 Solution

Write $\Delta =ad-bc>0$. For real $a,b,c,d$ and $z\in H$, the denominator cannot vanish: if $c\ne 0$ its zero $-d/c$ is real, and if $c=0$ then $d\ne 0$. Taking the imaginary part after multiplying by $\overline{cz+d}$ givesThus $f(H)\subseteq H$. Alsoso 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 isIts 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.