Does there exist an infinite sequence of complex numbers $\{ z_i \}_{i=1,2,\cdots}$ such that $$ \boxed{ \sqrt{1-z} = \prod_{i=1}^\infty \left( 1 - \frac{z}{z_i} \right) \qquad \textrm{(for } |z|<1) } \quad ? $$
If such a list exists, it is clear that $|z_i| \geq 1$.
If we take a log, we have $$ \frac{1}{2} \ln(1-z) = \sum_{i=1}^\infty \ln \left( 1 - \frac{z}{z_i} \right). $$ Using the series expansion $\ln(1-z) = - \sum_{k=1}^\infty \frac{z^k}{k}$ and assuming the convergence of the product is nice enough to be able to switch the limits, then by bringing everything to the same side, we obtain $$ \sum_{k=1}^\infty \left( \frac{1}{2} - \sum_{i=1}^\infty \frac{1}{z_i^k} \right) \frac{z^k}{k} = 0. $$ By uniqueness of power series, we thus arrive at a condition on the $z_i$: $$\boxed{ \forall k \in \mathbb N_+: \quad \frac{1}{2} = \sum_{i=1}^\infty \frac{1}{z_i^k} }\; .$$
Does a solution exist? It seems not if we demand that $\sum_{i=1}^\infty \frac{1}{z_i^k}$ is absolutely convergent, but perhaps there is hope beyond that?