= Solution
{c}
Write $\Delta=ad-bc>0$. For real $a,b,c,d$ and $z\in H$, the denominator cannot vanish: if $c\ne0$ its zero $-d/c$ is real, and if $c=0$ then $d\ne0$. Taking the <imaginary part> after multiplying by $\overline{cz+d}$ gives
$$
\boxed{\operatorname{Im}\frac{az+b}{cz+d}=\frac{\Delta\operatorname{Im}z}{|cz+d|^2}>0.}
$$
Thus $f(H)\subseteq H$. Also
$$
f'(z)=\frac\Delta{(cz+d)^2}\ne0,
$$
so it is <conformal>. Cross-multiplication shows that $f(z_1)=f(z_2)$ implies $\Delta(z_1-z_2)=0$, proving <injectivity>.
The inverse <Möbius transformation> is
$$
f^{-1}(w)=\frac{dw-b}{-cw+a}.
$$
Its coefficients are real and its <determinant> is again $\Delta>0$, so the same <imaginary part> identity shows that it maps $H$ into $H$. Therefore every $w\in H$ has a preimage in $H$, proving <surjectivity>. Hence $f$ maps $H$ one-to-one and onto itself.
Back to article page
