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For a nonidentity Möbius transformation whose two fixed points are antipodal under stereographic projection, conjugate by a sphere rotation sending them to $0,\infty$. The conjugated map fixes both, hence is $z\mapsto az$ with $a\ne 0$. If $|a|=1$ it is an axial rotation; if $|a|<1$ every finite orbit tends to zero; if $|a|>1$ every nonzero orbit tends to infinity. Undoing the rotation gives either a sphere rotation or one attracting fixed point with the other as the sole exceptional initial point.

The Möbius transformation $M(z)=(3z-1)/(3-z)$ fixes the antipodal pair $1,-1$ under stereographic projection. But $M^{\prime }(1)=2$, so it multiplies infinitesimal spherical lengths at that fixed point by two, as follows from the round line element $2|dz|/(1+|z{|}^2)$. It is not an isometry and hence not a sphere rotation. Fixing an antipodal pair is therefore necessary for a nontrivial rotational Möbius transformation, but is not sufficient.

Let a group $G$ of Möbius transformations of the unit disk have an orbit satisfying the Blaschke condition, counted once per distinct orbit point. For its Blaschke product $B$, invariance of the pseudohyperbolic distance and permutation of the orbit give $|B(Tz)|=|B(z)|$. The ratio $B(Tz)/B(z)$ extends across its zeros and has modulus one, so the open mapping theorem makes it a constant $\chi (T)$. Composition gives a group homomorphism $\chi :G\to \{\lambda :|\lambda |=1\}$. It is essential to compare moduli of the products: two arbitrary bounded holomorphic functions with identical zeros need not have constant quotient.

For $|w|<R$, this Möbius transformation maps the radius-$R$ disc to the unit disc, vanishes at $w$ and has modulus one on the boundary. Expanding gives $|R^2-\overline{w}z{|}^2=R^2|z-w{|}^2$ when $|z|=R$; its pole lies outside the disc. Dividing a holomorphic function by factors corresponding to its interior zeros removes those zeros without changing its boundary modulus. This gives a direct factorization proof of Jensen's formula.

This Möbius transformation maps the upper half-plane onto the unit disk. For $\mathrm{Im}z>0$, the identity $1-|C(z){|}^2=4\mathrm{Im}z/|z+i{|}^2$ puts the image inside the unit disk. Its inverse is $z=i(1+w)/(1-w)$, so the map is a conformal bijection. On the extended real boundary it maps onto the unit circle. It is the scalar complex-analytic version of a Cayley map, rather than the matrix parametrization used in Cayley transform.

Two disjoint ordinary circle boundaries can be taken to distinct concentric circles by a Möbius transformation. After their centres are put at $0,d$ with radii $a,b$, the map $(z-s)/(z-t)$ works when $st=a^2$ and $s+t=(d^2+a^2-b^2)/d$. Disjointness makes the quadratic have distinct real roots. Already concentric circles require no transformation.

Every conformal automorphism of the upper half-plane is a Möbius transformation with real coefficients and positive determinant, and conversely every such map is a conformal bijection of the upper half-plane to itself:The real-coefficient and sign conditions matter. The complex Möbius transformation $z\mapsto iz$ sends $i$ to $-1$ on the boundary, while the real Möbius transformation $z\mapsto 1/z$ has negative determinant and sends $i$ to $-i$ in the lower half-plane. ^[ref]

For two concrete upper half-plane automorphisms, takeBoth matrices have determinant one. Here $T(i)=1+i$, $S(i)=i$, and $S(1+i)=(-1+i)/2$. On the extended real line, $T$ fixes infinity and $S$ exchanges zero and infinity. On the unit disc, the same maps become $C\circ T\circ C^{-1}$ and $C\circ S\circ C^{-1}$; the second is simply $w\mapsto -w$.

Multiplying all four coefficients by a nonzero real scalar leaves $f$ unchanged. Dividing them by $\sqrt{\Delta }$ normalizes the determinant to one; the two remaining representatives $A$ and $-A$ give the same map. Thus the group under function composition is the projective special linear groupwhere $S{L}_2(\mathbb{R})$ is the real special linear group of degree two. Composition of the Möbius transformations corresponds to multiplication of their matrices. ^[ref]

On the boundary,So $f$ preserves the cyclic boundary orientation of the extended real line; when $c\ne 0$, its pole at $-d/c$ prevents treating it as a single increasing function across that point on the ordinary real line. Three distinct boundary images determine a Möbius transformation uniquely, but an upper half-plane automorphism can realize only triples with matching cyclic boundary orientation.

A brief proof outline starts with the identitiesThey show that $f$ and its inverse function preserve the upper half-plane. For completeness of the classification, conjugate an arbitrary conformal bijection by the Cayley transform. Normalize its image of zero with an automorphism of the unit disk, then apply the Schwarz lemma to the resulting map and its inverse function; both inequalities force a rotation. Conjugating back gives the real Möbius transformation above.

Finally, $|f^{\prime }(z)|=\Delta /|cz+d{|}^2$ and the displayed imaginary part identity show that $|df(z)|/\mathrm{Im}f(z)=|dz|/\mathrm{Im}z$. Every upper half-plane automorphism is therefore also an isometry of the Poincaré half-plane model. The same formulas describe both holomorphic coordinate symmetries and motions of the hyperbolic plane.

The mapmaps $\{z:|z|<1,\mathrm{Im}z>0\}$ conformally onto the upper half-plane. The Möbius transformation maps the half-disc onto the first quadrant, and squaring maps that quadrant onto the upper half-plane.

Two intersecting circles bound a lens. A Möbius transformation sending their intersection points to zero and infinity sends the two boundary arcs to rays, and the lens to a sector. A suitable power straightens the sector to a half-plane. The chosen sector and its branch must be identified by an interior test point; merely squaring an arbitrary boundary-straightening map need not be injective on the intended domain. A Cayley transform then gives a conformal bijection to a disc.

On $\mathbb{C}\setminus [-1,1]$, take the principal square root with positive real part. The Möbius transformation $(z-1)/(z+1)$ has image $(\mathbb{C}\setminus (-\infty ,0])\setminus \{1\}$. The principal square root therefore maps to the right half-plane with $1$ omitted, and the final Möbius transformation gives exactly $0<|w|<1$. The inverse is $z=-(w+w^{-1})/2$. Adding $\infty$ to the source on the Riemann sphere fills the missing value $w=0$ and gives a conformal map onto the full disk. Without $\infty$, a loop surrounding the segment has nonzero winding number, so the source is not a simply connected space and cannot be conformally equivalent to a disk.

The intersection of the unit disks centered at $1/2\pm i\sqrt{3}/2$ has endpoints zero and one. The Möbius transformation $w=z/(1-z)$ maps it onto the sector $|\arg w|<\pi /6$: the two disk conditions become $\mathrm{Re}(e^{\pm i\pi /3}w)>0$. Cubing gives the right half-plane, and the Cayley transform between the half-plane and disk gives the displayed holomorphic bijection to the unit disk. The sector width ensures that cubing is injective.

Normalize three distinct points on a generalized circle by a Möbius transformation $f$ taking them to $1,0,\infty$. The inverse image of the extended real line is their generalized circle, and $J=f^{-1}\circ c\circ f$, with $c(z)=\overline{z}$, fixes that circle pointwise and squares to the identity. Changing the three normalizing points replaces $f$ by $M\circ f$ with a real-coefficient Möbius map $M$. Since $M$ commutes with conjugation, $J$ is unchanged. Thus this construction depends only on the circle. It is inversion in a circle for an ordinary circle and Euclidean reflection for a straight line, each extended to the Riemann sphere.

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 $1-r>1$.

An even elliptic function descends through the degree-two quotient map $\wp$ to a rational function $R$ on the Riemann sphere. If the original function also has degree two, composition of map degrees gives $2=2\mathrm{deg}R$, so $R$ has degree one. Consequently it is a Möbius transformation with $ad-bc\ne 0$. The shift $\mathrm{sn}(z+K,k)$ is even and has degree two, providing an example for the common period lattice of the Jacobi elliptic sine and $\wp$.

For a regular nonconstant J-holomorphic sphere in a $2n$-dimensional target, the parametrized Fredholm index is $2n+2c_1(A)$. Removing the six-dimensional Möbius transformation reparametrization group gives the displayed local moduli dimension. The quotient is a manifold for a simple J-holomorphic sphere and may be an orbifold for a multiple cover with finite stabilizer. In dimension four the formula is $2c_1(A)-2$.

A real Möbius transformation represented by $A\in {\mathrm{SL}}_2(\mathbb{R})$ with $|\mathrm{tr}A|<2$. It fixes a point of the Poincaré half-plane model; an integral matrix of this type has trace $0$ or $\pm 1$. In the Poincare disc model, moving its fixed point to zero makes it a rotation about zero.

A nonidentity Möbius transformation of finite order $n$ is conjugate to multiplication by a primitive $n$th root of unity. Zero and infinity are fixed; for any other point, a return after $k$ iterates requires ${\mu }^k=1$, whose least positive solution is $n$. Thus its only orbit sizes are one and $n$. For the identity, every orbit has size one. This is a direct orbit computation, independent of an appeal to the orbit-stabilizer theorem.

Four positive-radius circles with pairwise disjoint interiors, successive external tangencies, and four distinct contact points have those contact points on a circle. Apply the reciprocal Möbius transformation $w=1/(z-T)$ at one contact $T$. The two circles through $T$ become disjoint parallel lines; the other two disks lie between them and avoid zero. Collinearity for tangent circles between parallel lines places the three finite transformed contacts on a straight line. That straight line misses zero: otherwise its segment across the strip would pass through zero, while that segment is contained in the two transformed closed disks. Its inverse is therefore a genuine circle through $T$, containing all four original contacts. Allowing internal tangency without the pictured disjoint-interior hypotheses admits collinear degeneracies instead.

The Weierstrass elliptic function has degree two on its period torus. Its three distinct half-period values $e_j$ have double local preimages, and its pole is double, so these four values are totally ramified. This also follows from ${\wp }^{\prime 2}=4{\prod }_j(\wp -e_j)$ and the principal part at the pole. A Möbius transformation with pole away from these values moves all four to finite values without changing local degrees.

For primary operators of holomorphic weights $h_i$ and a Möbius transformation invariant vacuum, infinitesimal translations, dilations and special conformal transformations give ${\sum }_i{\partial }_{i}G=0$, ${\sum }_i(z_i{\partial }_i+h_i)G=0$, and ${\sum }_i(z_i^2{\partial }_i+2h_i{z}_i)G=0$. Solving gives a constant one-point function only at $h=0$, a two-point function $C{z}_{12}^{-2h}$ only for equal weights, and a three-point function $C{z}_{12}^{h_3-h_1-h_2}{z}_{13}^{h_2-h_1-h_3}{z}_{23}^{h_1-h_2-h_3}$. Three-point covariance imposes no further equality of weights; dynamical selection rules may still force $C=0$. For more points, arbitrary functions of conformal cross-ratios remain.