在同一个阿尔冈图中，画出以下轨迹的示意图，并清晰地标出每个轨迹。

1. Identify the shape and key features of the locus.

The equation $|z| = 5$ represents the locus of points $z$ in the complex plane whose distance from the origin is $5$.

This forms a circle with centre $0$ and radius $5$.

2. Determine the intersections with the axes for the sketch.

Since the radius is $5$ and the centre is at the origin, the circle intersects the axes at:

• Real axis: $5$ and $-5$
• Imaginary axis: $5i$ and $-5i$

The sketch will be a circle passing through these points.

1. Express the given equation in Cartesian coordinates by substituting $z = x + iy$.

Let $z = x + iy$.
$|x + iy + 8| = |x + iy - 8i|$
$|(x + 8) + iy| = |x + i(y - 8)|$

2. Evaluate the moduli and simplify to find the Cartesian equation of the locus.

$\sqrt{(x + 8)^2 + y^2} = \sqrt{x^2 + (y - 8)^2}$
$(x + 8)^2 + y^2 = x^2 + (y - 8)^2$
$x^2 + 16x + 64 + y^2 = x^2 + y^2 - 16y + 64$
$16x = -16y$
$y = -x$

3. Identify and describe the locus for sketching.

The locus is the perpendicular bisector of the line segment joining $(-8, 0)$ and $(0, 8)$, which simplifies to the straight line $y = -x$ passing through the origin. This line should be drawn on the Argand diagram and labelled.

1. Express locus (ii) as a straight line

Let $z = x + iy$. Then $|z+8| = |z-8i|$ gives
$(x+8)^2 + y^2 = x^2 + (y-8)^2$
$x^2 + 16x + 64 + y^2 = x^2 + y^2 - 16y + 64$
$16x = -16y \implies y = -x.$

2. Find the intersection with the circle |z| = 5

Substitute $y = -x$ into $x^2 + y^2 = 25$:
$x^2 + x^2 = 25 \implies 2x^2 = 25 \implies x = \pm\frac{5}{\sqrt{2}} = \pm\frac{5\sqrt{2}}{2}.$
So the intersection points are
$\left(\tfrac{5\sqrt{2}}{2},\,-\tfrac{5\sqrt{2}}{2}\right) \quad\text{and}\quad \left(-\tfrac{5\sqrt{2}}{2},\,\tfrac{5\sqrt{2}}{2}\right).$

3. Write the two points as complex numbers and find their arguments

The two complex numbers are
$z_1 = \frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2}i \quad (\arg z_1 = -\tfrac{\pi}{4}),$
$z_2 = -\frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}i \quad (\arg z_2 = \tfrac{3\pi}{4}).$

4. Assign p and q using the argument condition

$\arg\!\left(\dfrac{p}{q}\right) = \arg p - \arg q > 0 \implies \arg p > \arg q.$
Hence $p$ is the point with the larger argument and $q$ the smaller:
$p = -\frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}i, \qquad q = \frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2}i.$

1. Use the geometric properties of the points on the Argand diagram to deduce the argument.

Since $p$ and $q$ are the intersections of the line $y = -x$ (passing through the origin) and the circle $|z| = 5$, they form the endpoints of a diameter of the circle.
The point $z = 5$ also lies on the circle $|z| = 5$ since its distance from the origin is $5$.
By the geometric property of an angle in a semicircle, the angle subtended by the diameter $pq$ at the point $5$ is a right angle ($\frac{\pi}{2}$ rad).
Thus, the vector $5-p$ is perpendicular to the vector $5-q$, which implies that $\frac{5-p}{5-q}$ is purely imaginary, and its argument is either $\frac{\pi}{2}$ or $-\frac{\pi}{2}$.

From the earlier parts, $p$ is in the second quadrant and $q$ is in the fourth quadrant.
The vector $5-q$ (pointing from $q$ to $5$) lies in the first quadrant (positive real and positive imaginary parts), while the vector $5-p$ (pointing from $p$ to $5$) lies in the fourth quadrant (positive real and negative imaginary parts).
Since the angle measured from $5-q$ to $5-p$ is clockwise, the argument difference is negative.
$\arg\left(\frac{5-p}{5-q}\right) = -\frac{\pi}{2}$