showing $f$ is entire Show that the function $f(z)$ is define by $f(0)=1$ and $f(z)=z^{-1}\sin z $ when $z\neq0$, is entire.

$\sin z=z-z^3/3!+z^5/5!-z^7/7!...$
(*) We can write $\sin z=z+z^2g_2(z)$   which $g_2(z)$ is analytic.
for $z\neq0$
$$\frac{\sin z}{z}=1+zg_2(z)$$
as $z\to0$ we have $f(z)\to1$. since we have also  $f(0)=1$, $f$ is entire.
is this correct? if it is, can you explain (*)?
 A: Since $f$ is analytic when $z \neq 0$ (product of two analytic functions), all that remains is to show that it is analytic at $z=0$.
For all $z$, we have $f(z)-f(0) -0 = \sum_{n=1}^\infty (-1)^n \frac{z^{2n}}{(2n+1)!} = z^2\sum_{n=1}^\infty (-1)^n \frac{z^{2(n-1)}}{(2n+1)!}$. For $|z|<\frac{1}{2}$, we have $|\sum_{n=1}^\infty (-1)^n \frac{z^{2(n-1)}}{(2n+1)!}| \leq \sum_{k=0}^\infty |z|^k=\frac{1}{1-|z|} \leq 2$, hence
$|f(z)-f(0) -0| \leq  2|z|^2$, from which it follows that $f$ is differentiable at $0$ with derivative $f'(0)=0$. Hence $f$ is entire.
A: You may find Morera's theorem applicable here. Any contour integral that does not contain the point $z = 0$ must give $0$ by Cauchy's Theorem, since $f$ is holomorphic there. If it does contain $z = 0$, then things are still fine because it can be very slightly deformed to one that doesn't, which will not affect the integral because $f$ is bounded near $0$. Hence all contour integrals give zero, hence Morera's theorem tells you $f$ is analytic.
The slight deformation involves making a detour: at some point around the contour, travel to $z = 0$, travel around that point, and then back along the way you came. The trip there and the trip back will cancel out, so the value of the integral is only changed by the tiny little journey around $z = 0$. Since $f$ is bounded, this journey contributes arbitrarily little to the integral. But the new contour doesn't contain $z = 0$, so by the previous argument, the integral around it is zero. So the integral around the original contour must be arbitrarily close to zero, and the only such number is zero itself.
This kind of (sketchy, but formalisable) argument will also show that in fact a function that is analytic everywhere but:


*

*an isolated collection of points

*a line

*an isolated collection of lines

*other more devious things


and continuous everywhere, must in fact be entire. Essentially because you can exclude the problematic points from your contour integrals without changing the integral very much, so in fact including them can't ruin it.
