptic, a half-turn in normalized coordinates. This is the book's analytic inversion, not the orientation-reversing circle reflection \(1/\bar z\).

Approach.

The image of infinity determines the remaining normal-form parameter.

(a) Set \(S(z)=(z-i)/(z+i)\). Since \(S(\infty)=1\) and \(S(0)=-1\), the multiplier is \(-1\). Solving gives \(T(z)=-1/z\).

(b) Put \(p=1+i\) and \(A=100-p=99-i\). The parabolic parameter is \(\beta=1/A\), giving \(T(z)=p+A(z-p)/(z-p+A)\). Expanding if desired gives \(T(z)=(100z-p^2)/(z+100-2p)\); it fixes only \(p\) and has the prescribed value at infinity.

(c) Set \(S(z)=(z-5)/(z-10)\). The multiplier is \(S(20)=3/2\). Solving \(S(Tz)=\tfrac32S(z)\) gives \(T(z)=(20z-50)/(z+5)\). Its determinant is \(150\ne0\), and it fixes \(5,10\).

Approach.

Study multiplication by \(\lambda^n\) first.

Write \(S(T^n z)=\lambda^nS(z)\), with \(S(p)=0,S(q)=\infty\). If \(|\lambda|<1\), the limit is \(p\) for every \(z\ne q\); the exceptional point \(q\) stays fixed. If \(|\lambda|>1\), the limit is \(q\) for every \(z\ne p\); \(p\) stays fixed. Limits are in the sphere, so a finite orbit can pass through infinity without invalidating the conclusion.

Approach.

Square the normal form.

A parabolic map is conjugate to \(w\mapsto w+\beta\) with \(\beta\ne0\); its square translates by \(2\beta\ne0\), so cannot be the identity. Thus a nonidentity involution has two fixed points and is conjugate to \(w\mapsto\lambda w\). Its square is the identity exactly when \(\lambda^2=1\). Since \(\lambda\ne1\), we have \(\lambda=-1\), an elliptic half-turn. The identity also satisfies \(T^2=I\) but is excluded from the asserted classification.

Approach.

Apply \(S\) to the fixed-point equation.

Apply the bijection \(S\) to \(Tz=z\): this is equivalent to \(STz=Sz\). Because \(ST=RS\), it is equivalent to \(R(Sz)=Sz\). Every implication is reversible, proving both directions, including sphere points at infinity.

Approach.

Use preservation of clines and incidence.

Each member is a cline through both \(z_1,z_2\). Its image is a cline through \(w_1,w_2\), hence belongs to the target family. Applying the same argument to the inverse map shows every member of the target family arises, so the mapping is onto the family.

Approach.

Normalize the pair so the first family consists of radial lines.

Send \(p,q\) to \(0,\infty\). The first family becomes all lines through zero. A circle perpendicular to a line has its center on that line. A circle perpendicular to every radial line must therefore have center zero; every circle centered at zero has the required perpendicularity. A straight line cannot be perpendicular to every radial direction, so gives no extra member. Pulling these circles back gives precisely the second-kind family.

Approach.

Use the orthogonality characterization.

By Exercise 12 a second-kind member is characterized as a cline perpendicular to every first-kind member. Möbius maps preserve clines, angles and the first-kind family (Exercise 11), so its image has that same characterization for the image pair. The inverse gives surjectivity. This proof avoids assuming Euclidean distance ratios themselves are Möbius invariants.

Approach.

Normalize one cline to a line and solve the two symmetry equations.

Move one cline to the real axis. The other cannot be a line, since all extended lines share infinity, so it is a circle with center \(a+ib\) and radius \(R\), where \(b^2>R^2\). A pair symmetric in the real axis has form \(z,\bar z\). Symmetry in the circle requires \[\bar z=a+ib+\frac{R^2}{\bar z-a+ib},\qquad(\bar z-a)^2=R^2-b^2.\] Thus the pair is exactly \(a\pm i\sqrt{b^2-R^2}\). These points are distinct and satisfy both symmetry formulas, proving existence and uniqueness. Transform back using invariance of symmetry.

Approach.

Normalize each incidence type separately.

(a) Send the two intersections to \(0,\infty\). The clines become distinct lines through zero. Common perpendicular clines are exactly circles centered at zero: a circle's center must lie on both lines, and no line is perpendicular to both. Pulling back yields the second-kind Steiner family for the intersections.

(b) Send their tangency point to infinity. They become distinct parallel lines. No circle can have its center on both lines, while every line perpendicular to them is a common perpendicular. Pull back this parallel family: it is a degenerate Steiner family of clines through the common tangency, mutually tangent there.

(c) Use Exercise 14 to send their common symmetric pair to \(0,\infty\). Each original cline is perpendicular to every cline through that symmetric pair, so the images are concentric circles. Common perpendiculars are exactly radial lines: a circle perpendicular to circles of radii \(r\ne R\) would require simultaneously \(|c|^2=\rho^2+r^2=\rho^2+R^2\), impossible. Pulling back yields the first-kind family through the common symmetric pair.

Approach.

Make the first two circles concentric.

Normalize the first two to concentric circles with distinct radii. By Exercise 15 their common perpendiculars are radial lines. If the third is a circle with center \(c\), then \(c\ne0\) (a concentric circle cannot meet either of the first circles), and exactly one radial line passes through \(c\); that line is perpendicular to the third circle. If the third is a straight line, exactly one radial line is perpendicular to it. In both cases this line is unique. Undo the normalization.

For generic finite z, \(1/T(z)=1/z+1\), hence \(1/T^n(z)=1/z+n\), giving \(T^n(z)=z/(nz+1)\). The same identity follows by induction as an equality of Möbius maps, so it remains valid even if an intermediate iterate is infinity. Its pole is \(-1/n\), infinity maps to \(1/n\), and solving \(z=z/(nz+1)\) gives \(nz^2=0\). Thus zero is its unique fixed point; infinity is not fixed.

Write \(T(z)=(az+b)/(cz+d)\) with nonzero determinant. If infinity is fixed, c=0, so \(T(z)=Az+B\) with A nonzero. Fixing two distinct finite points gives \((A-1)u+B=(A-1)v+B=0\); subtraction yields A=1 and then B=0. If infinity is not fixed, c is nonzero and all three fixed points are finite. Each satisfies \(cz^2+(d-a)z-b=0\), a nonzero quadratic that cannot have three distinct roots. This case is impossible. Thus only identity has the prescribed three fixed points.