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I know that the radius of convergence of any power series can be found by simply using the root test, ratio test etc.

I am confused as to how to find the radius of convergence for an analytic $f$ such as

$f(z)=\frac{4}{(z-1)(z+3)}$.

I can't imagine that I would have to find the power series representation of this, find the closed form, and then use one of the convergence tests. I am fairly certain that the radius of convergence would have to do with the singularities at $1$ and $-3$, however, I can't find a formula for the radius of convergence..

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  • $\begingroup$ The answer will depend on what the center of expansion is. If you’re looking for a series in powers of $x$, that’s one thing, but if you’re looking for powers of $x-a$ for some nonzero $a$, the answer will be different. $\endgroup$ – Lubin Dec 11 '13 at 21:34
  • $\begingroup$ @Lubin, I understand it will differ. I guess I am just very confused as to the relationship between the root test, ratio test etc. and the singularities. $\endgroup$ – user7090 Dec 11 '13 at 22:01
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It is very useful to remember that the radius of convergence of power series in the complex plane is basically the distance to nearest singularity of the function. Thus if a function has poles at $i$ and $-i$ and you do a power series expansion about the point $3+i$, then the radius of convergence will be $3$ since that is the distance from $3$ to $i$.

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  • $\begingroup$ This is correct, and follows from the proof (not the statement) of the theorem that holomorphic functions are analytic. Note that singularities do not have to be poles. They could be branch points, or stranger beasts see mathoverflow.net/questions/10831/… $\endgroup$ – Steven Gubkin Dec 11 '13 at 22:35
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The power series for $f = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^{n}$ is guaranteed tdo have a radius of convergence $r=1$. The power series diverges for $|z| > 1$ and converges absolutely for $|z| < 1$ because of the singularity at $z=1$.

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The theory tells us that the radius of the expansion about $z=0$ is $1$, but you can see this computationally as well. It turns out, just by direct long division of power series, that: $$ f(z)=-\sum_{i=1}^\infty\frac{3^i+(-1)^{i+1}}{3^i}z^{i-1}\,. $$

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