Quick Recall
Pick the correct option in each question. These cover the modulus-condition loci from the previous lesson.
The Wall
Modulus conditions tell us about distance — how far a point is from a fixed location. But what about direction?
A radar station at position $a$ detects an aircraft along a bearing of $45^\circ$. This means the aircraft lies somewhere along the ray starting at $a$ at an angle of $45^\circ$. How can we describe this set of points using complex numbers?
The Need
In navigation, air traffic control, and sonar, direction is just as important as distance. The combination of a distance constraint and a direction constraint defines a precise region — a sector of a circle. Today we learn how to write and sketch these direction-based loci using the argument of a complex number.
The half-line includes the point $a$ and extends infinitely in the direction of angle $\alpha$. The point $a$ itself is the endpoint of the ray.
All points lying between the two half-lines at angles $\alpha$ and $\beta$, forming an infinite wedge (sector) starting from $a$.
A region defined by both a modulus condition and an argument condition is the intersection of a circle (or disc) and a sector. The result is a circular sector or a portion of a disc.
For each of the following conditions, sketch the set of points on an Argand diagram. Label the key points and angles clearly.
- $\arg(z) = \dfrac{\pi}{4}$
- $\arg(z - 2\mathrm{i}) = 0$
- $\arg(z + 3) = \dfrac{\pi}{2}$
- $-\dfrac{\pi}{3} \leqslant \arg(z - 1) \leqslant \dfrac{\pi}{3}$
Think-First: Take 2 minutes to visualise each condition before sketching. What does $\arg(z) = \pi/4$ mean geometrically?
Half-line from the origin at $45^\circ$ to the positive real axis, extending into the first quadrant. The origin itself is the endpoint.
Half-line from $2\mathrm{i}$ (point $(0,2)$) at angle $0$, i.e. a horizontal ray to the right, parallel to the positive real axis.
Rewrite as $\arg(z - (-3)) = \pi/2$. Half-line from $(-3,0)$ vertically upward (angle $90^\circ$).
Sector from $1$ on the real axis, bounded by two half-lines at $-60^\circ$ and $60^\circ$ to the positive real axis. The region is the infinite wedge between these rays.
In groups of 3–4, work together to shade the regions defined by the combined conditions below. Each group member should take responsibility for one region, then explain their reasoning to the group.
Region 1: $|z - 2| \leqslant 3$ and $\arg(z - 2) \geqslant 0$
Region 2: $|z - 4\mathrm{i}| \leqslant 5$ and $-\dfrac{\pi}{2} \leqslant \arg(z - 4\mathrm{i}) \leqslant 0$
Region 3: $|z + 2 - \mathrm{i}| \leqslant 4$ and $\dfrac{\pi}{4} \leqslant \arg(z + 2 - \mathrm{i}) \leqslant \dfrac{3\pi}{4}$
Region 4: $|z - 1 - 2\mathrm{i}| \geqslant 3$ and $0 \leqslant \arg(z - 1 - 2\mathrm{i}) \leqslant \dfrac{\pi}{2}$
Peer discussion: Compare your shaded regions. Are there any ambiguities? What happens at the boundary points?
Disc centre $2$, radius $3$, intersected with the upper half-plane (including the horizontal radius). A semicircular disc above the real axis.
Disc centre $4\mathrm{i}$, radius $5$, intersected with the fourth-quadrant sector (angles between $-\pi/2$ and $0$). A quarter-circular disc in the lower-right quadrant relative to the centre.
Disc centre $(-2,1)$, radius $4$, intersected with the sector between $45^\circ$ and $135^\circ$. A quarter-circular disc in the upper-left quadrant relative to the centre.
Exterior (including boundary) of disc centre $(1,2)$, radius $3$, intersected with the sector between $0^\circ$ and $90^\circ$. The region outside the circle in the first quadrant relative to the centre.
Sketch the region satisfying both conditions:
$$|z - 8\mathrm{i}| \leqslant 4 \quad \text{and} \quad -\dfrac{\pi}{4} \leqslant \arg z \leqslant \dfrac{\pi}{4}$$
Hence find the greatest and least possible values of $\arg z$ for points in this region.
This is a closed disc (including boundary) with centre at $8\mathrm{i}$ (point $(0,8)$ on the imaginary axis) and radius $4$.
This is a sector (infinite wedge) from the origin, bounded by rays at $-45^\circ$ and $45^\circ$ to the positive real axis.
The region is the part of the disc $|z - 8\mathrm{i}| \leqslant 4$ that lies within the sector $-\pi/4 \leqslant \arg z \leqslant \pi/4$.
The disc centre is at $(0,8)$; the sector extends outward from the origin. The disc overlaps the sector in a curved region.
The extreme arguments occur at the points where the boundary of the disc intersects the boundary rays of the sector.
The disc has centre $(0,8)$ and radius $4$, so its equation is $x^2 + (y - 8)^2 = 16$.
The ray $\arg z = \pi/4$ has equation $y = x$ (with $x \geqslant 0$).
Substituting: $x^2 + (x - 8)^2 = 16 \implies x^2 + x^2 - 16x + 64 = 16 \implies 2x^2 - 16x + 48 = 0 \implies x^2 - 8x + 24 = 0$.
Discriminant: $64 - 96 = -32 \lt 0$, so the ray $y = x$ does not intersect the disc.
The ray $\arg z = -\pi/4$ has equation $y = -x$ (with $x \geqslant 0$).
Substituting: $x^2 + (-x - 8)^2 = 16 \implies x^2 + x^2 + 16x + 64 = 16 \implies 2x^2 + 16x + 48 = 0 \implies x^2 + 8x + 24 = 0$.
Discriminant: $64 - 96 = -32 \lt 0$, so this ray also does not intersect the disc.
Since the disc is entirely above the origin $(0,8)$, the extreme arguments are determined by the tangents from the origin to the disc.
The distance from origin to centre is $|8\mathrm{i}| = 8$. The radius is $4$. The half-angle of the tangent cone is $\sin \theta = \frac{4}{8} = \frac{1}{2}$, so $\theta = \frac{\pi}{6}$.
The centre is at argument $\pi/2$, so the extreme arguments are $\frac{\pi}{2} \pm \frac{\pi}{6}$.
However, the sector constraint $-\pi/4 \leqslant \arg z \leqslant \pi/4$ restricts us further.
The disc lies entirely above the origin. The ray $\arg z = \pi/4$ is the upper boundary of the sector. The disc intersects the region for arguments between $\pi/2 - \pi/6 = \pi/3$ and $\pi/2 + \pi/6 = 2\pi/3$.
But $\pi/3 \approx 60^\circ$ and $\pi/4 = 45^\circ$, so $\pi/3 > \pi/4$. The sector $-\pi/4 \leqslant \arg z \leqslant \pi/4$ does not intersect the disc at all!
Therefore, the region is empty — there are no points satisfying both conditions simultaneously.
No points satisfy both conditions simultaneously. The region is empty.
For each of the following, draw an Argand diagram showing the set of points $z$ for which the given condition is true.
(i) $\arg z = -\dfrac{\pi}{3}$
(ii) $\arg (z - 4\mathrm{i}) = 0$
Half-line from the origin at angle $-\pi/3$ ($-60^\circ$), extending into the fourth quadrant.
Half-line from $4\mathrm{i}$ (point $(0,4)$) at angle $0$, i.e. a horizontal ray to the right from $(0,4)$.
For each of the following, draw an Argand diagram showing the set of points $z$ for which the given condition is true.
(iii) $\arg (z + 3) \geqslant \dfrac{\pi}{2}$
(iv) $\arg (z + 1 + 2\mathrm{i}) = \dfrac{3\pi}{4}$
Rewrite as $\arg(z - (-3)) \geqslant \pi/2$. This is the region consisting of all half-lines from $(-3,0)$ at angles $\geqslant 90^\circ$, i.e. the entire upper half-plane to the left of (and including) the vertical ray from $(-3,0)$.
Rewrite as $\arg(z - (-1 - 2\mathrm{i})) = 3\pi/4$. Half-line from $(-1, -2)$ at angle $135^\circ$ to the positive real axis, extending into the second quadrant.
For each of the following, draw an Argand diagram showing the set of points $z$ for which the given condition is true.
(v) $\arg (z - 3 + \mathrm{i}) \leqslant -\dfrac{\pi}{6}$
(vi) $-\dfrac{\pi}{4} \leqslant \arg (z + 5 - 3\mathrm{i}) \leqslant \dfrac{\pi}{3}$
Rewrite as $\arg(z - (3 - \mathrm{i})) \leqslant -\pi/6$. The region of points $z$ such that the argument relative to $(3, -1)$ is at most $-30^\circ$. This is the region below and including the ray at $-30^\circ$ from $(3, -1)$.
Rewrite as $-\pi/4 \leqslant \arg(z - (-5 + 3\mathrm{i})) \leqslant \pi/3$. Sector from $(-5, 3)$ bounded by half-lines at $-45^\circ$ and $60^\circ$.
Find the least and greatest possible values of $\arg z$ if $|z - 8\mathrm{i}| \leqslant 4$.
$|z - 8\mathrm{i}| \leqslant 4$ is a closed disc centred at $8\mathrm{i}$ (point $(0,8)$) with radius $4$.
The extreme arguments occur at the points where the tangent lines from the origin touch the circle.
The distance from the origin to the centre is $8$. The radius is $4$.
The angle between the line from origin to centre and a tangent is $\theta$ where $\sin\theta = \dfrac{4}{8} = \dfrac{1}{2}$, so $\theta = \dfrac{\pi}{6}$.
The centre lies on the positive imaginary axis, so its argument is $\dfrac{\pi}{2}$.
The least possible argument is $\dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}$.
The greatest possible argument is $\dfrac{\pi}{2} + \dfrac{\pi}{6} = \dfrac{2\pi}{3}$.
$$\dfrac{\pi}{3} \leqslant \arg z \leqslant \dfrac{2\pi}{3}$$
It is given that $w = 2 + 4\mathrm{i}$ (shown in part (i)). Let $p$ be a real number such that $\dfrac{1}{4}\pi \leqslant \arg(w + p) \leqslant \dfrac{3}{4}\pi$.
Find the set of possible values of $p$.
💡 Hint 1
💡 Hint 2
$w = 2 + 4\mathrm{i}$ is represented by the point $(2, 4)$ on the Argand diagram.
$w + p = (2 + p) + 4\mathrm{i}$ shifts the point horizontally to $(2 + p, 4)$.
$\arg(w + p) \geqslant \pi/4$ means the point $(2 + p, 4)$ lies on or above the line $y = x$ (since $\tan(\pi/4) = 1$).
$\arg(w + p) \leqslant 3\pi/4$ means the point $(2 + p, 4)$ lies on or below the line $y = -x$ (since $\tan(3\pi/4) = -1$).
For $\arg(w + p) \geqslant \pi/4$: the point must satisfy $y \geqslant x$ (in the first quadrant) or be above the $y = x$ line.
$4 \geqslant 2 + p \implies p \leqslant 2$.
For $\arg(w + p) \leqslant 3\pi/4$: the point must satisfy $y \geqslant -x$ (i.e., be above the line $y = -x$ in the second quadrant), or equivalently $4 \geqslant -(2 + p)$.
$4 \geqslant -(2 + p) \implies 4 \geqslant -2 - p \implies p \geqslant -6$.
We also need $2 + p \gt 0$ for the point to be in the first or second quadrant (otherwise $\arg$ is not in $[\pi/4, 3\pi/4]$).
$2 + p \gt 0 \implies p \gt -2$.
Combining: $-2 \lt p \leqslant 2$.
$$-2 \lt p \leqslant 2$$
Exit Ticket
Check the learning objectives you feel confident about: