If $$|z - 3 + 2i| \leq 4$$ then the difference between the greatest value and the least value of $$|z|$$ is:

Let the complex number be written as $$z = x + iy,$$ where $$x$$ and $$y$$ are real numbers. In the Argand (complex) plane this point corresponds to the Cartesian point $$\,(x,\;y).$$

We are given the condition $$|\,z - 3 + 2i\,| \le 4.$$

First recall the distance formula for complex numbers: for any two complex numbers $$z_1$$ and $$z_2,$$ their distance in the plane is $$|\,z_1 - z_2\,|.$$ Hence the set of all points satisfying $$|\,z - (3 - 2i)\,| \le 4$$ is a closed circle of radius $$4$$ centred at the point $$3 - 2i.$$

Thus the centre of the circle is $$C(3,\,-2)$$ and its radius is $$r = 4.$$

We need to study the quantity $$|z|,$$ which is the distance of the point $$z(x,y)$$ from the origin $$O(0,0).$$ In Euclidean geometry the largest and the smallest distances from the origin to points on the same circle occur along the straight line passing through the origin and the centre of the circle.

Let $$d$$ denote the distance from the origin to the centre of the circle. We calculate:

$$d = \sqrt{\,3^{2} + (-2)^{2}\,} = \sqrt{9 + 4} = \sqrt{13}.$$

Once more recall a basic geometry fact: for a circle of radius $$r$$ whose centre is at distance $$d$$ from the origin,

• the greatest possible value of $$|z|$$ equals $$d + r,$$
• the least possible value of $$|z|$$ equals $$|\,d - r\,|.$$

Applying this, we obtain

Maximum value of $$|z|$$ :

$$|z|_{\max} = d + r = \sqrt{13} + 4.$$

Minimum value of $$|z|$$ :

$$|z|_{\min} = |\,d - r\,| = |\,\sqrt{13} - 4\,|.$$

Because $$4 > \sqrt{13},$$ the absolute value simply removes the sign, so

$$|z|_{\min} = 4 - \sqrt{13}.$$

The required difference between the greatest and the least values is therefore

$$|z|_{\max} - |z|_{\min} = \bigl(\sqrt{13} + 4\bigr) - \bigl(4 - \sqrt{13}\bigr) = \sqrt{13} + 4 - 4 + \sqrt{13} = 2\sqrt{13}.$$

Hence, the correct answer is Option B.