Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 / 4

Let $h=f^{-1}$, $w_0=f(z_0)$ and $z=h(w)$. Continuity of $h$ gives $z\to z_0$ as $w\to w_0$, and injectivity gives $z\ne z_0$ when $w\ne w_0$. ThereforeThe assumed nonvanishing derivative makes this limit well defined. It exists at every $w_0\in V$, so $h$ is holomorphic, with $\boxed{h^{\prime }(w)=1/f^{\prime }(h(w))}$.

Squaring maps the first quadrant bijectively onto the upper half-plane. For $\zeta$ in that half-plane, $|\zeta -i|<|\zeta +i|$, and the Cayley transform $(\zeta -i)/(\zeta +i)$ maps it bijectively onto the unit disc. Their composition isThe inverse is the unique first-quadrant square root of $i(1+w)/(1-w)$, and is holomorphic by the first argument. Hence $H(w)=g(F^{-1}(w))$ is holomorphic throughout the disc. Its Taylor series at zero converges for $|w|<1$, locally uniformly there. Substituting $w=F(z)$ gives the stated expansion, with $c_n=H^{(n)}(0)/n!$, convergent at every point of the quadrant and locally uniformly on it.

If a holomorphic bijection existed, its inverse would be an entire function taking values in the unit disc. Liouville theorem would make that inverse constant, contradicting bijectivity. Thus no holomorphic bijection exists.

A continuous bijection does exist, even a homeomorphism:Both maps preserve the argument and change the radius by the mutually inverse strictly increasing functions $r\mapsto r/(1-r)$ and $s\mapsto s/(1+s)$. The displayed formulas also prove continuity at zero and everywhere else.