On a single Argand diagram sketch the loci given by
(a) $|z - 12 - 12 ext{i}| = 12$,
(b) $| ext{i}z - 1 - 12 ext{i}| = 5$. Hence, or otherwise, find the complex numbers $z_1$ and $z_2$ that satisfy both (a) and (b). Given that $z_1$ and $z_2$ also satisfy $|z - 12 - 12 ext{i}| = |z - a - b ext{i}|$. Find the real values of $a$ and $b$. Find the least possible value of $\arg(w - 30 - 12 ext{i})$ given that $w$ satisfy $|w - 12 - 12 ext{i}| = 12$ and $\text{Re}(w) \ge 12$.

1. Identify the geometric properties of locus (a).

The equation is $|z - (12 + 12\text{i})| = 12$.
This represents a circle with centre $C_1$ at $12 + 12\text{i}$ (or coordinates $(12, 12)$) and radius $r_1 = 12$.
Because both the real and imaginary parts of the centre are equal to the radius, the circle is tangent to both the real and imaginary axes.

2. Simplify the equation for locus (b) to the standard modulus form.

$|\text{i}z - 1 - 12\text{i}| = 5$
Factor out $\text{i}$ inside the modulus:
$\left|\text{i}\left(z - \frac{1}{\text{i}} - 12\right)\right| = 5$
Since $|\text{i}| = 1$ and $\frac{1}{\text{i}} = -\text{i}$, we have:
$|1| \cdot |z - (-\text{i}) - 12| = 5$
$|z - (12 - \text{i})| = 5$

3. Identify the geometric properties of locus (b).

$|z - (12 - \text{i})| = 5$
This represents a circle with centre $C_2$ at $12 - \text{i}$ (or coordinates $(12, -1)$) and radius $r_2 = 5$.

4. Determine the relative position of the circles and describe the sketch.

Both centres lie on the vertical line $\text{Re}(z) = 12$.
The distance between the centres is $12 - (-1) = 13$.
The sum of their radii is $12 + 5 = 17$, and the difference is $12 - 5 = 7$.
Since $7 < 13 < 17$, the two circles intersect at two points.

Sketch requirements on a single Argand diagram:

• Real (Re) and Imaginary (Im) axes.
• Circle (a) in the first quadrant, touching the axes at $(12,0)$ and $(0,12)$, with centre $(12,12)$ marked.
• Circle (b) crossing the real axis, with centre $(12,-1)$ marked and radius $5$ (extending from $\text{Im}(z)=4$ to $\text{Im}(z)=-6$).
• The intersection of the two circles clearly shown.

1. Express the equation of the first locus in Cartesian form.

Let $z = x + y\text{i}$.
For $|z - 12 - 12\text{i}| = 12$:
$|(x - 12) + (y - 12)\text{i}| = 12$
$(x - 12)^2 + (y - 12)^2 = 144 \quad \text{--- (1)}$

2. Express the equation of the second locus in Cartesian form.

For $|\text{i}z - 1 - 12\text{i}| = 5$:
$|\text{i}(x + y\text{i}) - 1 - 12\text{i}| = 5$
$|\text{i}x - y - 1 - 12\text{i}| = 5$
$|(-y - 1) + (x - 12)\text{i}| = 5$
$(-y - 1)^2 + (x - 12)^2 = 25$
$(x - 12)^2 + (y + 1)^2 = 25 \quad \text{--- (2)}$

3. Solve the equations simultaneously to find $y$.

Subtracting (2) from (1) to eliminate $(x-12)^2$:
$(y - 12)^2 - (y + 1)^2 = 144 - 25$
$(y^2 - 24y + 144) - (y^2 + 2y + 1) = 119$
$-26y + 143 = 119$
$26y = 24 \implies y = \frac{12}{13}$

4. Find the corresponding values of $x$ and state the complex numbers.

Substitute $y = \frac{12}{13}$ into (2):
$(x - 12)^2 + \left(\frac{12}{13} + 1\right)^2 = 25$
$(x - 12)^2 + \left(\frac{25}{13}\right)^2 = 25$
$(x - 12)^2 = 25 - \frac{625}{169} = \frac{3600}{169}$
$x - 12 = \pm \frac{60}{13}$
$x = 12 \pm \frac{60}{13}$
$x = \frac{216}{13} \quad \text{or} \quad x = \frac{96}{13}$
Thus, the complex numbers are $z_1 = \frac{216}{13} + \frac{12}{13}\text{i}$ and $z_2 = \frac{96}{13} + \frac{12}{13}\text{i}$.

1. Find the Cartesian equation of the line passing through $z_1$ and $z_2$ using the loci from (i).

From part (i), we have two circles:
$C_1$: $|z - 12 - 12\text{i}|^2 = 144 \implies (x - 12)^2 + (y - 12)^2 = 144$
$C_2$: $|\text{i}z - 1 - 12\text{i}|^2 = 25 \implies |z - (12 - \text{i})|^2 = 25 \implies (x - 12)^2 + (y + 1)^2 = 25$

$z_1$ and $z_2$ are the intersections of $C_1$ and $C_2$. Subtracting the equations gives the common chord passing through them:
$((y - 12)^2 - (y + 1)^2 = 144 - 25$
$y^2 - 24y + 144 - y^2 - 2y - 1 = 119$
$-26y = -24 \implies y = \frac{12}{13}$

2. Interpret the given locus geometrically.

The locus of $z$ defined by $|z - 12 - 12\text{i}| = |z - a - b\text{i}|$ is the perpendicular bisector of the line segment joining $(12, 12)$ and $(a, b)$.
Since $z_1$ and $z_2$ both satisfy this equation, the straight line passing through $z_1$ and $z_2$ (which is $y = \frac{12}{13}$) must exactly be this perpendicular bisector.

3. Determine $a$ and $b$ using perpendicularity and the midpoint formula.

Since the bisector $y = \frac{12}{13}$ is a horizontal line, the line segment joining $(12, 12)$ and $(a, b)$ must be vertical. Thus, their real parts are equal:
$a = 12$

The midpoint of this vertical segment must lie on the bisector $y = \frac{12}{13}$:
$\frac{12 + b}{2} = \frac{12}{13}$
$12 + b = \frac{24}{13}$
$b = \frac{24}{13} - 12 = -\frac{132}{13}$

1. Set up the geometry and identify where the least argument occurs

$|w-12-12\mathrm{i}|=12$ is the circle centre $C(12,12)$, radius $r=12$ (with $\operatorname{Re}(w)\ge 12$ giving the right-hand half).

$\arg(w-30-12\mathrm{i})$ is the argument of the vector from the fixed point $P(30,12)$ to $w$. Distance
$PC = |30-12| = 18.$
Since $P$ is outside the circle and the circle lies to the left of $P$, the vector $w-P$ always points up-/down-left. The extreme (least and greatest) arguments occur when $Pw$ is tangent to the circle; the least argument corresponds to the upper tangent.

2. Compute the tangent angle to obtain the least argument

In the right-angled triangle formed by $P$, $C$ and the tangent point $T$ (right angle at $T$):
$\sin(\angle CPT) = \frac{r}{PC} = \frac{12}{18} = \frac{2}{3}, \qquad \angle CPT = \sin^{-1}\tfrac{2}{3} \approx 41.8^\circ.$
The direction from $P$ to $C$ has argument $180^\circ$ (pointing in the negative real direction). The upper tangent lies $41.8^\circ$ above this, so
$\arg(w-30-12\mathrm{i})_{\text{least}} = 180^\circ - \sin^{-1}\tfrac{2}{3} \approx 138.2^\circ \ (\approx 2.41\text{ rad}).$