= Solution
Choose $0<\arg z<2\pi$, so $\sqrt z=\sqrt R\,e^{i\theta/2}$ on the arc. Its clockwise orientation gives
$$
I(R)=i\sqrt R\int_{3\pi/2}^{\pi/2}f(Re^{i\theta})e^{i\theta/2}\,d\theta.
$$
Because $f$ is an <analytic function> at zero, $f(Re^{i\theta})=f(0)+O(R)$ uniformly on the small arc. Integration then gives
$$
I(R)=2\sqrt R f(0)\big(e^{i\pi/4}-e^{3i\pi/4}\big)+O(R^{3/2})=2\sqrt{2R}\,f(0)+O(R^{3/2}).
$$
Therefore
$$
\boxed{\lim_{R\to0}I(R)=0.}
$$
The <estimation lemma> also gives $|I(R)|\leq\pi\sqrt R\max_C|f|$. Hence a sufficient large-arc condition is
$$
\boxed{\max_C|f(z)|=o(R^{-1/2})\quad(R\to\infty).}
$$
For example $f(z)=O(|z|^{-1/2-\eta})$ uniformly on the arc, with $\eta>0$, suffices. Under the stronger literal assumption that $f$ is analytic at infinity, its Laurent expansion is $f(z)=f(\infty)+O(1/z)$; the necessary and sufficient condition is then $f(\infty)=0$. A nonzero constant term would instead give the growing contribution $2\sqrt{2R}f(\infty)$.
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