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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.

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  • $\begingroup$ Note that just from $g(0) = 0$you cannot concluded that you can write $g(z) = zg_1(z)$ with $g_1(0) \neq 0$; try it, for example, with $g(z) = z^2$, or more extremely, with $g(z) = 0$. $\endgroup$ – Mees de Vries Jun 27 '18 at 15:20
  • $\begingroup$ Mmm, so when i can? I mean, im just using the Taylor Theorem because it's an entire function. $\endgroup$ – Lecter Jun 27 '18 at 15:31
  • $\begingroup$ You can always do it with no further restriction on $g_1$, i.e. allowing $g_1(0)$ to take on any value. $\endgroup$ – Mees de Vries Jun 27 '18 at 15:34
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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.

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  • $\begingroup$ Didn't know about this theorem, thanks. $\endgroup$ – Lecter Jun 27 '18 at 15:27
  • $\begingroup$ No problem! If you want a reference, it's theorem 10.21 in Rudin's Real and Complex Analysis. The remarkable feature of this result is how dramatically different each possibility is. $\endgroup$ – Alex Nolte Jun 27 '18 at 15:32

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