Is $\sin z/z$ analytic at the origin? For $z\in\Bbb C$ let
$$
f(z) = \frac{\sin z}{z}
$$
Along the real line this is well behaved, and approaches $1$ as $z\to 0$.
But is $f(z)$ analytic at the origin ($z=0$)?
I tried explicitly checking the Cauchy conditions but that gets ugly (unless I am missing something).
The function I was originally interested in is
$$
g(z) = z\,\sin\left( \frac{1}{z} \right)
$$
which my gut tells me must not be analytic at the origin, but is the singularity essential or a pole, or the end of a branch cut or something even uglier?
 A: As defined, no, as it isn't well-defined at $z=0$.  It does, however, have an analytic continuation to the entire plane.  What you can observe is that $\sin(z)$ has a zero of order $1$ at $z=0$, and therefore the singularity of $\sin(z)/z$ has order 0 at $z=0$, meaning it is removable and an analytic continuation at $z=0$ exists (and is given by the limit).  Equivalently, $$\lim_{z\rightarrow 0} \sin(z)/z = 1$$ holding over complex numbers means that the singularity is removable.
As for your $g(z)$ function, we can write $g(z)=f(1/z)$ and $2\sin(z) = e^{iz}-e^{-iz}$, and so for $x\in\mathbb R$ and $n\in\mathbb N$ in the following we have
$$\lim_{x\rightarrow 0}|x^n g(x)| = \lim_{x\rightarrow\pm\infty} |f(x)/x^n| = 0$$
whereas
$$\lim_{x\rightarrow 0}|x^n g(ix)| = \lim_{x\rightarrow\pm\infty} |f(ix)/x^n| = \infty,$$
which means that $g$ has an essential singularity at $z=0$
A: Since
$$
\sin z=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{2k+1}
$$
we have, for $z\ne0$,
$$
\frac{\sin z}{z}=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{2k}
$$
Since this power series has radius of convergence infinite, it defines an entire function, which is an analytic continuation of $(\sin z)/z$ at $0$, where it has the value $1$.
So, as written, $(\sin z)/z$ is not analytic at $0$, because it's not defined there, but $0$ is a removable singularity.
For $z\sin(1/z)$ you can use the same series:
$$
z\sin\frac{1}{z}=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{-2k}
$$
which holds for $z\ne0$. This is the Laurent series for the analytic function $z\sin(1/z)$ at $0$ and it shows the singularity is essential, because the Laurent series at a pole has only a finite number of terms with $z$ having negative exponent.
