On a single Argand diagram, sketch the locus of points representing the complex number $z$ such that
$|z-4-2i| \le 2\quad\text{and}\quad |z-3| \le |z-5|.$ Find the greatest and least possible values of (a) $|z|$, (b) $\arg(z)$. Find the greatest and least possible values of $|z|$. Find the greatest and least possible values of $\arg(z)$.

1. Identify the region defined by the first inequality.

The inequality $|z - (4 + 2\mathrm{i})| \le 2$ represents the set of points inside and on the boundary of a circle with centre $4 + 2\mathrm{i}$ and radius $2$.

Since the imaginary part of the centre is $2$ and the radius is $2$, the circle touches the real axis at $(4,0)$.

2. Identify the region defined by the second inequality.

The inequality $|z - 3| \le |z - 5|$ represents the set of points whose distance to $3$ is less than or equal to their distance to $5$.

This is the half-plane on and to the left of the perpendicular bisector of the segment joining $(3,0)$ and $(5,0)$. The perpendicular bisector is the vertical line $\operatorname{Re}(z) = 4$. Therefore, the region is $\operatorname{Re}(z) \le 4$.

3. Sketch the combined locus on the Argand diagram.

The required locus is the intersection of these two regions.

On a single Argand diagram:

1. Draw the real ($\operatorname{Re}$) and imaginary ($\operatorname{Im}$) axes.
2. Sketch the circle with centre $(4,2)$ passing through $(2,2)$, $(6,2)$, $(4,4)$, and touching the real axis at $(4,0)$.
3. Draw the vertical line $\operatorname{Re}(z) = 4$ passing through the centre.
4. Shade the region satisfying both inequalities, which is the left semicircle (including all its boundaries).

1. Determine the least possible value of $|z|$ using the distance from the origin to the centre of the circle.

The least value of $|z|$ corresponds to the minimum distance from the origin $O(0,0)$ to the shaded region.
For the full disc given by $|z - (4+2i)| \le 2$, the centre is $C(4, 2)$ and the radius is $r = 2$.
The distance from $O$ to $C$ is:
$OC = |4 + 2i| = \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5}$
The point on the disc closest to $O$ lies on the line segment $OC$. Since $O$ is at $x=0$ and $C$ is at $x=4$, this closest point must have an $x$-coordinate less than $4$. Therefore, it satisfies the condition $\operatorname{Re}(z) \le 4$ and lies within the allowed region.
$\text{Least value of } |z| = OC - r = 2\sqrt{5} - 2$

2. Determine the greatest possible value of $|z|$ by evaluating the valid points on the boundary line.

The greatest possible value of $|z|$ is the maximum distance from $O$ to the region.
The point on the full disc furthest from $O$ lies on the ray $OC$ extended past $C$. This point has an $x$-coordinate $> 4$, placing it outside our allowed region ($x \le 4$).
Thus, the furthest point in our bounded region must lie on the vertical boundary line $x = 4$.
The intersection of the disc with $x = 4$ is given by:
$(4-4)^2 + (y-2)^2 \le 4 \implies (y-2)^2 \le 4 \implies 0 \le y \le 4$
For points $z = 4 + yi$ on this segment, the distance to the origin is $\sqrt{4^2 + y^2}$.
This distance is strictly increasing with $y$ and is maximised when $y$ is as large as possible, which is at the intersection point $(4, 4)$.
$\text{Greatest value of } |z| = |4 + 4i| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$

1. Find the minimum argument by inspecting the lowest point of the locus.

The region is defined by the intersection of the closed disc $(x-4)^2 + (y-2)^2 \le 4$ and the half-plane $x \le 4$.
The lowest point of the full circle is $(4,0)$, which satisfies $x \le 4$ and is therefore inside the region.
Since $y \ge 0$ for all points in the circle, the minimum possible value of $\arg(z)$ is the argument of $z=4$.
$\text{Least } \arg(z) = 0$

2. Express the maximum argument condition algebraically by substituting a line $y=mx$ into the circle equation.

To find the greatest possible argument, we want to maximize the gradient $m = \frac{y}{x}$ for points in the region.
Substitute the line $y = mx$ into the boundary of the circle $(x-4)^2 + (y-2)^2 = 4$:
$(x-4)^2 + (mx-2)^2 = 4$
$x^2 - 8x + 16 + m^2x^2 - 4mx + 4 = 4$
$(1+m^2)x^2 - 4(2+m)x + 16 = 0$
For the line to intersect the circle, the quadratic in $x$ must have real roots, so its discriminant must be non-negative:
$\Delta = [-4(2+m)]^2 - 4(1+m^2)(16) \ge 0$

3. Solve the discriminant inequality, verify the intersection point is within the region, and calculate the maximum argument.

$16(4 + 4m + m^2) - 64(1+m^2) \ge 0$
$64 + 64m + 16m^2 - 64 - 64m^2 \ge 0$
$64m - 48m^2 \ge 0 \implies 16m(4 - 3m) \ge 0$
This inequality gives $0 \le m \le \frac{4}{3}$. Thus, the maximum possible gradient is $m = \frac{4}{3}$.
When $m = \frac{4}{3}$, the discriminant is $0$ and the $x$-coordinate of the point of tangency is the repeated root:
$x = \frac{-b}{2a} = \frac{4(2 + 4/3)}{2(1 + (4/3)^2)} = \frac{40/3}{50/9} = 2.4$
Since $x = 2.4 \le 4$, the point of tangency lies well within the allowed half-plane $x \le 4$.
The greatest possible value of $\arg(z)$ is the angle of this tangent line:
$\text{Greatest } \arg(z) = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927 \text{ rad}$