Radius of convergence of entire function

Question

Let $f$ be an entire function on the complex plane.

Is the radius of convergence of $f$ around any point $z_0$ infinite? If so, why?

Thank you.

Answer 1

A function $f: \mathbb{C} \to \mathbb{C}$ is said to be entire if it is holomorphic in the all complex plane, but using the Cauchy's Integral Formula, we have that a holomorphic function in a region is also analytic in this same region. Therefore, an entire function $f$ is analytic in the all complex plane. Furthermore, we also have that the radius of convergence of a series centred at a point $z_0$ is the distance between $z_0$ and the nearest singularity of the series, in that case, it will be infinite.

EDIT (10 years later, time really flies) If I remember correctly, this is how the argument goes:

The radius of convergence is precisely defined as $$ R(f,z_0):= \sup\left\{r \ge 0: \left\|\sum_{n=0}^\infty \frac{f^(n)(z_0)}{n!} (z-z_0)^n\right\| < \infty, \forall z\in B(z_0,r)\right\}. $$ Let us also define another value $$ \tilde{R}(f,z_0):= \sup\left\{r \ge 0: f'(z) \text{ exists for all } \forall z\in B(z_0,r)\right\}. $$

Showing that $R(f,z_0)\le \tilde{R}(f,z_0)$ is an mere exercise of differentiating power series. We will now show that $R(f,z_0)=\tilde{R}(f,z_0)$. In particular, for entire functions, we know that $\tilde{R}(f,z_0)$ is infinity, and we want to show that $R(f,z_0)$ is also infinity.

For all $z \in B(z_0,r)$ for any $r \in (0,\tilde{R}(f,z_0))$, we can use the Cauchy integral formula to get $$ (\star)\qquad f(z)=\frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{z-w}dw $$ where $\gamma(t):= z+r e^{it}$.

We can then use the Taylor series $$ \frac{1}{w-z}=\frac{1}{(w-z_0)-(z-z_0)}=\frac{1+\frac{z-z_0}{w-z_0}+\left(\frac{z-z_0}{w-z_0}\right)^2+\cdots}{w-z_0}.$$ And show that the series $$\sum_{n=0}^\infty \frac{f(w)(z-z_0)^n}{(w-z_0)^{n+1}}$$ converges absolutely (also a simple exercise).

This allows us to exchange the order of the series and integral to rewrite the right-hand side of $(\star)$ as $$ \sum_{n=0}^\infty \frac{1}{2 \pi i} (z-z_0)^n \oint_\gamma f(w) \frac{1}{(w-z_0)^{n+1}} dw. $$

Well, but because of $(\star)$, this coincides with $f(z)$. But now we have a power series representation of $f(z)$ that works anywhere in $B(z_0,r)$. Once again, by differentiating this power series, it is easy to show that this must be the Taylor series of $f$. As $r \le \tilde{R}(f,z_0)$ was arbitrary, we get $R(f,z_0)\ge\tilde{R}(f,z_0)$, therefore, $R(f,z_0)=\tilde{R}(f,z_0)$.