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

Clearly, $\left|z-\left(4+3 i\right)\right|=2$ represents a circle with centre $\left(4, 3\right)$ and radius $2$ units and $\left|z\right|+\left|z-4\right|=6$ represents an ellipse with foci $\left(0, 0\right)$ and $\left(4, 0\right)$ and vertices are $\left(5, 0\right)$ and $\left(-1, 0\right)$ .

Observe that, $z=\left(4+i\right)$ lies on circle and $\left|4+i\right|+\left|4+i-4\right|=1+\sqrt{17}<6$.

Hence, the point $z=\left(4+i\right)$ lies inside the ellipse.

So, the graphs of the curves are as shown in figure.

It is clear from the above diagram, there exist two complex numbers which satisfy both the given curves.