$z \sin z \sin 1/z$ does not have singularities, why?

So wolfram tells me, that $f(z) = z \sin(z) \sin(\frac{1}{z})$ does not have any poles (and thus no residue).

How? I thought, $1/z$ is not defined at $z=0$.

-
Try to compute the limit! $\sin(\frac{1}{z})$ is not defined at $z=0$, but it is certainly bounded in a neighborhood. While $z$ and $\sin(z)$... – Giovanni De Gaetano Jan 23 '13 at 9:21
@GiovanniDeGaetano: What about $z=i/1000$? – Rahul Jan 23 '13 at 9:27
Ooops! Despite the use of the variable, the words "pole" and "residue" and the tag I was looking at the real situation... I'm sorry! – Giovanni De Gaetano Jan 23 '13 at 9:30

$z=0$ is certainly a singularity of $f$, but it's an essential singularity, not a pole.
The residue of $f$ at $z=0$ is pretty tricky to work out. The Laurent series of $f$ will be
$$\left( z^2 - \frac{z^4}{3!} + \frac{z^6}{5!} - \cdots \right) \left(\frac{1}{z} - \frac{1}{3!z^3} + \frac{1}{5!z^5} - \cdots \right)$$ and we want to extract the coefficient of $z^{-1}$ in this.
The result will be $$-\frac{1}{3!} - \frac{1}{3!5!} - \frac{1}{5!7!} - \cdots$$
With a little help of Maple, this sum evaluates to $-\frac12(I_0(2)+J_0(2)-I_1(2)-J_1(2))$, were $J_0$ and $J_1$ are the Bessel functions of the first kind, and $I_0$ and $I_1$ are modified Bessel functions of the first kind. Numerically this is $\approx -0.1680572095415502424521646140963393221$ and the inverse symbolic calculator doesn't recognize this number.