Mobius Transformations

Technique: algebraic characterization of generalized circles via a single real quadratic equation, then verification under the elementary generators of the Möbius group. Step 1: Algebraic form of a generalized circle. A subset $\Gamma \subset \mathbb{C}$ (before adjoining $\infty$) is an ordinary circle or a line if and only if it is the solution set of an equation

where $A, C \in \mathbb{R}$, $B \in \mathbb{C}$, and $|B|^2 > AC$, with the further conditions:

• if $A \neq 0$, $(\star)$ describes a circle (complete the square: dividing by $A$, $(\star)$ becomes $|z + B/A|^2 = |B|^2/A^2 - C/A > 0$, a genuine circle since the right side is positive);
• if $A = 0$, $(\star)$ becomes $Bz + \bar B \bar z + C = 0$, i.e. $2\,\mathrm{Re}(\bar B z) = -C$, which is precisely the equation of a straight line (nondegenerate since $B \neq 0$, forced by $|B|^2 > AC = 0$).

Conversely, every circle $|z-p|=r$ expands to $z\bar z - \bar p z - p \bar z + (|p|^2 - r^2) = 0$, matching $(\star)$ with $A=1$, $B = -p$, $C = |p|^2-r^2$, and $|B|^2 - AC = r^2 > 0$. Every line $\mathrm{Re}(\bar\beta z) = t$ ($\beta \neq 0$, $t\in\mathbb R$) matches $(\star)$ with $A = 0$, $B = \beta/2$... wait, normalize: with $A=0$, $B = \beta$, $C = -2t$, we need $|B|^2 > AC$, i.e. $|\beta|^2 > 0$, true. So $(\star)$ with the stated sign condition is exactly the class of circles and lines. We adjoin $\infty$ to the line-solutions, and by convention a "generalized circle" is the solution set of $(\star)$ in $\widehat{\mathbb{C}}$, with $\infty$ included exactly when $A = 0$. Step 2: Reduce to generators. By the factorization in the Intuition section, every Möbius transformation $T$ with $c \neq 0$ is a composite

and if $c = 0$ then $T(z) = \frac{a}{d} z + \frac{b}{d}$ is already affine (a composite of a scaling/rotation and a translation). Since the class of generalized circles is preserved under composition of maps that individually preserve it, it suffices to check three types of generators: (a) translations $z \mapsto z + t$, (b) scalings/rotations $z \mapsto \lambda z$ ($\lambda \neq 0$), and (c) inversion $z \mapsto 1/z$. Step 3: Translations and rotation-scalings. Let $w = z + t$. Substituting $z = w - t$ into $(\star)$:

which expands to $A w\bar w + (B - At)w + \overline{(B-At)}\bar w + C' = 0$ for a suitable real constant $C'$ (direct computation: $C' = C - B t - \bar B \bar t + A|t|^2$, real since $A,C \in \mathbb R$ and the middle terms are conjugates of each other up to the real coefficient $A$). This is again of form $(\star)$ with the same $A$, new $B' = B - At$; the discriminant condition $|B'|^2 > AC'$ is preserved because a Euclidean translation cannot change whether a curve is a genuine circle or line (this can also be checked directly, but geometric preservation of circles/lines under translation is elementary). So translations preserve the class, with $A$ unchanged — in particular preserving whether $\infty$ is in the generalized circle. Similarly for $w = \lambda z$ with $\lambda \ne 0$: substitute $z = w/\lambda$ into $(\star)$ to get

multiply through by $|\lambda|^2$ to get $A w\bar w + (B\bar\lambda) w + (\bar B \lambda)\bar w + C|\lambda|^2 = 0$, again of form $(\star)$ with $A$ unchanged (up to an overall positive rescale which doesn't affect the sign structure), $B'' = B\bar\lambda$, $C'' = C|\lambda|^2$, and one checks $|B''|^2 - A C'' = |\lambda|^2(|B|^2 - AC) > 0$ is preserved. So rotation-scalings preserve the class and preserve whether $A = 0$. Step 4: Inversion. Let $w = 1/z$, i.e. $z = 1/w$ (for $z, w \neq 0$). Substitute into $(\star)$:

Multiply both sides by $w \bar w = |w|^2 \neq 0$:

i.e.

This is exactly of the form $(\star)$ in the variable $w$ with new coefficients $A^* = C$, $B^* = \bar B$, $C^* = A$ (all real/complex-conjugate-compatible as required since $A, C$ real), and the discriminant condition becomes $|B^*|^2 > A^*C^*$, i.e. $|\bar B|^2 > CA$, i.e. $|B|^2 > AC$ — identical to the original condition. So inversion maps the solution set of $(\star)$ (minus possibly the point $z=0$ or $w=0$) to the solution set of the transformed equation, again a generalized circle. We must also handle $z=0$ and $w=\infty$, $z=\infty$ and $w=0$: if $0$ satisfies $(\star)$, that means $C = 0$, so the original curve passes through the origin; the transformed equation has $A^* = C = 0$, meaning the image curve is a line (through the point at infinity, i.e. genuinely unbounded) — consistent with the map $z\mapsto 1/z$ sending a circle/line through $0$ to a line. Conversely if $A = 0$ (original is a line, containing $\infty$), then $A^* = C$; the image equation has "circle indicator" $A^* = C$, and by symmetry the image is a line iff $C = 0$ iff the original line passes through the origin. This matches the geometric fact that inversion sends a line through the origin to itself (a line) and a line missing the origin to a circle through the origin. Step 5: Conclusion. Since $(\star)$ exactly characterizes generalized circles (Step 1), and each generator (Steps 3–4) transforms a solution set of an equation of form $(\star)$ into the solution set of another equation of the same form with the same nondegeneracy condition, by induction (Step 2) every Möbius transformation $T$ carries generalized circles to generalized circles. The "circle $\to$ line" case occurs exactly when $A^{\text{final}} = 0$; tracing through the composition, this happens precisely when $T^{-1}(\infty)$ lies on the original curve, since $\infty$ enters the image curve exactly when the coefficient of $z\bar z$ vanishes after transformation, which by Step 4's inversion step corresponds to the pole of $T$ lying on the source curve. $\blacksquare$

Technique: direct algebraic computation, reduced to the three generator types, using that the cross-ratio of $T z_1,\dots,Tz_4$ is a rational function of $z_1,\dots,z_4$ we can simplify directly for a general $T$. Write $T(z) = \frac{az+b}{cz+d}$, $ad-bc\ne0$. For any $i \ne j$ with $z_i, z_j \ne \infty$,

Expand the numerator:

Hence

Now form the cross-ratio of the images, using $(\dagger)$ for each of the four factors $(Tz_1-Tz_3)$, $(Tz_2-Tz_4)$, $(Tz_1-Tz_4)$, $(Tz_2-Tz_3)$:

Every factor $(cz_k+d)$ for $k=1,2,3,4$ appears exactly once in the numerator's denominator-product and once in the denominator's denominator-product (each $z_k$ appears in exactly two of the four original differences, once "up" and once "down" combinatorially — check: $z_1$ appears in $(z_1-z_3)$ and $(z_1-z_4)$, contributing $(cz_1+d)$ to both numerator and denominator of the big fraction). Concretely, all four factors $(cz_1+d), (cz_2+d), (cz_3+d), (cz_4+d)$ appear exactly once in the numerator's denominator and exactly once in the denominator's denominator, so they cancel completely. The two factors of $(ad-bc)$ in the numerator cancel against the two in the denominator (nonzero since $ad-bc \ne 0$). We are left with

If one of the $z_i$ equals $\infty$, or one of the $Tz_i$ involves the pole of $T$, the identity persists by continuity: both sides are continuous (indeed rational) functions of $z_1,\dots,z_4$ on $\widehat{\mathbb{C}}^4$ minus diagonals, and the identity holds on a dense open set, hence everywhere by continuity of both sides. $\blacksquare$

(a) Existence. Define $S(z) = (z, z_1, z_2, z_3)$, i.e.

$S$ is a Möbius transformation (its coefficients form an invertible matrix because $z_1,z_2,z_3$ are distinct — direct check: the determinant of the associated matrix is a nonzero multiple of $(z_1-z_2)(z_1-z_3)(z_2-z_3)\neq 0$), and $S(z_1) = 1$... let us verify directly: $S(z_1) = \frac{(z_1-z_2)(z_1-z_3)}{(z_1-z_3)(z_1-z_2)} = 1$, and one checks $S(z_2) = 0$ (numerator vanishes) and $S(z_3) = \infty$ (denominator vanishes). Similarly define $R(w) = (w,w_1,w_2,w_3)$ with $R(w_1)=1, R(w_2)=0,R(w_3)=\infty$. Then $T := R^{-1}\circ S$ is Möbius (composition and inverse of Möbius maps are Möbius, since $\mathrm{M\ddot ob}$ is a group under composition, matching matrix multiplication and inversion in $\mathrm{PGL}(2,\mathbb C)$), and $T(z_i) = R^{-1}(S(z_i))= w_i$ for $i=1,2,3$ by construction. Uniqueness. Suppose $T_1, T_2$ both send $(z_1,z_2,z_3)\mapsto(w_1,w_2,w_3)$. Then $U := T_2^{-1}\circ T_1$ fixes $z_1,z_2,z_3$. A nonidentity Möbius transformation $U(z) = \frac{az+b}{cz+d}$ has at most two fixed points in $\widehat{\mathbb C}$: solving $U(z)=z$ gives $cz^2 + (d-a)z - b = 0$ if $c\ne 0$ (a quadratic, at most 2 roots, and $\infty$ is not a fixed point since $c\ne0$ means $U(\infty)=a/c\ne\infty$), or if $c=0$, $U(z) = \frac{a}{d}z+\frac bd$ which fixes $\infty$ and at most one finite point (if $a\ne d$) or is the identity (if $a=d, b=0$). Either way a non-identity Möbius map has at most 2 fixed points. Since $U$ fixes three distinct points $z_1,z_2,z_3$, $U$ must be the identity, so $T_1 = T_2$. (b) By the cross-ratio invariance lemma, applying $S(z) = (z,z_1,z_2,z_3)$ from part (a)'s construction, we have $(z_1,z_2,z_3,z_4) = (Sz_1,Sz_2,Sz_3,Sz_4) = (1,0,\infty, Sz_4) = Sz_4$ (direct check of the degenerate cross-ratio with entries $1,0,\infty$ collapses to the fourth argument itself: $(1,0,\infty,w) = \lim_{Z\to\infty}\frac{(1-\infty)(0-w)}{(1-w)(0-\infty)} = \frac{0-w}{-(1-w)}\cdot\frac{-1}{-1}$; computing the limit carefully by clearing $Z=\infty$ symbolically gives $(1,0,\infty,w) = w$). So $(z_1,z_2,z_3,z_4) = S(z_4)$. Since $S$ is a Möbius transformation with $S(z_1)=1, S(z_2)=0,S(z_3)=\infty$, and by Möbius transformations preserve generalized circles $S$ maps the generalized circle $\Gamma$ through $z_1,z_2,z_3$ (unique such circle, since three distinct points determine a unique circle or line — standard Euclidean fact, or apply the theorem's Step 1 characterization: three points impose three real linear conditions on the 3 real parameters $(A,B,C)$ up to scale) onto a generalized circle through $1, 0, \infty$. A generalized circle through $\infty$ is a line; the line through $0$ and $1$ is $\mathbb{R}$ itself. So $S(\Gamma) = \widehat{\mathbb{R}} = \mathbb{R}\cup\{\infty\}$. Therefore $z_4 \in \Gamma$ if and only if $S(z_4) \in \widehat{\mathbb R}$, i.e. iff $(z_1,z_2,z_3,z_4) = S(z_4) \in \mathbb R \cup \{\infty\}$; excluding the case $S(z_4)=\infty$ (which happens iff $z_4 = z_3$, excluded since points are distinct), we get $(z_1,z_2,z_3,z_4)\in\mathbb R$ iff $z_4 \in \Gamma$, as claimed. $\blacksquare$