I am following the proof of theorem 1.3 above until at the point the author finds the recurrence relationship $a_n = a_{n-k}e^{b - 2\pi ni\tau}$.
When $k = 0$ we have $a_n = a_{n}e^{b - 2\pi ni\tau} \implies a_n(1 -e^{b - 2\pi ni\tau})
= 0$. Therefore $a_n = 0$, that I understand.
My question is why is $a_n \neq 0 $ for at most one $n$? Isn't $a_n = 0$ for all $n$?
My second question is why in this case we have $f(z) = e^{2\pi niz}$? . My guess is this is one of the possibility of the case when $a_n \neq 0 $, specifically when $a_1 = 1$.
