If $f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n$ and $g(z) = \sum_{n=0}^\infty b_n(z-z_0)^n$ are complex power series whose radii of convergence are $R_1$ and $R_2$ correspondently, then the complex power series of $f(z)g(z)$ has a radius of convergence of what?

I believe I have seen before that for two complex power series, the radius of convergence for $f(z)g(z)$ is $R = min({R_1, R_2})$. I have also seen before where the radius of convergence of $f(z)g(z)$ is $R \ge min({R_1,R_2})$. I believe the latter one is correct, but I am unsure. I assume the latter one is correct because of a simple example like this:

Let $f(z) = \frac{1-z}{1+z}$ and $g(z) = \frac{1+z}{1-z}$. Both have a radius of convergence of $|z| < 1$. Yet $f(z)g(z) = 1$ and its series expansion has a radius of convergence of $\infty$.

So which one is correct? Is one of them correct only in the case of a series expansion in real variables while one is only correct for complex variables? For whichever one is correct, how would you prove the result?

(First question asked, so sorry if anything is sloppy or formatted incorrectly)

  • $\begingroup$ latter one is correct. $\endgroup$
    – Koro
    Sep 13 '15 at 4:45
  • $\begingroup$ And it is correct for both real and complex power series? How would we go about showing that? $\endgroup$
    – Psymon
    Sep 13 '15 at 5:26
  • $\begingroup$ ok@achille hui,no problem.Would you please have a look at my question and attempt to answer ...here is the link<a>math.stackexchange.com/questions/1431794/…<\a> $\endgroup$
    – Koro
    Sep 13 '15 at 9:01
  • $\begingroup$ @Psymon,yes it is true both real and complex power series. $\endgroup$
    – Koro
    Sep 13 '15 at 9:03

For sure, it's truth for both real and complex power series.

Consider two functions $f$ and $g$ which are analytic at least within balls of radii $r$ and $d$ respectively, so they have power series about the origin within those balls. Now, the product of two analytic functions is analytic, so $fg$ is analytic at least within a ball of radius $s=min{(r,d)}$. This implies $fg$ also has power series expansion about zero.

Now assume that radius of convergence of $fg$ can never be greater than $s$, then your example gives a contradiction and hence proved!

PS: Wherever I used the term "ball" it is assumed that the ball is centered at the origin $(0,0)$. I have considered a power series about zero but you can always transform this to the power series about $z_0$ case.

  • $\begingroup$ Yeah, that makes sense when you consider where the product of two functions would be analytic. That helps show how the radius of convergence couldn't be less than the min. I got it from here, thank you :) $\endgroup$
    – Psymon
    Sep 13 '15 at 11:15

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