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Let $f:U\rightarrow \mathbb{C}$ be analytic. Suppose $\overline{A}_{r,R}(z_0)\subset U$ where $r<R$ and $r,R \in (0,\infty)$. Also, $|f(z)|\leq M$ in $\overline{A}_{r,R}(z_0)$. I need to prove the following. $$|a_n|\leq MR^{-n},n\geq 0$$ $$|a_n|\leq Mr^{-n},n< 0$$ This is my attemt. Let $r\leq s\leq R$. Let $\zeta$ be such that $|\zeta-z_0|=s$.

Then the Laurent series coefficient is, $$a_n=\frac{1}{2\pi i}\int_{|\zeta-z_0|=s}^{}\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}dz,n\in \mathbb{Z} $$Also, $$\frac{1}{2\pi }|\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}|\leq \frac{1}{2\pi }\frac{M}{s^{n+1}} $$ By the ML estimate $$|a_n|\leq \frac{M}{s^n}=Ms^{-n}$$

When $n<0$ and $r\leq s\leq R$ $\implies$ $r^{-n}\leq s^{-n}\leq R^{-n}\implies|a_n|\leq MR^{-n}$

When $n\geq 0$ and $r\leq s\leq R$ $\implies$ $R^{-n}\leq s^{-n}\leq r^{-n}\implies|a_n|\leq Mr^{-n}$.

What I get is not what I need to prove. Maybe my question should be fixed or my answer is wrong. Any help will be much appreciated. Thanks alot

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Note that you have got $|a_n| \le M s^{-n}$, but $s$ can be any arbitrary real number in $(r,R)$.

So for $n < 0$, the tightest bound for $|a_n|$ is achieved by setting $s$ close to $r$. Hence $|a_n| \le M r^{-n}$.

For $n \ge 0$, similarly, you should let $s \to R$, so $|a_n| \le MR^{-n}$.

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  • $\begingroup$ So n<0 and r<s implies $r^{-n}>s^{-n}$? $\endgroup$ – Heisenberg Nov 23 '14 at 11:39
  • $\begingroup$ You are not even trying to understand. I say that $s$ is any real number in $(r,R)$. and $|a_n| \le M s^{-n}$ for all such $s$. Now let $s \to r$. $\endgroup$ – Shubhodip Mondal Nov 23 '14 at 11:42
  • $\begingroup$ Chill :) I get your point now $\endgroup$ – Heisenberg Nov 23 '14 at 11:44
  • $\begingroup$ but s can be any number in [r,R] right? $\endgroup$ – Heisenberg Nov 23 '14 at 11:59
  • $\begingroup$ Only if the closure of the annulus is contained in the domain of holomorphicity. $\endgroup$ – Shubhodip Mondal Nov 23 '14 at 12:03

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