I want to compute a contour integral say: $\oint_{C_z} \frac{dz}{S(z)-a}$ or $\oint_{C_z} g(S(z))\,dz$. The problem is that I don't have an explicit form for S(z); rather I have the implicit equation $z=f(S)$. $f(S)$ turns out to be a multivalued function with branch cuts (in my case $f(S)$ involves arctan and square roots). I wanted to calculate the integral instead as: $\oint_{C_S} g(S(z))\frac{\partial z}{\partial S} \,dS $ since it seems much simpler (and actually doable). Questions: How limited is the implicit function theorem in this case?Are there any theorems on how the path should be mapped? How do I deal with branch cuts that possibly arise on the path? Can the integral be computed without explicitly knowing the path in $S$ coordinates?
EDIT: As pointed out below, the question boils down to whether the contour in $S$ is closed (or how to make it a closed one) and how to apply the residue theorem (or whether it is possible or not).
There are a couple of issues here. When you know the path $C_S$ in $S$ space, say parametrized as $S(t)$, $0\leq t\leq 1$ then $z(t)=f(S(t))$ and $dz=f'(S(t))S'(t)dt$ so you may insert in e.g. your first contour integral to get
$$ \oint_{C_z} \frac{dz}{S(z)-a}= \int_0^1 \frac{f'(S(t)) S'(t) dt}{S(t) -a} = \int_{C_S} \frac{f'(S)dS}{S-a}$$ where $C_S$ is the path in $S$-space. When $f$ is 'multi-valued' I suspect that $C_S$ need not be closed (even though your contour $C_z$ is). So be careful if you want to apply e.g. a residue theorem. Due to singularities the integral in $S$ form may also pick up residues when changing contour so it is likely necessary to know more about the contour than just the end-points, say. There should be no branch cuts on the contours so you will also need to do analytic continuation along the path (like e.g. the logarithmic 'spiral' if you want to integrate $\log z$ when winding around zero [no cut!]).
As long as derivatives are bounded away from zero and infinity I don't think you have to worry about the implicit function theorem?