# Radius of Convergence for analytic functions

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..

• 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. Dec 11, 2013 at 21:34
• @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. Dec 11, 2013 at 22:01

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 $i$.
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 $z=1$.
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: $$f(z)=-\sum_{i=1}^\infty\frac{3^i+(-1)^{i+1}}{3^i}z^{i-1}\,.$$