For a power series $f(z) = \sum\_{i=0}^{\infty} a\_i z^i$ with $a\_1$ nonzero, [Lagrange's inversion formula](http://en.wikipedia.org/wiki/Lagrange_inversion_theorem) gives an explicit way to compute the Taylor coefficients of the inverse function.


Is there any analogous formula for Laurent series?



Inversion of Laurent series
The Lagrange inversion formula is meant to give you the Taylor series expansion of $f^{-1}$ at the point $f(0)$. If $f$ has a Laurent series instead, then it means that $f(0) = \infty$ and that $f$ is meromorphic. The Taylor series at $\infty$ of $f^{-1}$ then doesn't particularly mean anything unless you change to a different coordinate patch on the Riemann sphere, for instance $\zeta = 1/z$. So you can first switch to the function $1/f$, which has a usual Taylor series, and then use the standard Lagrange inversion formula for $(1/f)^{-1}$.


(If I have understood the question correctly. Maybe this answer is too straightforward to address the real question.)


