= Conformal automorphism of the upper half-plane
{title2=$\phi(z)=(az+b)/(cz+d)$}
= Upper half-plane automorphism
{synonym}
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:
$$
f(z)=\frac{az+b}{cz+d},\qquad a,b,c,d\in\mathbb R,\qquad \Delta=ad-bc>0.
$$
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> $z\mapsto1/z$ has negative <determinant> and sends $i$ to $-i$ in the <lower half-plane>. https://math.ucdavis.edu/~romik/data/uploads/teaching/math205a-2018/mat205a-2018-hw4.pdf[Romik, Complex Analysis, Homework 4, page 3, automorphism statement 5]{ref}
For two concrete <upper half-plane automorphisms>, take
$$
T(z)=z+1,\quad A_T=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad S(z)=-\frac1z,\quad A_S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
$$
Both <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 group>
$$
\operatorname{Aut}(\mathbb H)\cong PSL_2(\mathbb R)=SL_2(\mathbb R)/\{I,-I\},
$$
where $SL_2(\mathbb R)$ is the <real special linear group of degree two>. Composition of the <Möbius transformations> corresponds to multiplication of their <matrices>. https://warwick.ac.uk/fac/sci/maths/people/staff/caroline_series/hyperbolic_geometry_ma448_lecture_notes.pdf[Series, section 1.3, Theorem 1.27]{ref}
On the <boundary>,
$$
f'(x)=\frac{\Delta}{(cx+d)^2}>0\qquad(cx+d\ne0).
$$
So $f$ preserves the cyclic <boundary orientation> of the extended <real line>; when $c\ne0$, 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 identities
$$
\operatorname{Im}f(z)=\frac{\Delta\operatorname{Im}z}{|cz+d|^2},\qquad f^{-1}(w)=\frac{dw-b}{-cw+a}.
$$
They 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, $|f'(z)|=\Delta/|cz+d|^2$ and the displayed <imaginary part> identity show that $|df(z)|/\operatorname{Im}f(z)=|dz|/\operatorname{Im}z$. 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>.
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