Series expansion of $\dfrac{z^2}{1+z^2}$

Question

Consider the complex function $$f(z) = \dfrac{z^2}{1+z^2}$$

It has two isolated singularities at $z=\pm i$. So when we consider the series expansion at $z_0 = 0$, then the radius of convergence is $1$. Doesn't this mean that if $|z| < 1$, then
$$f(z) = \sum_{k=0}^{\infty} c_n z^k$$
converges absolutely? But on the other hand
$$f(z) = \dfrac{1}{1-\left(-\dfrac{1}{z^2}\right)} = \sum_{k=0}^{\infty}(-1)^k\dfrac{1}{z^{2k}}$$
if $\left|\dfrac{1}{z^2}\right| < 1 \Leftrightarrow |z^2| > 1 \Leftrightarrow |z| > 1 $.

I don't understand why are two different series expansion at $z_0 = 0$ with different radii of convergence. I thought that the series expansion and the radius of convergence at $z_0 = 0$ is unique in some neighborhood of $z_0$. I would be grateful if someone could clarify what I am misunderstanding.

Answer 1

I thought that the series expansion and the radius of convergence at $z_0 = 0$ is unique in some neighborhood of $z_0$. I would be grateful if someone could clarify what I am misunderstanding.

The Laurent serie doesn't tell you anything about the neighbourhood of $z_0=0$ because it doesn't converge there.