= Solution
{c}
The <quadratic formula> gives two algebraic possibilities. To select one consistent <holomorphic square root>, set
$$
S(\zeta)=\sqrt{\zeta-2}\,\sqrt{\zeta+2},\qquad \operatorname{Im}\zeta>0,
$$
where each factor is the <principal square root>. Both arguments lie in the <upper half-plane>, so neither crosses its <branch cut>. This choice satisfies $S(\zeta)^2=\zeta^2-4$ and $S(\zeta)\sim\zeta$ at large $\zeta$ in the <upper half-plane>. The required <inverse function> is
$$
\boxed{z=g^{-1}(\zeta)=\frac{\zeta+S(\zeta)}2
=\frac{\zeta+\sqrt{\zeta-2}\sqrt{\zeta+2}}2}.
$$
For $\zeta=iR$, $R>0$, this gives $z=i(R+\sqrt{R^2+4})/2$, whose modulus exceeds one. As $\zeta$ varies continuously in the <upper half-plane>, this root cannot cross $|z|=1$, since that would make $\zeta$ real. Hence the selected root always lies in the fluid region. The other root is its reciprocal and lies inside the unit <disk>. Simply writing the <principal square root> of $\zeta^2-4$ would not specify this branch correctly throughout the <upper half-plane>.
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