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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.

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  • $\begingroup$ What are your thoughts on this question? $\endgroup$
    – abiessu
    Jun 22, 2014 at 18:43

1 Answer 1

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

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    $\begingroup$ Here, there is a link which explains that in detail: en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions . $\endgroup$
    – Kernel
    Jun 22, 2014 at 18:51
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    $\begingroup$ Thank you, that's what I wanted. $\endgroup$
    – u1571372
    Jun 22, 2014 at 18:52
  • $\begingroup$ We know from the definition of an analytic function that in some neighborhood of every point in it's domain it can be represented by power series. So for an entire function it means that for every point $a \in \mathbb{C}$ we have the equality $f(z)=\sum_{k \ge 0} c_k \cdot (z-a)^k$ in some neighborhood of $a$. I don't understand why do we get the radius is $\infty$. Can you please elaborate on that a little bit more? The link you provided doesn't answer the question actually :(( $\endgroup$ Apr 4, 2022 at 16:48
  • $\begingroup$ Notice that the proof did not depend on the radius of the disk $D$, so it can be taken arbitrarily large. Does that make sense? $\endgroup$
    – Kernel
    Apr 6, 2022 at 8:57

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