My problem is that I don't understand why functions like $f(z)=\frac {e^{iz}-1}z$ are entire in $\mathbb{C}$. The only thing we have to prove in this case is that the function is continuos at $0$. If we do the limit to 0, we get that it exist and it's finite. Is this enought to say that is continuos at $0$?
Another argument i thought is this one:
If we write that $g(z)=e^{iz}-1$ then we know that this function has a zero on $z_0=0$, then we can write (as far as this one is entire) $g(z)=zg_1(z)$ with $g_1(0)\neq0$ and entire. Then our function is $f(z)=g_1(z)$ and problem solved.
I would like to know how to proceed in the first argument.
This is enough to show that the function is continuous at $0$. If a function $f$ is holomorphic on $\Omega - \{p\}$, where $\Omega$ is an open set, then $f$ either has a removable singularity, pole, or essential singularity at $p$. In particular, if $f$ is bounded in some neighborhood of $p$, the singularity is removable. So filling in $\frac{e^{iz} - 1}{z}$ with the limit at $0$ will make $\frac{e^{iz} - 1}{z}$ entire.