The number of elements in the set $$\{z = a + ib \in C : a, b \in \mathbb{Z}$$ and $$1 < |z - 3 + 2i| < 4\}$$ is ______

We need to find the number of Gaussian integers $$z = a + ib$$ (with $$a, b \in \mathbb{Z}$$) such that $$1 < |z - 3 + 2i| < 4$$.

This means $$1 < |(a-3) + i(b+2)| < 4$$, i.e.:

$$1 < \sqrt{(a-3)^2 + (b+2)^2} < 4$$

$$1 < (a-3)^2 + (b+2)^2 < 16$$

Let $$u = a - 3, v = b + 2$$. We need integer pairs $$(u, v)$$ with $$1 < u^2 + v^2 < 16$$, so $$u^2 + v^2 \in \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$$.

Total lattice points with $$u^2 + v^2 < 16$$: we enumerate for $$u^2 + v^2 \leq 15$$.

For each value of $$u^2 + v^2$$:

$$u^2+v^2 = 0$$: $$(0,0)$$ — 1 point

$$u^2+v^2 = 1$$: $$(\pm 1, 0), (0, \pm 1)$$ — 4 points

$$u^2+v^2 = 2$$: $$(\pm 1, \pm 1)$$ — 4 points

$$u^2+v^2 = 4$$: $$(\pm 2, 0), (0, \pm 2)$$ — 4 points

$$u^2+v^2 = 5$$: $$(\pm 1, \pm 2), (\pm 2, \pm 1)$$ — 8 points

$$u^2+v^2 = 8$$: $$(\pm 2, \pm 2)$$ — 4 points

$$u^2+v^2 = 9$$: $$(\pm 3, 0), (0, \pm 3)$$ — 4 points

$$u^2+v^2 = 10$$: $$(\pm 1, \pm 3), (\pm 3, \pm 1)$$ — 8 points

$$u^2+v^2 = 13$$: $$(\pm 2, \pm 3), (\pm 3, \pm 2)$$ — 8 points

Values $$u^2+v^2 = 3, 6, 7, 11, 12, 14, 15$$: no integer solutions (can be verified).

Total with $$u^2 + v^2 < 16$$: $$1 + 4 + 4 + 4 + 8 + 4 + 4 + 8 + 8 = 45$$

Exclude $$u^2 + v^2 \leq 1$$: $$1 + 4 = 5$$ points.

Answer: $$45 - 5 = 40$$

Hence the answer is $$\boxed{40}$$.