Does Stirling formula work with the branch of $\log$ other than the principal branch?

Question

I think the Stirling formula $\Gamma(z) \sim \sqrt{2\pi} z^{z - 1/2} e^{-z} (1 + O(1/z))$ uses the principal branch of $\log$ for $z^{z - 1/2}$. If we take different branch of the function, the formula may give different value for the same $z$, but the gamma function does not depend on the choice of the branch. So the formula works only with the principal branch of $\log$. Is it right?

Answer 1

Yes, this is right. It can be traced to the integral definition of the gamma function,

$$ \Gamma(z)=\int_0^\infty x^{z-1}\mathrm e^{-x}\mathrm dx\;, $$

in which $x^{z-1}$, a power with real base and complex exponent, is defined as $\mathrm e^{(z-1)\log x}$, with the standard real logarithm $\log x$, which corresponds to the principal branch of the complex logarithm.