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I am self studying Analytic number theory from Tom M Apostol and I cannot think about a deduction in proof of a theorem in Chapter 6 ( Modular forms with multiplicative coefficients) .

In section 6.15 - Estimates for the Fourier coefficients of Entire forms

Apostol assumes f to be an entire form with Fourier expansion

f($\tau$) = $\sum_{n=0}^\infty = c(n) x^n $ . ----(1)

Then later in same page Apostol mentions

The series in (1) converges absolutely if |x| <1 .

I am unable to think why f($\tau$) converges absolutely for |x|<1 although there is no restriction on coefficients c(n) ? Why does value of coefficients c(n) doesn't matters and what theorem or result is Apostol actually using to deduce this.

My second doubt is -- Apostol later writes If $\tau $ is in fundamental region of $\Gamma$ then $\tau$ = u +iv with v $\ge$ $\frac {√3} {2} $

Can somebody please tell why v couldn't be less than $\frac {√3} {2} $ ?

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The word "entire form" doesn't have any meaning. If $f(z)$ is analytic and $1$-periodic on $\Im(z) > 0$ then $F(q)=f(\frac{\log q}{2i\pi }) $ is analytic on $0<|q|< 1$, if $f$ is bounded on $\Im(z) > 1$ then $q^2 F(q)$ is holomorphic thus analytic on $|q|<1$ thus (Cauchy integral formula) it is equal to its Taylor series $F(q)=\sum_{n\ge 0} c(n)q^n$ which converges, thus absolutely, for $|q|<1$ and $f(z)=\sum_{n\ge 0} c(n) e^{2i\pi nz}$. Without the boundedness assumption we have the Laurent series $F(q)=\sum_{n\in \Bbb{Z}} c(n)q^n$ which converges absolutely for $0<|q|< 1$ and $f(z)=\sum_{n\in \Bbb{Z}} c(n) e^{2i\pi nz}$.

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  • $\begingroup$ @ reuns can you please tell what does symbol in 3rd line of your answer after periodic on... stands for? I followed a book on complex analysis which never mentioned such a symbol $\endgroup$
    – user775699
    Dec 14, 2019 at 8:34
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    $\begingroup$ imaginary part ${}{}{}{}$ $\endgroup$
    – reuns
    Dec 15, 2019 at 13:38
  • $\begingroup$ can you please explain how F(q) = f(log q/ 2πi) is analytic in 0<|q|<1. I got F(q) = c (n) exp(( logq) n) . But how to deduce that it is analytic in 0<|q|<1? $\endgroup$
    – user775699
    Dec 16, 2019 at 13:07
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    $\begingroup$ What do you not understand exactly ? And $\sqrt{3}/2$ is because of the fundamental region (for $SL_2(\Bbb{Z)\setminus H}$) we chose to be $|\Re(z)|\le 1/2, |z|\ge 1$. There are other books than Apostol (Diamond and Shurman). $\endgroup$
    – reuns
    Dec 17, 2019 at 5:13
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    $\begingroup$ If $F$ is bounded around $0$ then $g(q)=q^2F(q) = O(q^2)$ thus $g'(0)=0$ thus $g$ is holomorphic on $|q|<1$ thus (Cauchy integral formula) it is analytic, $F(q)=g(q)/q^2$ doesn't have a pole because it is bounded thus it is analytic. The last step is the Cauchy integral formula showing its Taylor series converges on the whole unit disk. A power series converges absolutely on its disk of convergence. $\endgroup$
    – reuns
    Dec 17, 2019 at 5:46

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