Fourier series expanding of holomorphic functions Let $f:\mathbb{C}\to \mathbb{C}$ be a holomorphic function with $f(z+1)=f(z)$. Then how to expand it into Fourier series $$f(z)=\sum\nolimits_{n\in \mathbb{Z}} a_n e^{2\pi inz}?$$
In another word, what is $a_n$?
Thanks a lot!
 A: The restriction of $f$ to $\mathbb R$ is periodic, so the $a_n$ are given by the usual formula $$a_n=\int_0^1e^{-2\pi int}f(t)\,dt.$$
Edit: Now the OP asks how we know the series converges to $f$ in the plane. That's a different question. 
(Edit${}^2$: To be fair, reading the question again I suppose this question really is part of the original post - I was concentrated on the last sentence of the question.)
There are at least two ways to prove this, one by elementary Fourier analysis and one by not-quite-so-elementary complex analysis. They both seem worthwhile:
Proof by Fourier analysis:
First recall a bit of elementary Fourier analysis:


If $f\in C(\mathbb R)$ is periodic and has an absolutely convergent Fourier series then $f$ is equal to its Fourier series.


Sketch: Say $g$ is the sum of the Fourier series. Then $\hat g(n) = \hat f(n)$ (since the Fourier series converges uniformly), hence by some argument or other $g=f$.


Cor If $f\in C^2(\mathbb R)$ is periodic then $f$ equals its Fourier series.


Proof: Integration by parts shows that $|\hat f(n)|\le c/n^2$.
Now suppose that $f$ is entire and $f(z+1)=f(z)$. Applying the Corollary to $f(.+iy)$ shows that for every $y\in\mathbb R$ there exist  coefficients $a_n(y)$ such that $$f(x+iy)=\sum a_n(y)e^{2\pi i nx};$$in fact $$a_n(y)=\int_0^1f(t+iy)e^{-2\pi i nt}\,dt.$$For $z\in\mathbb C$ define $$g_n(z)=\int_0^1f(z+t)e^{-2\pi int}\,dt,$$so that in particular $$a_n(y)=g_n(iy).$$
It's easy to see that $g_n$ is an entire function (differentiate under the integral, or apply Morera and Fubini.) If $x\in\mathbb R$ then periodicity shows that $$\begin{aligned}g_n(x)&=\int_0^1f(x+t)e^{-2\pi int}
\\&=\int_x^{x+1}f(t)e^{-2\pi in(t-x)}\,dt
\\&=e^{2\pi inx}\int_x^{x+1}f(t)e^{-2\pi int}\,dt
\\&=e^{2\pi inx}a_n(0).\end{aligned}$$Sine $g_n$ and the exponential are both entire it follows that $$g_n(z)=e^{2\pi inz}a_n(0),$$or in particular
$$a_n(y)=g_n(iy)=e^{-2\pi ny}a_n(0).$$
So $$a_n(y)e^{2\pi inx}=e^{2\pi in(x+iy)}a_n(0),$$ so the Fourier expansion of $f(.+iy)$ above reduces to $$f(z)=\sum a_n(0)e^{2\pi inz}.$$
Note of course a similar result holds, with the same proof, for functions holomorphic in any horizontal strip $A<y<B$.
The proof by complex analysis seems quicker, but it's perhaps less elementary or more subtle:
Proof by complex analysis:
Say $\mathbb C'=\mathbb C\setminus\{0\}$. Since $f(z+1)=f(z)$ there exists $g\in H(\mathbb C')$ such that $$f(z)=g(e^{2\pi iz}).$$
(Why? I don't want to talk about it in detail - we already have another proof, and in fact this follows from the result itself. The existence of $g$ is clear from basic facts about covering maps. Or, more or less equivalently, you can use a branch of $\log(z)$ to show there exists such a $g$ locally and then show it has an analytic continuation to all of $\mathbb C'$.
Not to cite myself, but the only reference I happen to know for the covering map argument is Theorem 19.0.9 in Complex Made Simple - also note Corollary 19.0.10.)
Now expand $g$ in a Laurent series and you're done.
