= Solution
An <analytic branch of a square root> determined by the branch $\ell$ is
$$
\boxed{\psi(z)=\exp\!\left(\frac12\ell(z)\right)}.
$$
If $\psi_1$ and $\psi_2$ are two such branches, their ratio $h=\psi_1/\psi_2$ is analytic and satisfies $h^2=1$. Since a domain is connected and $h$ takes values in the discrete set $\{1,-1\}$, $h$ is constant. Thus $\psi_1=\psi_2$ throughout $D$ or $\psi_1=-\psi_2$ throughout $D$.
For $z=re^{i\theta}$, the <principal square root> is
$$
\sigma_1(z)=\sqrt r\,e^{i\theta/2},
\qquad -\pi<\theta<\pi.
$$
On $D_2$, one may instead take
$$
\sigma_2(z)=\sqrt r\,e^{i\theta/2},
\qquad 0<\theta<2\pi.
$$
The respective removed half-axes are their <branch cuts>.
On $\mathbb C\setminus[-1,1]$, define
$$
\boxed{\varphi(z)=-iz\,\sigma_1\!\left(1-\frac1{z^2}\right)}.
$$
It is analytic there, its square is $1-z^2$, and
$$
\varphi(2i)=-i(2i)\sqrt{1+\frac14}=\sqrt5,
$$
so it is the required branch. Substitution gives, for $0<|z|<1$,
$$
\varphi(1/z)=-\frac{i}{z}\sigma_1(1-z^2).
$$
The <binomial series> yields
$$
\sigma_1(1-z^2)=1-\frac12z^2-\frac18z^4+O(z^6),
$$
hence the first three terms of the <Laurent series> are
$$
\boxed{\varphi(1/z)=-\frac{i}{z}+\frac{i}{2}z+\frac{i}{8}z^3+O(z^5)}.
$$
Since
$$
g(z)=\frac{\varphi(1/z)}{1+z^2},
$$
its <residue> at zero is $-i$. Under the change of variable $z=1/\zeta$, the two orientation reversals cancel, and the <residue theorem> gives
$$
\boxed{\int_{|z|=2}f(z)\,dz
=\int_{|\zeta|=1/2}g(\zeta)\,d\zeta
=2\pi i(-i)=2\pi}.
$$
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