Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $\Omega\subseteq \mathbb{C}$ be open and $f\in H(\Omega)$ ($f$ analytic on $\Omega$). If $ C(z_{0},R)\subseteq\Omega$ (where $C(z_0,R)$ is the circle with origin $z_0$ and radius $R$), then we can represent $f$ on $C(z_{0},R)$ as a power series with convergence radius $\geq R$.


Let $0<r<R$ and $\gamma:[0,2\pi]\rightarrow\Omega$ be the path $\gamma(t)=z_{0}+re^{it}$.

Because $C(z_{0},R)$ is convex and $\gamma$ is in this circle, we have the following from the Cauchy integral formula: $$ f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in C(z_{0},\ r)) $$

$\color{red}{\text{(1) Why not directly on } C(z_{0},\ R)? }$

Because $C(z_{0},\ r)\subseteq \mathbb{C}\backslash \gamma^{*}$ it follows that $f(z)=\displaystyle \sum_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ for a power series with convergence radius $\geq r$. Because $ r\in (0,\ R)$ is chosen arbitrarily and because the the coefficients $c_{n}$ are determined by the differentials $f^{(n)}(z_{0})$, the power series $\displaystyle \Sigma_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ has the convergence radius $\geq R$.

$\color{red}{\text{(2) Why are the coefficients calculated via }f^{(n)}(z_{0})? }$

$\color{red}{\text{(3) Why does it follow that the convergence radius is } \geq R? }$

share|cite|improve this question
Chris, per Mex's comment to my answer, do you mean that $C(z_0,R)$ denotes the open ball of radius $R$ centered at $z_0$, instead of, as you stated, the circle (i.e., boundary) of the same center and radius? – J. Loreaux Jun 7 '12 at 0:16
up vote 1 down vote accepted

For (1), we choose $r<R$ because the path $\gamma_R(t)=z_0+Re^{it}$ for $t\in[0,2\pi]$ may not lie inside the region $\Omega$. This is because while $C(z_o,R)\subset\Omega$, its boundary, which is $\gamma_R^*$, may not be contained within $\Omega$. Now, to see why the integral formula only holds when $z\in C(z_0,r)$, and not on its boundary (which is $\gamma^*$ in your statement), this is simply due to the statement of the Cauchy Integral Formula. That is, it only holds for points that do not lie on $\gamma$.

For (2), the coefficients $c_n$ are determined by the differentials $f^{(n)}(z_0)$ because of the Taylor formula (which we can obtain from the finite Taylor formula if need be).

For (3), we have shown that the power series has radius of convergence $r$ where $r$ is an arbitrary positive number less than $R$. Thus, the radius of convergence is at least $\sup_{0<r<R} r= R$. So the radius of convergence is greater than or equal to $R$.

share|cite|improve this answer
@Loreaux, what do you mean by $\gamma_R(t)$ may not lie inside the region $\Omega$? given that $\Omega$ is an open set and $C(z_0,R)\subseteq \Omega$. So his No.1 question is justified and I am still confused what will be the problem if he apply cauchy formula on $C(z_0,R)$? – Un Chien Andalou Jun 6 '12 at 4:10
@Mex: So, perhaps I misunderstood the notation, but I thought that $C(z_0,R)$ denoted the open ball of radius $R$ centered at $z_0$. I see that the OP said it was the circle of radius $R$ (i.e., the boundary of the set I just mentioned), but that doesn't seem to make sense, for example, because the OP claimed that $C(z_0,R)$ is convex. Anyway, given that it denotes the open ball, it stands to reason if we take $\Omega:=C(z_0,R)$ then certainly $\gamma_R*=\partial C(z_0,R)$ which lies entirely outside $\Omega$. – J. Loreaux Jun 7 '12 at 0:14
It was my mistake, you are right J. Loreaux! – Chris Jun 7 '12 at 0:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.