在同一个阿甘图中，画出以下轨迹的草图，并清晰地标出每个轨迹。

1. Identify the geometric meaning of the equation.

The modulus $|z|$ represents the distance of the complex number $z$ from the origin $(0,0)$. Thus, the equation $|z| = 5$ represents all points on the Argand plane that are exactly $5$ units away from the origin.

2. State the features of the locus to be sketched.

The locus is a circle with its centre at the origin and a radius of $5$. Sketch this circle on the Argand diagram ensuring it intersects the real axis at $5$ and $-5$, and the imaginary axis at $5i$ and $-5i$, and label it $|z| = 5$.

1. Substitute the Cartesian form $z = x + iy$ into the locus equation.

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

2. Use the definition of modulus to find the Cartesian equation of the locus.

$(x + 8)^2 + y^2 = x^2 + (y - 8)^2$
$x^2 + 16x + 64 + y^2 = x^2 + y^2 - 16y + 64$
$16x = -16y$
$y = -x$

3. Describe the required sketch on the Argand diagram.

The equation $y = -x$ represents a straight line passing through the origin with a gradient of $-1$.

Draw this line on the Argand diagram and label it as $|z + 8| = |z - 8i|$. (Note: Geometrically, this line is the perpendicular bisector of the line segment joining $(-8, 0)$ and $(0, 8)$)..

1. Find the Cartesian equation of locus (ii)

Let $z = x + iy$. Then
$|z+8| = |z-8i| \Rightarrow (x+8)^2 + y^2 = x^2 + (y-8)^2.$
Expanding:
$x^2+16x+64+y^2 = x^2+y^2-16y+64 \Rightarrow 16x = -16y \Rightarrow y = -x.$
Locus (ii) is the line $y=-x$ (the perpendicular bisector of $(-8,0)$ and $(0,8)$).

2. Solve the loci simultaneously

Locus (i) is $|z|=5 \Rightarrow x^2+y^2 = 25$. Substituting $y=-x$:
$x^2 + x^2 = 25 \Rightarrow 2x^2 = 25 \Rightarrow x = \pm\frac{5}{\sqrt2} = \pm\frac{5\sqrt2}{2}.$
Hence the intersection points are
$z = \frac{5\sqrt2}{2} - \frac{5\sqrt2}{2}i \quad\text{and}\quad z = -\frac{5\sqrt2}{2} + \frac{5\sqrt2}{2}i.$

3. Assign $p$ and $q$ using $\arg(p/q)>0$

$\arg\!\left(\dfrac{p}{q}\right) = \arg p - \arg q > 0 \Rightarrow \arg p > \arg q.$

The point $-\dfrac{5\sqrt2}{2}+\dfrac{5\sqrt2}{2}i$ lies in the 2nd quadrant with $\arg = \dfrac{3\pi}{4}$, while $\dfrac{5\sqrt2}{2}-\dfrac{5\sqrt2}{2}i$ lies in the 4th quadrant with $\arg = -\dfrac{\pi}{4}$.

Since we need $\arg p > \arg q$:
$p = -\frac{5\sqrt2}{2} + \frac{5\sqrt2}{2}i, \qquad q = \frac{5\sqrt2}{2} - \frac{5\sqrt2}{2}i.$

1. Identify the geometrical relationship forming a right angle

From parts (i) and (ii), the complex numbers $p$ and $q$ lie on the line $y = -x$ and the circle $|z| = 5$. This makes $p$ and $q$ diametrically opposite points on the circle.
The point $z = 5$ also lies on the circumference of the circle $|z| = 5$.
By the 'angle in a semicircle' property, the line segments connecting $z=5$ to $p$ and $q$ are perpendicular. Hence, the angle between the vectors $5-p$ and $5-q$ is $\frac{\pi}{2}$.

2. Determine the exact directed angle

We require the directed angle $\arg\left(\frac{5-p}{5-q}\right) = \arg(5-p) - \arg(5-q)$.
From part (iii), $p$ is in the second quadrant and $q$ is in the fourth quadrant.
The vector $5-p$ (from $p$ to $5$) goes down and right, so it lies in the fourth quadrant, giving it a negative argument.
The vector $5-q$ (from $q$ to $5$) goes up and right, so it lies in the first quadrant, giving it a positive argument.
Since $\arg(5-p) < 0$ and $\arg(5-q) > 0$, the difference $\arg(5-p) - \arg(5-q)$ must be negative.
$\arg\left(\frac{5-p}{5-q}\right) = -\frac{\pi}{2}$