I have a function with a relationship between $f(z)$ and $f(1/z)$ that, when interpreted naively using the series for $f$, gives wrong results. I'm going to give this naive argument in the hopes that someone can point out very clearly where the problem is. I realize that there are subtle issues of convergence and several implicit assumptions here, but I'm not sure which one fails.
Suppose that I have a function $f$ that is analytic in a neighborhood of $z=0$, so
$$ f(z) = \sum_{n=0}^\infty f_n z^n $$
I'm omitting Considering $f(1/z)$, we get
$$ f\left( \frac{1}{z} \right) = \sum_{n=0}^\infty f_n z^{-n} = \sum_{n=-\infty}^0 f_{-n} z^n $$
The first equality here is a purely symbolic substitution, so I think it is okay. The second is a bit fishy because it rearranges the terms in the sum, but I think that the uniqueness of the expansion here should make this okay. (Please point out if this is wrong.)
Now consider a particular case where there is a relationship between $f(z)$ and $f(1/z)$. A simple example is
$$ f(z) = \left( \frac{1-z}{1+z} \right)^\alpha $$
Assuming we do branch cuts from $\pm 1$ to $\pm \infty$, then this is analytic near zero and has a radius of convergence of $1$. Now consider that
$$ f\left(\frac{1}{z}\right) = (-1)^\alpha f(z) $$
(Invoking $(-1)^\alpha$ is slightly fishy here because it is on the branch cut, but we are not using $z=-1$, so this may be okay.)
If we put some of the equations above together, we get
$$ \sum_{n=0}^\infty f_n z^n = \sum_{n=-\infty}^0 (-1)^\alpha f_{-n} z^n $$
Now I will equate these series term-by-term. This is a very fishy operation, but could be justified by saying that both of these expressions could be interpreted as the Laurent series of $f(z)$, and Laurent series are supposed to be unique. Obviously, equating them gives the result
$$ f_n = (-1)^\alpha f_{-n} $$
for $n \ge 0$, which is clearly wrong because in actuality, for this case, $f_n = 0$ for $n < 0$ and $f_n \ne 0$ for $n \ge 0$.
I've identified a few fishy steps above, and if someone could explain exactly why they fail, or if there is another implicit assumption that I have not considered, that would be great.
Ultimately, my goal in thinking about this was to try to prove whether a different function $f(z)$ was a polynomial (i.e., the series terminates) or not by looking at a relationship with $f(1/z)$. If anyone has any thoughts on this, please feel free to share them too, but in the meantime, this "contradiction" became interesting in its own right.
You consider Laurent series of $f$ on the punctured disk $0<|z|<1$ and write it as $\sum_{n=0}^{\infty}f_nz^n$. Then, it is true that $f(1/z) = \sum_{n=0}^{\infty}f_nz^{-n}$, but not on the punctured disk, but the annulus $|z|>1$! So, even if it is true that $f(1/z) = (-1)^\alpha f(z)$, you can't equate $\sum_{n=0}^\infty f_n z^n = \sum_{n=-\infty}^0 (-1)^\alpha f_{-n} z^n$ since they don't converge on the same annulus.