Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 6 / b

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.

The prescribed pole lets us choose $c=4$ and $d=-5-3i$. Write $f(z)=(Az+B)/(4z-5-3i)$. Fixing $1/2$ and $2$ givesso $A=5-3i$ and $B=-4$. ThusIts determinant is $-18\ne 0$, so it is a genuine Möbius transformation. Direct substitution gives $f(1/2)=1/2$, $f(2)=2$, and a pole at $(5+3i)/4$; the numerator at that pole is $9/2\ne 0$.

To identify the image of the unit disc, putThen $|\alpha {|}^2=8/17<1$, $|\lambda |=1$, andIn particular, its pole is outside the closed unit disc. The identityshows $f(D)\subseteq D$. For any $|w|<1$, solve $w=\lambda (z-\alpha )/(1-\overline{\alpha }z)$ to getIts denominator is nonzero, and the same modulus identity for this inverse gives $|z|<1$. Thus $f(D)=D$, and global injectivity of a Möbius transformation proves the required one-to-one correspondence.