A-Level Further Mathematics – Chapter 2: Argand Diagram and Loci
The Argand diagram gives complex numbers a geometric home. Once we represent $z = a + b\mathrm{i}$ as the point $(a, b)$ in the plane, the algebra of complex numbers takes on vivid geometric meaning: addition becomes vector addition, the modulus becomes Euclidean distance, the argument becomes a directed angle, and multiplication becomes a simultaneous scaling and rotation. This chapter develops those geometric ideas into the theory of loci — curves and regions traced out by complex numbers satisfying algebraic conditions — and concludes with an introduction to complex plane transformations.
Specification Note
All content in this chapter is Further level and is assessed in the Further Pure 1 component. Questions on loci appear in nearly every paper; practise sketching as well as finding Cartesian equations.
Contents
2.1 The Argand Diagram Further
Definition 2.1 — Argand Diagram
The Argand diagram (or complex plane) represents the complex number $z = a + b\mathrm{i}$ as the point $(a, b)$ in the Cartesian plane, where the horizontal axis is the real axis and the vertical axis is the imaginary axis.
Complex numbers may also be represented as position vectors $\overrightarrow{OZ}$, in which case addition corresponds to the parallelogram law of vector addition.
Geometric Addition and Subtraction
If $z_1$ and $z_2$ are represented by vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ from the origin, then $z_1 + z_2$ is represented by $\mathbf{v}_1 + \mathbf{v}_2$ (the diagonal of the parallelogram). The difference $z_1 - z_2$ is represented by the vector from $z_2$ to $z_1$.
Conjugate Symmetry
The conjugate $\bar{z} = a - b\mathrm{i}$ is the reflection of $z$ in the real axis. This observation makes it immediate that real polynomials have conjugate-symmetric root sets (as established in Chapter 1).
Example 2.1 — Plotting and Adding
Plot $z_1 = 2 + 3\mathrm{i}$ and $z_2 = -1 + \mathrm{i}$ on an Argand diagram and show geometrically that $z_1 + z_2 = 1 + 4\mathrm{i}$.
Plot the points $(2, 3)$ and $(-1, 1)$. The parallelogram with one diagonal from the origin to $(2,3)$ and another from the origin to $(-1,1)$ has its fourth vertex at $(2 + (-1),\; 3 + 1) = (1, 4)$, confirming $z_1 + z_2 = 1 + 4\mathrm{i}$.
2.2 Geometric Interpretation of Modulus and Argument Further
Theorem 2.1 — Distance and Angle Interpretations
- $|z_1 - z_2|$ equals the Euclidean distance between the points representing $z_1$ and $z_2$ in the Argand diagram.
- $\arg(z - z_0)$ equals the angle that the line segment from $z_0$ to $z$ makes with the positive real direction.
- Multiplying $z$ by $\mathrm{i}$ rotates it by $\frac{\pi}{2}$ anticlockwise about the origin (since $\arg(\mathrm{i}) = \frac{\pi}{2}$).
Example 2.2 — Distance Between Two Complex Numbers
Find the distance between $z_1 = 3 + \mathrm{i}$ and $z_2 = -1 + 4\mathrm{i}$.
$|z_1 - z_2| = |(3+\mathrm{i}) - (-1+4\mathrm{i})| = |4 - 3\mathrm{i}| = \sqrt{16 + 9} = \sqrt{25} = 5.$
Example 2.3 — Rotation by $\mathrm{i}$
Describe geometrically the effect of multiplying $z = 3 + \mathrm{i}$ by $\mathrm{i}$.
$\mathrm{i}(3 + \mathrm{i}) = 3\mathrm{i} + \mathrm{i}^2 = -1 + 3\mathrm{i}$.
The point $(3, 1)$ has been rotated anticlockwise by $90°$ about the origin to $(-1, 3)$. Verify: $|z| = \sqrt{10}$ and $|\mathrm{i}z| = \sqrt{1+9} = \sqrt{10}$; the modulus is preserved.
2.3 Loci of the Form $|z - a| = r$ Further
The condition $|z - a| = r$ (where $a \in \mathbb{C}$ and $r > 0$ is real) says that the distance from the point $z$ to the fixed point $a$ is constantly equal to $r$. This is precisely the definition of a circle.
Theorem 2.2 — Circle Locus
The locus $|z - a| = r$, where $a = p + q\mathrm{i}$ and $r > 0$, is the circle with centre $(p, q)$ and radius $r$ in the Argand diagram. Its Cartesian equation is $(x - p)^2 + (y - q)^2 = r^2$.
Example 2.4 — Circle: Cartesian Equation
Sketch the locus $|z - (3 + 2\mathrm{i})| = 4$ and find its Cartesian equation.
Step 1 The centre is at $(3, 2)$ and the radius is $4$.
Step 2 Write $z = x + y\mathrm{i}$: $|z - 3 - 2\mathrm{i}| = |(x-3) + (y-2)\mathrm{i}| = \sqrt{(x-3)^2 + (y-2)^2} = 4$.
Step 3 Cartesian equation: $(x-3)^2 + (y-2)^2 = 16$.
Example 2.5 — Converting Cartesian Equation to Complex Form
Express the circle $x^2 + y^2 - 6x + 4y - 3 = 0$ in the form $|z - a| = r$.
Step 1 Complete the square: $(x-3)^2 - 9 + (y+2)^2 - 4 - 3 = 0$, so $(x-3)^2 + (y+2)^2 = 16$.
Step 2 Centre $(3, -2)$ corresponds to $a = 3 - 2\mathrm{i}$, radius $r = 4$.
Step 3 Complex form: $|z - (3 - 2\mathrm{i})| = 4$.
Figure 2.1 — The circle $|z - (2+\mathrm{i})| = 3$ drawn in the Argand diagram. The centre $2 + \mathrm{i}$ is marked in red and the radius of $3$ units is shown.
2.4 Loci of the Form $\arg(z - a) = \theta$ Further
The condition $\arg(z - a) = \theta$ fixes the direction of the vector from $a$ to $z$. This traces out a half-line (ray) starting at $a$ (but not including $a$ itself, since $\arg(0)$ is undefined) making angle $\theta$ with the positive real direction.
Definition 2.2 — Half-Line Locus
The locus $\arg(z - a) = \theta$, where $a = p + q\mathrm{i}$ and $\theta \in (-\pi, \pi]$, is the half-line starting at (but excluding) the point $(p, q)$ and making angle $\theta$ with the positive real direction.
Its Cartesian equation is $y - q = \tan\theta\,(x - p)$ with the constraint that $(x, y)$ lies in the correct half-plane (the direction from $(p,q)$ must match $\theta$).
Example 2.6 — Half-Line: Cartesian Equation
Find the Cartesian equation of the locus $\arg(z - (1 + 2\mathrm{i})) = \dfrac{\pi}{4}$ and state any restrictions.
Step 1 The half-line starts at $(1, 2)$ and makes angle $\pi/4$ with the positive real direction.
Step 2 Gradient: $\tan(\pi/4) = 1$. Line equation: $y - 2 = 1 \cdot (x - 1)$, i.e. $y = x + 1$.
Step 3 Restriction: since $\theta = \pi/4$ is in the first quadrant relative to the centre, we need $x > 1$ (and $y > 2$).
Example 2.7 — Half-Line in the Third Quadrant
Sketch and find the Cartesian equation of $\arg(z - 3) = -\dfrac{3\pi}{4}$.
The half-line starts at $(3, 0)$ and makes angle $-3\pi/4$ with the positive real direction. Since $-3\pi/4$ points down-left, the half-line extends into the third quadrant relative to the fixed point.
Gradient: $\tan(-3\pi/4) = \tan(\pi/4) = 1$ (since $\tan$ has period $\pi$). So $y - 0 = 1 \cdot (x - 3)$, i.e. $y = x - 3$, with $x < 3$.
2.5 Loci of the Form $|z - a| = |z - b|$ Further
The condition $|z - a| = |z - b|$ says that $z$ is equidistant from the fixed points $a$ and $b$. In Euclidean geometry, the set of all points equidistant from two given points is the perpendicular bisector of the segment joining them.
Theorem 2.3 — Perpendicular Bisector Locus
The locus $|z - a| = |z - b|$, where $a, b \in \mathbb{C}$ and $a \neq b$, is the perpendicular bisector of the line segment joining the points $a$ and $b$ in the Argand diagram.
Example 2.8 — Perpendicular Bisector
Find the Cartesian equation of the locus $|z - (1 + \mathrm{i})| = |z - (5 - \mathrm{i})|$.
Step 1 Write $z = x + y\mathrm{i}$. Then: $$\sqrt{(x-1)^2 + (y-1)^2} = \sqrt{(x-5)^2 + (y+1)^2}.$$
Step 2 Square both sides: $(x-1)^2 + (y-1)^2 = (x-5)^2 + (y+1)^2$.
Step 3 Expand: $x^2 - 2x + 1 + y^2 - 2y + 1 = x^2 - 10x + 25 + y^2 + 2y + 1$.
Step 4 Simplify: $-2x - 2y + 2 = -10x + 2y + 26$, so $8x - 4y = 24$, giving $2x - y = 6$.
Check: The midpoint of $(1,1)$ and $(5,-1)$ is $(3, 0)$; substituting: $2(3) - 0 = 6$ ✓. The gradient of the segment is $(-1-1)/(5-1) = -\frac{1}{2}$; the perpendicular gradient is $2$, and $y = 2x - 6$ has gradient $2$ ✓.
Figure 2.2 — The circle $|z - (2+\mathrm{i})| = 3$ (blue), the half-line $\arg(z - (1+2\mathrm{i})) = \pi/4$ (green ray), and the perpendicular bisector of $(0,0)$ and $(4,2)$ (red dashed), illustrated together in the Argand diagram.
2.6 Regions and Intersections Further
Inequalities involving $|z - a|$, $\arg(z - a)$, or combinations of these define regions in the Argand diagram. Exam questions frequently ask for a sketch of a shaded region, for the Cartesian inequalities defining it, or for the algebraic coordinates of intersection points.
Definition 2.3 — Regions
- $|z - a| \leq r$: the closed disc of radius $r$ centred at $a$ (interior plus boundary circle).
- $|z - a| \geq r$: the exterior of the disc plus boundary.
- $\alpha \leq \arg(z - a) \leq \beta$: the sector between the half-lines at angles $\alpha$ and $\beta$ from $a$.
- $|z - a| \leq |z - b|$: the half-plane on the $a$-side of the perpendicular bisector of $ab$.
Example 2.9 — Describing a Region
Sketch and describe in words the region $R$ satisfying both $|z - 2| \leq 3$ and $0 \leq \arg(z) \leq \dfrac{\pi}{2}$.
The first condition is the disc centred at $(2, 0)$ with radius $3$: it covers the range $-1 \leq x \leq 5$.
The second condition is the first quadrant (including the positive real and imaginary axes).
The region $R$ is the part of the disc that lies in the first quadrant — a sector-like shape bounded by an arc of the circle and portions of the coordinate axes.
Example 2.10 — Finding an Intersection Point
The loci $|z - 2| = |z - 4\mathrm{i}|$ and $|z| = 5$ intersect. Find all intersection points.
Step 1 The perpendicular bisector of $(2, 0)$ and $(0, 4)$: write $z = x + y\mathrm{i}$, then $(x-2)^2 + y^2 = x^2 + (y-4)^2$, giving $-4x + 4 = -8y + 16$, so $4x - 8y = -12$, i.e. $x - 2y = -3$, or $x = 2y - 3$.
Step 2 Substitute into $|z| = 5$, i.e. $x^2 + y^2 = 25$: $(2y-3)^2 + y^2 = 25$, so $4y^2 - 12y + 9 + y^2 = 25$, giving $5y^2 - 12y - 16 = 0$.
Step 3 Discriminant: $144 + 320 = 464$. $y = \dfrac{12 \pm \sqrt{464}}{10} = \dfrac{12 \pm 4\sqrt{29}}{10} = \dfrac{6 \pm 2\sqrt{29}}{5}$.
Corresponding $x = 2y - 3 = \dfrac{-3 \pm 4\sqrt{29}}{5}$. The two intersection points are $z = \dfrac{-3 + 4\sqrt{29}}{5} + \dfrac{6 + 2\sqrt{29}}{5}\mathrm{i}$ and $z = \dfrac{-3 - 4\sqrt{29}}{5} + \dfrac{6 - 2\sqrt{29}}{5}\mathrm{i}$.
Exam Tip — Sketching Regions
Always label: (i) the centre and radius for circle loci; (ii) the starting point and angle for half-lines; (iii) the midpoint and gradient for perpendicular bisectors. Indicate the boundary clearly (solid line for $\leq$, dashed for $<$) and shade the correct region. Losing marks for missing labels is avoidable.
2.7 Transformations in the Complex Plane Further
A complex transformation is a function $w = f(z)$ mapping the complex plane (the $z$-plane) to another copy of the complex plane (the $w$-plane). The image of a curve or region under $f$ can be found by substituting $z = f^{-1}(w)$ and simplifying.
Translation: $w = z + c$
Every point is shifted by the vector $c \in \mathbb{C}$. The image of the circle $|z - a| = r$ under $w = z + c$ is the circle $|w - (a + c)| = r$.
Scaling and Rotation: $w = kz$, $k \in \mathbb{C}$
Multiplication by $k = |k|e^{\mathrm{i}\alpha}$ scales distances from the origin by $|k|$ and rotates by $\alpha = \arg(k)$. Circles centred at the origin map to circles centred at the origin; circles not centred at the origin generally map to circles of different radius and centre.
Conjugate Reflection: $w = \bar{z}$
This reflects every point in the real axis. A circle $|z - a| = r$ maps to $|w - \bar{a}| = r$.
Example 2.11 — Image of a Circle Under Translation
Find the image of $|z - 3| = 2$ under the transformation $w = z + \mathrm{i}$.
Write $z = w - \mathrm{i}$. Substitute: $|w - \mathrm{i} - 3| = 2$, i.e. $|w - (3 + \mathrm{i})| = 2$.
The image is the circle centred at $3 + \mathrm{i}$ with radius $2$ — the original circle translated by $\mathrm{i}$ (one unit upwards).
The Transformation $w = 1/z$
The reciprocal transformation maps $z = re^{\mathrm{i}\theta}$ to $w = \frac{1}{r}e^{-\mathrm{i}\theta}$: it inverts the modulus and negates the argument. Circles and lines (which may be regarded as circles of infinite radius) generally map to circles or lines.
Example 2.12 — Image Under $w = 1/z$
Find the image of the line $\operatorname{Re}(z) = 2$ (i.e. $x = 2$) under $w = 1/z$.
Step 1 Write $z = x + y\mathrm{i}$ and $w = u + v\mathrm{i}$. Then $z = 1/w$, so $x + y\mathrm{i} = \dfrac{u - v\mathrm{i}}{u^2 + v^2}$.
Step 2 Real part of $z$: $x = \dfrac{u}{u^2 + v^2} = 2$, giving $u = 2(u^2 + v^2)$.
Step 3 Rearrange: $2u^2 - u + 2v^2 = 0$, i.e. $u^2 - \frac{1}{2}u + v^2 = 0$, so $\left(u - \frac{1}{4}\right)^2 + v^2 = \frac{1}{16}$.
The image is a circle in the $w$-plane centred at $\frac{1}{4}$ (on the real axis) with radius $\frac{1}{4}$.
Example 2.13 — Image of a Half-Line Under Rotation
Find the image of the half-line $\arg(z) = \dfrac{\pi}{3}$ under the transformation $w = 2\mathrm{i}z$.
Write $z = w/(2\mathrm{i})$. Then $\arg(z) = \arg(w) - \arg(2\mathrm{i}) = \arg(w) - \pi/2$.
Setting this equal to $\pi/3$: $\arg(w) = \pi/3 + \pi/2 = 5\pi/6$.
The image is the half-line $\arg(w) = 5\pi/6$ (pointing into the second quadrant of the $w$-plane). The modulus is scaled by $|2\mathrm{i}| = 2$, so the half-line extends to all $|w|$ values.
Practice Problems
Problem 1 — Distance
Find the distance between $z_1 = -2 + 5\mathrm{i}$ and $z_2 = 4 - 3\mathrm{i}$.
Show solution
$|z_1 - z_2| = |(-2-4) + (5-(-3))\mathrm{i}| = |-6 + 8\mathrm{i}| = \sqrt{36 + 64} = \sqrt{100} = 10$.
Problem 2 — Circle Locus
Sketch the locus $|z + 3 - \mathrm{i}| = 5$. State the centre and radius, and find the Cartesian equation.
Show solution
Write as $|z - (-3 + \mathrm{i})| = 5$. Centre $(-3, 1)$, radius $5$.
Cartesian equation: $(x + 3)^2 + (y - 1)^2 = 25$.
Problem 3 — Half-Line Locus
Find the Cartesian equation of $\arg(z - 2 - 3\mathrm{i}) = \dfrac{2\pi}{3}$, stating the domain restriction.
Show solution
Starting point $(2, 3)$, angle $2\pi/3$. Gradient $= \tan(2\pi/3) = -\sqrt{3}$.
Line: $y - 3 = -\sqrt{3}(x - 2)$, i.e. $y = -\sqrt{3}\,x + 2\sqrt{3} + 3$.
The direction $2\pi/3$ points into the second quadrant relative to $(2,3)$, so the restriction is $x < 2$.
Problem 4 — Perpendicular Bisector
Find the Cartesian equation of the locus $|z - 4| = |z + 2\mathrm{i}|$.
Show solution
$(x-4)^2 + y^2 = x^2 + (y+2)^2$.
$x^2 - 8x + 16 + y^2 = x^2 + y^2 + 4y + 4$.
$-8x + 16 = 4y + 4$, so $-8x - 4y = -12$, giving $2x + y = 3$.
Problem 5 — Region Description
Describe and sketch the region satisfying $|z - \mathrm{i}| \leq 2$ and $\operatorname{Re}(z) \geq 0$.
Show solution
$|z - \mathrm{i}| \leq 2$ is the closed disc centred at $(0, 1)$ with radius $2$.
$\operatorname{Re}(z) \geq 0$ is the right half-plane including the imaginary axis.
The region is the right semicircle: the part of the disc with $x \geq 0$. It is bounded by the arc of the circle in the right half-plane and by the chord of the circle on the imaginary axis (from $(0, -1)$ to $(0, 3)$).
Problem 6 — Intersection of Loci
Find the points of intersection of $|z - 3\mathrm{i}| = 5$ and $|z| = |z - 6|$.
Show solution
The second locus is the perpendicular bisector of $0$ and $6$ (i.e. $0$ and $6+0\mathrm{i}$): $x^2 + y^2 = (x-6)^2 + y^2$, giving $x^2 = x^2 - 12x + 36$, so $x = 3$.
Substitute $x = 3$ into $|z - 3\mathrm{i}| = 5$: $(3)^2 + (y-3)^2 = 25$, so $(y-3)^2 = 16$, giving $y = 7$ or $y = -1$.
Intersection points: $z = 3 + 7\mathrm{i}$ and $z = 3 - \mathrm{i}$.
Problem 7 — Rotation by i
The point $z = 1 + 2\mathrm{i}$ is multiplied successively by $\mathrm{i}$ four times. List the four results and describe the geometric effect.
Show solution
$z_1 = \mathrm{i}(1+2\mathrm{i}) = \mathrm{i} - 2 = -2 + \mathrm{i}$.
$z_2 = \mathrm{i}(-2+\mathrm{i}) = -2\mathrm{i} - 1 = -1 - 2\mathrm{i}$.
$z_3 = \mathrm{i}(-1-2\mathrm{i}) = -\mathrm{i} + 2 = 2 - \mathrm{i}$.
$z_4 = \mathrm{i}(2-\mathrm{i}) = 2\mathrm{i} + 1 = 1 + 2\mathrm{i} = z$.
Each multiplication rotates the point by $90°$ anticlockwise about the origin; after four such rotations, we return to $z$. The four points form a square.
Problem 8 — Image Under Translation
Find the image of the circle $|z - 1| = 3$ under the transformation $w = z - 2 + 4\mathrm{i}$.
Show solution
$z = w + 2 - 4\mathrm{i}$. Substitute: $|w + 2 - 4\mathrm{i} - 1| = 3$, i.e. $|w + 1 - 4\mathrm{i}| = 3$, or $|w - (-1 + 4\mathrm{i})| = 3$.
The image is the circle centred at $-1 + 4\mathrm{i}$ with radius $3$.
Problem 9 — Image Under $w = 1/z$
Find the image of the circle $|z| = 2$ under the transformation $w = 1/z$.
Show solution
Write $z = 1/w$. Then $|1/w| = 2$, i.e. $1/|w| = 2$, giving $|w| = 1/2$.
The image is the circle $|w| = 1/2$, centred at the origin with radius $1/2$.
Problem 10 — Composite Locus
Sketch the set of points $z$ satisfying $|z - 2| \leq |z + 2|$ and $\arg(z - \mathrm{i}) \leq \dfrac{\pi}{4}$. State the boundary of each constraint in Cartesian form.
Show solution
Constraint 1: $|z - 2| \leq |z + 2|$ is the half-plane on the side of $+2$ of the perpendicular bisector of $-2$ and $2$. The bisector is $x = 0$ (the imaginary axis). So the region is $x \leq 0$ (the left half-plane, including the boundary $x = 0$).
Constraint 2: $\arg(z - \mathrm{i}) \leq \pi/4$ means the argument of $z - \mathrm{i}$ is at most $\pi/4$. This gives the region lying on or below (in terms of argument) the half-line $y - 1 = x - 0$, i.e. $y = x + 1$ for $x \geq 0$. Combined with the first constraint ($x \leq 0$), the boundary from Constraint 2 does not intersect the region from Constraint 1 except at $(0, 1)$.
The combined region is: the entire left half-plane $x \leq 0$, together with points in the right half-plane satisfying $\arg(z - \mathrm{i}) \leq \pi/4$.