= Solution
Use the reciprocal <Möbius transformation>
$$
\boxed{F(z)=\frac1z.}
$$
It is a <holomorphic function> on $U_1$ by part (b). For $z\neq0$,
$$
|F(z)+i|=\frac{|1+iz|}{|z|}=\frac{|z-i|}{|z|}.
$$
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 inverse
$$
\boxed{F^{-1}(w)=\frac1w.}
$$
The inverse is also holomorphic, so this is a <biholomorphism>.
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