Why $f(z+1)=f(z)$ implies $f$ can be expressed as a function of $e^{2\pi iz}$ I am reading modular forms from J.P.Serre's book, where I came across a complex function which satisfies property $f(z+1)=f(z)$. Then, it is mentioned that we can express $f$ as a function of $e^{2\pi iz}$. I can see that any function expressed as a function of $e^{2\pi iz}$ always satisfies the above property, but how the converse is true?
 A: Assuming $f$ is holomorphic. This holds because you can represent the function as a Fourier series (which is a sum of powers of the nome $q = e^{2 \pi i z}$), due to its periodicity.
A: Functions that are $1$-periodic like this can be considered as functions on $\Bbb{C} / \sim$, where $\sim$ is the equivalence relation
$$z \sim w \iff z - w \in \Bbb{Z}.$$
If you like, $\Bbb{C}/ \sim$ is the quotient of $(\Bbb{C}, + )$ by the normal subgroup $(\Bbb{Z}, +)$.
Let $g(z) = e^{2\pi i z}$ is a well-defined injection of $\Bbb{C} / \sim$ into $\Bbb{C}$. It is well-defined because it is $1$-periodic. It is injective because
$$e^{2 \pi i z} = e^{2\pi i w} \iff e^{2\pi i (z - w)} = 1 \iff z - w \in \Bbb{Z} \iff z \sim w.$$
As such, there must exist a left inverse
$$h : \Bbb{C} \to \Bbb{C} / \sim,$$
so that $h \circ g$ is the identity on $\Bbb{C} / \sim$.
Suppose $f$ is $1$-periodic. Then $f$ can be considered as function from $\Bbb{C} / \sim$ to $\Bbb{C}$. We can then compose $f \circ h$ to get a map from $\Bbb{C}$ to $\Bbb{C}$. This tells us that
$$f = f \circ \operatorname{Id}_{\Bbb{C} / \sim} = f \circ (h \circ g) = (f \circ h) \circ g,$$
implying $f$ is a function of $g$, as required.
Note that no continuity or analyticity assumptions were assumed, nor concluded. This works purely algebraically.
