Write $z \sin z = z^2 g(z)$, where $g(z) = \frac{\sin(z)}{z}$. Since $g$ has a removable singularity at $0$ by defining $g(0) = 1$, we can define a single valued square root of $g(z)$ in a neighborhood of $0$. Thus we get $f(z) = z \,\sqrt{g(z)}$.
Write $z \sin z = z^2 g(z)$, where $g(z) = \frac{\sin(z)}{z}$. Since $g$ has a removable singularity at $0$ by defining $g(0) = 1$, we can define a single valued square root of $g(z)$ in a neighborhood of $0$. Thus we get $f(z) = z \,\sqrt{g(z)}$.