I am self studying analytic number theory from Tom M Apostol Modular functions and Dirichlet series in number theory and I am unable to think about a step in a proof which uses complex analysis but I am not able to recall how to think about it

Can somebody please give some hint. I am not writing here whole proof, just the step which I can't understand but I will make sure it is absolutely clear to reader.

Step is - Apostol mentions a function f(x) and proves it analytic in D= { x : 0 < |x| <1 } which I can understand. Now since f is analytic, so it must have a Laurent expansion about 0 . So, f(x) = $\sum_{n= -\infty}^\infty a(n)x^n $ where assume x=$ exp(2πit) $ ( it is mentioned in proof why x is equal to this particular value)

Now, I can't understand this statement which appears next in text - ** f(x) is absolutely convergent for each x in D ** .

I tried looking for some related theorem in my complex analysis textbook but couldn't find any.

Can somebody please explain this step.

  • 1
    $\begingroup$ See the discussion preceding proposition $4.1$ at users.drew.edu/capelian/… $\endgroup$
    – saulspatz
    Dec 3, 2019 at 17:33
  • 1
    $\begingroup$ A Laurent series converges absolutely on its (open) annulus of convergence (on the boundary it may diverge or converge at any point). Go back to the Cauchy integral formula to see what the Laurent series and its annulus of convergence come from $$2i \pi f(z) = \int_{|s|=1-\epsilon} -\int_{|s|=\epsilon} \frac{f(s)}{s-z}ds$$ expanding the denominator in geometric series gives the result. $\endgroup$
    – reuns
    Dec 4, 2019 at 17:09
  • $\begingroup$ @reuns I think there has been a typo as 1 integrand is blank $\endgroup$
    – user775699
    Dec 4, 2019 at 17:18
  • $\begingroup$ No this is a common notation in complex analysis : the integrand is the same over each circle but they are traversed in opposite direction $\endgroup$
    – reuns
    Dec 4, 2019 at 17:37
  • $\begingroup$ @reuns ohh, sorry i didnt knew about it $\endgroup$
    – user775699
    Dec 4, 2019 at 19:30

1 Answer 1


The Laurent series of $f$ has the form

$$f(z)=\sum_{n=-\infty}^{\infty} a_nz^n,\,\,0<|z|<1.$$

You can write $z=re^{it},0<r=|z|<1,t\in \mathbb R.$ In any decent treatment of Laurent expansions it is shown that such a series converges absolutely at each such $z.$ Also shown: the series converges uniformly on compact subsets of $\{0<|z|<1\}.$


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