Radius of Convergence for $f(z) = \dfrac{1}{1+z^2+z^4}$ at $\dfrac{1}{2}$ I am practicing some qualifying problems, and I cannot compute the following:
Find the radius of convergence $R$ of the Taylor series of $f(z) = \dfrac{1}{1+z^2+z^4}$ centered at $\dfrac{1}{2}$.
I'm thinking I am having just a technical issue by not seeing the trick for finding the Taylor series for this function. 
I tried rewriting $f(z)$ and got to $f(z) = -\dfrac{4}{3} \dfrac{1}{1-\frac{4}{3}\left(\left(z-\frac{1}{2}\right)^2+z+\frac{1}{4}\right)^2}$ although I couldn't find the Taylor series from here.
 A: The radius of convergence is $\sqrt3/2$.  It's just the distance from $z=1/2$ to the nearest zero(es) of $1+z^2+z^4$ (i.e., the nearest pole(s) of $1/(1+z^2+z^4)$.  If you note that $(1-z^2)(1+z^2+z^4)=1-z^6$, you see that those zeroes are sixth roots of unity, i.e., $\pm{1\over2}\pm{\sqrt3\over2}i$, so the nearest zeroes are at ${1\over2}\pm{\sqrt3\over2}i$.
A: If a function is holomorphic on a disk centered at $\frac12$, then it has a convergent power series on that disk centered at $\frac12$, and therefore the radius of convergence is at least as big as the radius of the disk.  Conversely, in the disk centered at $\frac12$ whose radius is the radius of convergence of the power series centered at $\frac12$, the function is holomorphic.
Thus the radius of convergence is the radius of the largest disk centered at $\frac12$ on which the function is holomorphic.  In this case, that means the distance from $\frac12$ to the nearest pole, a.k.a. the nearest zero of the polynomial $1+z^2+z^4$.  This polynomials is quadratic in $z^2$, so it is easy to explicitly calculate its roots and determine their distances to $\frac12$.
