Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2006 / 6 / Solution

A holomorphic function with $f^{\prime }(z)\ne 0$ at every point is locally a conformal map. Write $a=z_0$. Since $|a|<1$ and $|z|<1$, $1-\overline{a}z\ne 0$, andThus the specified Möbius transformation is conformal throughout the disk. Expanding the two squared absolute values givesso $f$ maps the open unit disk into itself. Solving $w=(a-z)/(1-\overline{a}z)$ gives $z=(a-w)/(1-\overline{a}w)=f(w)$. Equivalently $f(f(z))=z$. Hence $f$ is its own inverse, proving both injectivity and surjectivity on the disk. Nonvanishing derivative establishes the local angle-preserving property; the explicit inverse supplies the separate global bijection claim.