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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?

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    $\begingroup$ 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.). $\endgroup$ Jul 31, 2021 at 20:13
  • $\begingroup$ @SangchulLee why do you say this is well defined? Is it following from f(z+1)=f(z)? $\endgroup$ Aug 7, 2021 at 8:17

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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.

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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.

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