= Solution
Parametrize the continued branch by $t=re^{i\theta}$, $0\le\theta\le2\pi$. Then $t^{z-1}dt=ir^z e^{iz\theta}d\theta$, giving
$$
\boxed{f(z)=r^z\frac{e^{2\pi iz}-1}{z}\quad(z\ne0),\qquad f(0)=2\pi i.}
$$
The apparent singularity at zero is removable, so $f$ is an <entire function>. The factor $r^z=e^{z\log r}$ shows its dependence on the contour radius; for instance $f(1/2)=-4\sqrt r$. The integrand is continued along the contour and generally returns on a different branch, so this is not an ordinary integral of a single-valued holomorphic function around a closed loop.
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