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Since $\Gamma$ is not one to one over the complex domain, Is it possible to define some principal values ( analogues to Principal Roots for the Root function ) so we can have a $\Gamma^{-1} (z)$ (inverse $\Gamma$ function)?

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    $\begingroup$ If $f$ is analytic, and $f'(a)\ne 0$, there is a neighborhood of $a$ where $f$ is injective. An appropriate version of the inverse function theorem tells us that $f^{-1}$ is analytic in that neighborhood. Any domain of holomorphy of this function would be an inverse of $\Gamma$. You probably want/need to specify something additional. $\endgroup$ – Andrés E. Caicedo Jan 20 '11 at 23:23
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A short sketch besides the request of Andres' for further specifications:
if you use the inverse-series for $ \Gamma(1+x)-1 $ (in Pari/GP: serreverse (gamma(1+x)-1) ) you get something like

$ f(x) = -1.73245471460*x + 5.14288192485*x^2 - 22.3588922658*x^3 + 120.586032684*x^4 $
$ - 732.743269181*x^5 + 4785.68759665*x^6 - 32793.0682929*x^7 + O(x^8) $

By 64 terms of the series it is not clear, whether the series has finite radius of convergence $ \rho $, possibly it is something like $ \rho \sim 1/9 $. If we define $ g(x) = f(x-1)+1 $ and use Euler-/Noerlund-sums for acceleration of convergence it seems we can arrive at meaningful values in a small range; examples

g(1.5)                  = 0.595332094501  // Noerlund-sum
gamma(  0.595332094501) = 1.500000000
g(1.8)                  = 0.492222531811  // Noerlund-sum
gamma(  0.492222531811) = 1.800000000
g(2.0)                  = 0.442877396485  // Noerlund-sum
gamma(  0.442877396485) = 2.000000000
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