Automorphisms of the upper half-plane are real Mobius maps

The Cayley transform $C(z)=(z-i)\mathrm{/}(z+i)$ is Möbius by [F4], hence biholomorphic by [F3]; the identities $\mathrm{\mid }z+i{\mathrm{\mid }}^2-\mathrm{\mid }z-i{\mathrm{\mid }}^2=4\mathrm{Im}z$ and $C^{-1}(\zeta )=i(1+\zeta )\mathrm{/}(1-\zeta )$ show that $C$ maps $\mathbb{H}$ biholomorphically onto $\mathbb{D}$.

Assume $f\in \mathrm{Aut}(\mathbb{H})$. Let $\alpha =f^{-1}(i)=u+iv$ with $v>0$, and define $m(z)=(z-u)\mathrm{/}v$; this is a real Möbius automorphism of $\mathbb{H}$ with $m(\alpha )=i$, so $h:=f\circ m^{-1}$ is an automorphism of $\mathbb{H}$ fixing $i$.

The map $G:=C\circ h\circ C^{-1}$ is an automorphism of $\mathbb{D}$ fixing $0$, so [F2] gives $G(\zeta )=e^{i\theta }\zeta$ for some real $\theta$. Conjugating back and simplifying with $1\pm e^{i\theta }=2e^{i\theta \mathrm{/}2}(\cos (\theta \mathrm{/}2)\mp i\sin (\theta \mathrm{/}2))$ gives $h(z)=(\cos (\theta \mathrm{/}2)z+\sin (\theta \mathrm{/}2))/(-\sin (\theta \mathrm{/}2)z+\cos (\theta \mathrm{/}2))$, which has real coefficients and determinant $1$; since $m$ also has real coefficients and positive determinant $v$, the composition $f=h\circ m$ is a real Möbius map with positive determinant.

Conversely, if $f(z)=(az+b)\mathrm{/}(cz+d)$ with $a,b,c,d\in \mathbb{R}$ and $ad-bc>0$, then $\mathrm{Im}f(z)=((ad-bc)\mathrm{Im}z)\mathrm{/}\mathrm{\mid }cz+d{\mathrm{\mid }}^2>0$ for $z\in \mathbb{H}$, so $f[\mathbb{H}]\subseteq \mathbb{H}$; its inverse $(dz-b)\mathrm{/}(-cz+a)$ has the same form with real coefficients and positive determinant, so $f\in \mathrm{Aut}(\mathbb{H})$.