The number of points of intersection $$|z - (4+3i)| = 2$$ and $$|z| + |z-4| = 6, z \in C$$ is

Given,

$$|z-(4+3i)|=2$$

This represents a circle with centre

$$(4,3)$$

and radius

$$2$$

Also,

$$|z|+|z-4|=6$$

represents an ellipse with foci

$$(0,0)\quad \text{and}\quad (4,0)$$

and major axis length

$$6$$

Hence,

$$2a=6\Rightarrow a=3$$

Distance between foci is

$$2c=4\Rightarrow c=2$$

Therefore,

$$b^2=a^2-c^2=9-4=5$$

So, equation of ellipse is

$$\frac{(x-2)^2}{9}+\frac{y^2}{5}=1$$

Now, the circle is

$$ (x-4)^2+(y-3)^2=4 $$

The topmost point of the ellipse is

$$\left(2,\sqrt5\right)$$

and the lowest point of the circle is

$$ (4,1) $$

Since

$$1<\sqrt5$$

the circle overlaps the ellipse.

Also, the centre of the circle lies outside the ellipse, so the circle cuts the ellipse at two distinct points.

Hence, the number of intersection points is

$$\boxed{2}$$