I came up with the following idea but I don't have a proof for it.
If $f $ is a meromorphic function on $\mathbb{C}$ with finite poles and zeros such that $\lim_{z\to 0} f(1/z) $ exists on $\mathbb{C}$, $\oint_{| z|=R} \frac{f'}{f}\;dz =0$ for a big enough $R>0 $.
Even though I don't have a proof for this fact, I have good ideas to think it is true. From a fact that I'm trying to prove, the meromorphic functions $ f$ such that $\lim_{z\to 0} f(1/z) $ exists, are only the reational functions which are actually meromorphic functions on $\overline{\mathbb{C}} $. So the previous limit can be thought as $f(\infty) $.
The argument principle told us that $\oint_{| z|=R} \frac{f'}{f}\;dz =N-P$ where $P $ is the number of poles and $ N$ is the number of zeros of $ f$ (with multiplicity), for a big enough $R>0 $. Given that $ f$ will turn out to be a quotient of polynomials, $ N$ will be the degree of the polynomial in the numerator and $P $ the degree of the denominator. Then the only way for $\lim_{z\to 0} f(1/z) $ to be complex is $N=P$ .
This is kind of a proof but I don't want to use the fact of $f $ being rational since that is what I want to prove. So what I'm asking you guys is to prove it with out using that fact, perhaps Laurent series will be useful.
Thank you all in advance