在同一个阿冈图（Argand diagram）上，画出以下轨迹的草图：

1. Identify the shape, centre, and radius of the locus

The equation is in the standard form $|z - z_0| = r$, where $z_0 = 3$ and $r = 5$.
This represents a circle in the Argand plane with centre at $3$ (the point $(3, 0)$) and radius $5$.

2. Determine key intersections to sketch the circle accurately

Intersections with the real axis (where the imaginary part is $0$):
$|x - 3| = 5 \implies x - 3 = 5 \text{ or } x - 3 = -5$
$\implies x = 8 \text{ or } x = -2$

Intersections with the imaginary axis (where the real part is $0$):
$|iy - 3| = 5 \implies \sqrt{(-3)^2 + y^2} = 5 \implies 9 + y^2 = 25 \implies y^2 = 16 \implies y = \pm 4$

Draw axes, mark the centre $(3,0)$ and intercepts $-2, 8, 4i, -4i$, and sketch the circle passing through them.

1. Identify the base point and the shape of the locus.

The equation $\arg(z - 8) = \alpha$ represents a locus that is a half-line starting from the point corresponding to $z = 8$. On the Argand diagram, this is the point $(8, 0)$.

2. Find the gradient and a reference point to sketch the half-line accurately.

The angle $\pi - \tan^{-1}\left(\frac{1}{2}\right)$ indicates that the half-line points towards the second quadrant relative to $(8,0)$.

The gradient of the half-line is $\tan\left(\pi - \tan^{-1}\left(\frac{1}{2}\right)\right) = -\frac{1}{2}$.

The line equation from $(8,0)$ with gradient $-\frac{1}{2}$ is $y - 0 = -\frac{1}{2}(x - 8)$, which simplifies to $y = -\frac{1}{2}x + 4$.

The half-line passes through the imaginary axis at $y = 4$, which corresponds to the point $(0, 4)$.

3. State the sketching instructions.

Sketch a straight half-line originating from an empty (unshaded) circle at $(8, 0)$, passing through $(0, 4)$, and continuing infinitely.

1. Identify the geometric meaning of the given locus equation.

The equation $|z + 2| = |z - 3 + 5i|$ can be rewritten as $|z - (-2)| = |z - (3 - 5i)|$.
This represents the perpendicular bisector of the line segment joining the points $A(-2, 0)$ and $B(3, -5)$.

2. Determine the Cartesian equation of the locus to identify key passing points for an accurate sketch.

The midpoint of $A$ and $B$ is $\left(\frac{-2+3}{2}, \frac{0-5}{2}\right) = \left(\frac{1}{2}, -\frac{5}{2}\right)$.
The gradient of the line segment $AB$ is $\frac{-5 - 0}{3 - (-2)} = -1$.
Therefore, the gradient of the perpendicular bisector is $1$.
The equation of the perpendicular bisector is:
$y - \left(-\frac{5}{2}\right) = 1\left(x - \frac{1}{2}\right)$
$y = x - 3$
The locus is a straight line passing through $(3, 0)$ and $(0, -3)$.

3. Sketch the locus on the Argand diagram.

Draw a straight line passing through the real axis at $3$ and the imaginary axis at $-3$. Note that this line passes exactly through the centre $(3, 0)$ of the circle from part (i).

1. Formulate Cartesian equations for the loci described in (i) and (ii).

Let $z = x + iy$.
Locus (i) is a circle: $|(x-3) + iy| = 5 \implies (x-3)^2 + y^2 = 25$.
Locus (ii) is a half-line: $\arg((x-8) + iy) = \pi - \tan^{-1}\left(\frac{1}{2}\right)$.
This gives $\frac{y}{x-8} = -\frac{1}{2}$ with $x < 8$ (since the angle is in the second quadrant relative to $(8,0)$).
So, $y = -\frac{1}{2}(x-8)$ for $x < 8$.

2. Solve the equations simultaneously to find the intersection points.

Substitute the equation of the half-line into the circle's equation:
$(x-3)^2 + \left(-\frac{1}{2}(x-8)\right)^2 = 25$
$(x-3)^2 + \frac{1}{4}(x-8)^2 = 25$
$4(x^2 - 6x + 9) + (x^2 - 16x + 64) = 100$
$4x^2 - 24x + 36 + x^2 - 16x + 64 = 100$
$5x^2 - 40x = 0$
$5x(x-8) = 0$
$x = 0$ or $x = 8$.

3. Apply the domain restriction to identify the valid intersection point and state $p$.

Since the half-line is restricted to $x < 8$, we reject $x = 8$.
For $x = 0$, $y = -\frac{1}{2}(0-8) = 4$.
Therefore, the point of intersection is $(0, 4)$, which corresponds to the complex number $p = 0 + 4i$.

1. Interpret the first inequality geometrically.

The inequality $|w - 3| \le 5$ refers to the set of points with a distance of $5$ or less from the point $3$. This represents the region inside and on the boundary of the circle with centre $3$ and radius $5$.

2. Interpret the second inequality geometrically.

The inequality $\arg(w - 8) \le \pi - \tan^{-1}\left(\frac{1}{2}\right)$ refers to angles smaller than or equal to the angle of the half-line in part (ii). This corresponds to the region swept clockwise from the half-line starting at $z = 8$.

3. Interpret the third inequality geometrically.

The inequality $|w + 2| \ge |w - 3 + 5i|$ requires the distance from $w$ to $-2$ to be greater than or equal to the distance from $w$ to $3-5i$. The boundary is the perpendicular bisector from part (iii). The valid region is the half-plane containing the point $3-5i$.

4. Identify and describe the final region to be shaded.

The locus of $w$ is the intersection of these three regions. On the Argand diagram, shade the area that is simultaneously inside the circle, below the half-line, and on the $3-5i$ side (below/right) of the perpendicular bisector.