# Regarding Fourier coefficients of entire modular forms and

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 $$\frac {√3} {2}$$ ?

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 $$f(z)=\sum_{n\in \Bbb{Z}} c(n) e^{2i\pi nz}$$.
• imaginary part ${}{}{}{}$ Dec 15, 2019 at 13:38
• 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). Dec 17, 2019 at 5:13
• 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. Dec 17, 2019 at 5:46