We do not require that the domain of convergence for a power series will be disk shaped, it just happens to be the case. This is quite a difference between general series of functions and power series.
There is a theorem that states that every power series $\sum_{i=1}^{\infty} a_n (z - z_0)^n$ comes with an associated number $0 \leq R \leq \infty $ called the radius of convergence. It satisfies the following properties:
- The power series converges for all $z$ such that $|z - z_0| < R$ (all $x$'s whose distance from $z_0$ is less than $R$).
- The power series diverges for all $z$ such that $|z - z_0| > R$.
- When $|z - z_0| = R$, the series might converge or might diverge. This must be checked on a case-to-case basis.
You can write various explicit formulas for the radius of converge in terms of the coefficients $a_n$. For example, the Cauchy-Hadamard formula for the radius of convergence is $ R = \frac{1}{\mathrm{lim \;sup}_{n \rightarrow \infty} |a_n|^{\frac{1}{n}}} $
So, given the power series $\sum_{i=0}^{\infty} (-1)^i z^{2i}$ around $0$, if you plug $z = 2$, you can see that the terms of the series are $(-1)^i 2^{2i}$, and so they don't even tend to $0$ and the series diverges. If you plug $z=1$, you see that the series $1 - 1 + 1 - 1 + ...$ also diverges. This means that the radius of convergence must be $1$ or less. You can prove directly, or using the formula, that $R$ is exactly $1$.
Everything said above is independent of whether you know to what function the power series converges. We didn't care above that the power series converged to $1/(1 + z^2)$. Denote by $B = \{ z \; | \; |z| < 1 \}$ the unit disc. What happens is that while the power series converges only when $|z| < 1$, the function it converges to, $u(z) = \frac{1}{1 + z^2} : B \rightarrow \mathbb{C}$, defined a priori only on $B$, can be extended to a function $\tilde{u}(z) = \frac{1}{1 + z^2} : \mathbb{C} \setminus \{ \pm i \} \rightarrow \mathbb{C}$ defined on a larger (maximal) domain of definition. But the power series won't converge to that function - it won't converge to anything when $|z| > 1$.
If you start the other way, from $\tilde{u}$, a function defined on $\mathbb{C} \setminus \{ \pm i \}$, and try to develop it into a Taylor series around $z = 0$, then what happens is that you get your series, the series has radius of convergence $R = 1$, and it converges on $|z| < 1$ to $\tilde{u}\restriction_B = u$.
Its not always the case that the Taylor expansion of a function converges to the function itself. If you knew in advance that the Taylor series must converge to $\tilde{u}$, you could deduce that its radius of convergence can't be more than $R = 1$, as $\tilde{u}$ is not defined at $z = \pm i$. And indeed, because the function $u$ is rational (more generally, holomorphic on its domain of definition), there is a theorem that states the Taylor series of $\tilde{u}$ around $z_0$ must converge to the function $\tilde{u}$ on the maximal possible (open) disc around $z_0$. For $z_0 = 0$, we have $|z| < 1$.
I hope I didn't confuse you too much, but somehow behind such "innocent" claims lies a lot of difficult / interesting / subtle mathematics which is worth being familiar with.