# Radius of convergence of power series development of $\frac{1}{\cos(z)}$

What is the radius of convergence of the power series development of $f(z) = \frac{1}{\cos(z)}$ at $z_0=i$?

The function $f$ is defined on $D=\{z\in \Bbb{C} : \cos(z)\neq 0\}$. The largest open disk with center $z_0=i$ which is contained in $D$ is the ball with radius $r=|i - \pi/2|=\sqrt{\pi^2/4+1}$. The radius of convergence is certainly $\geq r$. Is it $=r$?

The closest singularities to the origin are the simple poles in $z=\pm\frac{\pi}{2}$, hence the radius of convergence of the Taylor series of $f(z)=\frac{1}{\cos z}$ in $z=0$ is $\rho=\frac{\pi}{2}$.
That also follows from the Weierstrass product for the cosine function. $$\cos(z)=\prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2\pi^2}\right)\tag{1}$$ leads to: $$\frac{1}{\cos(z)}=\prod_{n\geq 0}\sum_{m\geq 0}\left(\frac{2z}{(2n+1)\pi}\right)^{2m}\tag{2}$$ and the radius of convergence of: $$\sum_{m\geq 0}\left(\frac{2z}{(2n+1)\pi}\right)^{2m}\tag{3}$$ is trivially $\rho=\frac{(2n+1)\pi}{2}$.
• The problem concerns the power series development of $f$ in $z=i$, not the origin. Is it true in general that the radius of convergence will be the distance to the nearest singularity? Commented May 20, 2015 at 12:38
• @iwriteonbananas: yes it is, so the radius of convergence in $z=i$ is $\frac{1}{2}\sqrt{4+\pi^2}$ as you claimed. Commented May 20, 2015 at 12:41