Learning about singularities, just to be sure that I grasp the content correctly I will ask the following.
Find the radius of convergence of the following function.
$$ \frac{cosz}{z^2-\pi^2/4}, $$
about $z=0$.
Since about $z=0$, the power series of the denominator is just a geometric series, then the singularities are removable, right? The radius of convergence is then infinite.
On the other hand, if I expand for example around $4$, then the singularities become poles, right? In this case the radius of convergence should be $3$??
Why a function can be analytic on different domains depending on where we decide to expand it? Is not counterintuitive?
Is is an overkill, but the Weierstrass product for the cosine function gives $$ \cos(z)=\prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2 \pi^2}\right) \tag{1} $$ hence $$ \frac{\cos(z)}{z^2-\frac{\pi^2}{4}} = -\frac{4}{\pi^2}\prod_{n\geq 1}\left(1-\frac{4z^2}{(2n+1)^2 \pi^2}\right) \tag{2} $$ is an entire function and the RHS of $(2)$ is uniformly convergent over any compact subset of $\mathbb{C}$.