Analytic continuation by contour deformation

If a parameter-dependent contour integral has a moving pole, deforming the integration contour continuously so that the pole never crosses it preserves a holomorphic branch. Returning the contour to its original path after crossing a pole adds or subtracts the corresponding residue contribution.

For a figure-eight contour around $1$ anticlockwise and $-1$ clockwise, start $(t^2-1{)}^z$ at $t=0$ with argument $-\pi$. Shrinking the two loops gives the displayed identity. If the interval integral instead uses $(t^2-1{)}^z=e^{-i\pi z}(1-t^2{)}^z$, the corresponding identity includes the extra factor $e^{i\pi z}$. Omitting this phase changes the value, rather than merely the locations of the meromorphic continuation's poles.