Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Analysis / 4 / d / Solution

Use the reciprocal Möbius transformationIt is a holomorphic function on $U_1$ by part (b). For $z\ne 0$,Part (c) now gives $z\in U_1$ if and only if $|F(z)+i|>1$, so $F(U_1)\subseteq U_2$. Conversely, $0\notin U_2$ because $|0+i|=1$. For any $w\in U_2$, put $z=1/w$; the same identity gives $|z-i|>|z|$, so $z\in U_1$ and $F(z)=w$. Finally, reciprocation is injective on nonzero complex numbers. This proves bijectivity, with inverseThe inverse is also holomorphic, so this is a biholomorphism.