# 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?

• This is essentially because $\tilde{f}(z) = f(\frac{1}{2\pi i}\log z)$ is well-defined. This also shows that $\tilde{f}$ will inherit any nice properties of $f$ (such as continuity, analyticity, etc.). Jul 31, 2021 at 20:13
• @SangchulLee why do you say this is well defined? Is it following from f(z+1)=f(z)? Aug 7, 2021 at 8:17

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.
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.