Does a branch cut discontinuity determine a function near the branch point? Suppose $g(z)$ is analytic on a disc centered at the origin, except along the negative real axis where it has a branch cut discontinuity. Also assume that $g(0)=0$. Let $h(x)$ give the discontinuity across the branch cut of $g(z)$ at $z =-x$.  The question is to what extent does $h(x)$ determine $g(z)$.  
It is clear that adding a function that is analytic everywhere on the disc does not affect $h(x)$, since the discontinuity for such a function would be zero.  But is $g(z)$ otherwise determined by $h(x)$, up to this ambiguity?  
I made a little progress thinking about integrating $g(z)$ around a contour that follows a circle $C(r)$ around the origin from $r e^{-i \pi}$ to $r e^{i\pi}$, and then follows the branch cut to the origin, and then goes back along the lower half of the branch cut.  This then gives the relation
$$\int_{C(r)}  g(z) dz = \int_0^r h(x) dx$$ 
but I was unable to see if I could then invert this relation to get just $g(z)$.  
 A: I believe I was able to construct the answer, and yes, it appears that the branch cut discontinuity determines the function up to addition of analytic pieces.  
This can be shown by applying Cauchy's integral formula.  For a point $a$ in the disc minus the negative real axis, we write 
$$g(a) = \frac{1}{2\pi i} \oint dz \frac{g(z)}{z-a}$$
where the contour circles $a$ counterclockwise while remaining in the disc and avoiding the negative real axis.  We can then deform the contour into the following keyhole configuration (the orientation in this image is opposite our choice): 

By shrinking the inner circle to a small radius and pushing the horizontal segments all the way to the real axis, this formula then defines an analytic function of $a$ in the disc minus the branch cut.  Expanding out the different terms in this expression, we have
$$g(a) = \frac{1}{2\pi i}\left(-\int_0^rdx \frac{h(x)}{x+a} + \int_{C_r} dz \frac{g(z)}{z-a}\right)$$
where $r$ is the radius of the disc, $h(x)$ gives the branch cut discontinuity, and the contour $C_r$ in the second integral is the outer circular contour at radius $r$, going from $-\pi<\arg(z)<\pi$.  
The final step is to realize that the $C_r$ integral gives an analytic function of $a$ over the entire disc (including the negative real axis), since the contour avoids the pole at $z=a$.  Hence, we arrive at the desired formula for $g(a)$:
$$g(a) = -\frac{1}{2\pi i}\int_0^r dx \frac{h(x)}{x+a} + f(a)$$
where $f(a)$ is an arbitrary function that is analytic in the disc.  
