Show that an analytic function in a strip has a complex Fourier expansion I am a self-studier, and this is a homework problem from a course in Complex Analysis. 
First, let me give a plug for the course as it is outstanding. Taught by Jerry Shurman at Reed. Great lecture notes and illustrative hw problems. http://people.reed.edu/~jerry/311/lec.html
Suppose $f(z)$ is analytic in the strip $a<y<b$ (where $z=x+iy$) and satisfies $f(z+1)=f(z)$.
Show that $f(z)$ has a complex Fourier expansion:
$$f(z)= \sum_{n=-\infty}^{+\infty} a_{n}e^{2\pi inz}$$
converging at all points of this strip where
$$a_n = e^{\pi n(a+b)}\int_{x=0}^{1}e^{-2\pi inx}f(x+\frac{a+b}{2}i)dx$$
Suggestion: Consider the mapping $w=e^{2\pi iz}$. Show that f(z) determines an analytic function in the appropriate region, and use the Laurent expansion. Indicate where the hypothesis $f(z+1)=f(z)$ is needed.
I have looked at several of the posts here relating Laurent and Fourier series, but would still appreciate help.
What I have done so far, which also includes the partial answer below:
-- Show that $w=e^{2\pi iz}$ satisfies $f(z+1)=f(z)$
$w=e^{2\pi i (z+1)}+e^{2\pi i(x+1+iy)}=e^{2\pi i}e^{2\pi i (x+iy)}= e^{2\pi iz}$
-- Using the summation for $f(z)$ as above, it can be represented in an annulus about the origin bounded by $|w_a| = |e^{2\pi i(x+ia)}|=e^{-2\pi a}$ and $|w_b|= e^{-2\pi b}$ where $|w_a|>|w_b|$. But assuming this shows $f(z)$ determines an analytic function, I am not sure how to then correctly apply the Laurent series as suggested.
 A: Thanks to Gamelin, "Complex Analysis," pages 182-183, here is a solution to the first part establishing the power series for $f(z)$.
Let $w=e^{2\pi iz}$ then $z= \frac{1}{2\pi i}\log|w|+\frac{\arg w}{2\pi}$
Set $g(w)=f(z)$ with $z$ as above. Since $f(z+1)=f(z)$, $g(w)$ does not depend on the choice of $\arg w$. Then $g(w)$ is analytic on the annulus as established in the question: $e^{-2\pi b}<|w|<e^{-2\pi a}$.
Thus can expand $g(w)$ as a Laurent series $\sum a_n w^n$, giving the desired $f(x)=\sum_{n=-\infty}^{+\infty}a_n e^{2\pi i n z}$.
To obtain $a_n$ by means of the integral as required by the question, let the path of integration be the horizontal line $y=x+\frac{a+b}{2}$ which runs along the middle of the strip. Due to the cyclicality $f(z+1)=f(z)$, the integral need only be taken from $x=0$ to $x=1$.
Multiplying the Laurent series for $f(z)$, now $f(x+i\frac{(a+b)}{2})$ by $e^{-2\pi inz}$ which equals $e^{2\pi n\frac{(a+b)}{2}}e^{-2\pi inx}$ and integrating (actually switching the order of integration and summation which is allowed due to uniform convergence) and integrating, gives the desired result, since the integral of each term where the index does not equal $n$ is zero.
