Argand Diagrams

The Argand diagram turns algebra into geometry. By plotting $z = a + bi$z=a+bi as the point $(a, b)$(a,b), every complex number becomes a point — or a vector from the origin — in a plane, and the operations you learned algebraically acquire vivid geometric meaning: addition becomes vector addition, conjugation becomes reflection, and multiplication by $i$i becomes a quarter-turn. This geometric picture is the foundation for modulus–argument form, loci, and the rotational magic of De Moivre's theorem. Named after Jean-Robert Argand (who popularised it in 1806), the diagram is what made complex numbers feel real to nineteenth-century mathematicians — once you can see $i$i as "the direction perpendicular to the reals," the mystery largely evaporates and is replaced by ordinary plane geometry.

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Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2. Plotting and interpreting Argand diagrams sits across AO1 (plot points, compute distances) and AO2 (interpret operations geometrically, describe regions and justify them). It is the visual launch-pad for modulus and argument, for loci (circles, perpendicular bisectors, half-lines), and ultimately for the transformation geometry of complex multiplication. Diagrams are explicitly creditworthy — a clearly labelled sketch frequently earns marks in its own right. Because so much of the later strand (loci, roots of unity, De Moivre) is most naturally understood pictorially, the time you invest in being fluent with the Argand plane now pays compounding dividends: a student who can instantly visualise "multiply by $i$i" as a quarter-turn, or "$|z - z_0| = r$∣z−z0​∣=r" as a circle, will reason about the harder topics far more quickly than one who only manipulates symbols.

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Core theory: complex numbers as points and vectors

An Argand diagram is a Cartesian plane in which the horizontal axis is the real axis ($\operatorname{Re}$Re) and the vertical axis is the imaginary axis ($\operatorname{Im}$Im). The number $z = a + bi$z=a+bi is the point $(a, b)$(a,b), or equivalently the position vector $\begin{pmatrix} a \\ b \end{pmatrix}$(ab​) from the origin.

$z$z	Point	Lies on
$3 + 2i$3+2i	$(3, 2)$(3,2)	—
$-1 + 4i$−1+4i	$(-1, 4)$(−1,4)	—
$5$5	$(5, 0)$(5,0)	real axis
$-3i$−3i	$(0, -3)$(0,−3)	imaginary axis
$0$0	$(0,0)$(0,0)	origin

Real numbers lie on the real axis; purely imaginary numbers lie on the imaginary axis. The origin represents $0$0, and the two axes meet there at right angles — capturing the idea that "real" and "imaginary" are perpendicular, independent directions.

Operations made geometric

• Addition is vector addition (the parallelogram rule): $z_1 + z_2$z1​+z2​ is the diagonal of the parallelogram with sides $\vec{z_1}$z1​​ and $\vec{z_2}$z2​​.
• Subtraction: $z_1 - z_2$z1​−z2​ is the vector from $z_2$z2​ to $z_1$z1​; its length $|z_1 - z_2|$∣z1​−z2​∣ is the distance between the points.
• Conjugation: $\bar z$zˉ is the reflection of $z$z in the real axis — $(a, b) \mapsto (a, -b)$(a,b)↦(a,−b).
• Negation: $-z$−z is a rotation by $180^\circ$180∘ about the origin — $(a, b) \mapsto (-a, -b)$(a,b)↦(−a,−b).
• Multiplication by $i$i: a $90^\circ$90∘ anticlockwise rotation about the origin, since $i(a + bi) = -b + ai$i(a+bi)=−b+ai, i.e. $(a, b) \mapsto (-b, a)$(a,b)↦(−b,a). Repeating: multiply by $i^2 = -1$i2=−1 for a $180^\circ$180∘ turn, by $i^3 = -i$i3=−i for $270^\circ$270∘ (a $90^\circ$90∘ clockwise turn), and by $i^4 = 1$i4=1 to return home — the powers-of-$i$i cycle realised as quarter-turns of the plane.

These geometric readings are not decoration: they let you see answers. The distance interpretation of $|z_1 - z_2|$∣z1​−z2​∣, in particular, is the basis of every locus question in the strand — a circle is "points at fixed distance from a centre," a perpendicular bisector is "points equidistant from two given points," and so on.

The figure below plots $z_1 = 3 + 2i$z1​=3+2i and $z_2 = -1 + 4i$z2​=−1+4i, their sum $z_1 + z_2 = 2 + 6i$z1​+z2​=2+6i (parallelogram rule), and the conjugate $\bar z_1 = 3 - 2i$zˉ1​=3−2i as a reflection of $z_1$z1​ in the real axis.

Distance, circles and regions

The distance between $z_1 = a + bi$z1​=a+bi and $z_2 = c + di$z2​=c+di is the modulus of their difference:

$|z_1 - z_2| = \sqrt{(a - c)^2 + (b - d)^2},$∣z1​−z2​∣=(a−c)2+(b−d)2​,

exactly the Euclidean distance formula. This one idea generates the standard families of sets you will be asked to recognise and sketch:

Set	Description
${z :	z - z_0
${z :	z - z_0
${z :	z - z_0
${z :	z - z_1
$\{z : \operatorname{Re}(z) > k\}${z:Re(z)>k}	half-plane right of the line $x = k$x=k
$\{z : \operatorname{Im}(z) = k\}${z:Im(z)=k}	horizontal line $y = k$y=k

To read a circle off an expression, always rewrite it in the canonical form $|z - z_0| = r$∣z−z0​∣=r: for instance $|z - 3 + 2i|$∣z−3+2i∣ is $|z - (3 - 2i)|$∣z−(3−2i)∣, so the centre is $(3, -2)$(3,−2), not $(3, 2)$(3,2). To convert a circle to Cartesian form, substitute $z = x + iy$z=x+iy: $|x + iy - (3 - 2i)| = 4$∣x+iy−(3−2i)∣=4 becomes $\sqrt{(x - 3)^2 + (y + 2)^2} = 4$(x−3)2+(y+2)2​=4, i.e. $(x - 3)^2 + (y + 2)^2 = 16$(x−3)2+(y+2)2=16.

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Worked examples (with mark scheme)

Example 1 — distance between two points

Find the distance between $z_1 = 3 + i$z1​=3+i and $z_2 = -1 + 4i$z2​=−1+4i.

$|z_1 - z_2| = |(3 + i) - (-1 + 4i)| = |4 - 3i| \quad (\text{M1 form the difference})$∣z1​−z2​∣=∣(3+i)−(−1+4i)∣=∣4−3i∣(M1 form the difference) $= \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5. \quad (\text{M1 modulus, A1})$=42+(−3)2​=25​=5.(M1 modulus, A1)

(M1 for the difference; M1 for the modulus formula; A1 for $5$5. This is the Euclidean distance $\sqrt{(a-c)^2 + (b-d)^2}$(a−c)2+(b−d)2​.)

Example 2 — rotation by $i$i

Verify geometrically that multiplying $z = 3 + i$z=3+i by $i$i is a $90^\circ$90∘ anticlockwise rotation.

$iz = i(3 + i) = 3i + i^2 = -1 + 3i. \quad (\text{M1})$iz=i(3+i)=3i+i2=−1+3i.(M1)

The point moves $(3, 1) \mapsto (-1, 3)$(3,1)↦(−1,3). Both have modulus $\sqrt{10}$10​, and the argument increases by $90^\circ$90∘ (the position vector turns a quarter-turn anticlockwise). (A1 image; A1 interpretation.)

Example 3 — describing a region

Describe the set $\{z : |z - 3 + 2i| < 4\}${z:∣z−3+2i∣<4}.

$|z - 3 + 2i| = |z - (3 - 2i)| < 4. \quad (\text{M1 rewrite in } |z - z_0| \text{ form})$∣z−3+2i∣=∣z−(3−2i)∣<4.(M1 rewrite in ∣z−z0​∣ form)

This is the interior of the circle centred at $(3, -2)$(3,−2) with radius $4$4, boundary excluded (strict inequality). (A1 centre and radius; A1 "interior, open" — draw the circle dashed.)

Example 4 — a perpendicular bisector

Find the Cartesian equation of the locus $|z - 1| = |z - 3i|$∣z−1∣=∣z−3i∣ and describe it.

Geometrically, this is the set of points equidistant from $1$1 (the point $(1, 0)$(1,0)) and $3i$3i (the point $(0, 3)$(0,3)) — the perpendicular bisector of the segment joining them. Algebraically, put $z = x + iy$z=x+iy:

$|(x - 1) + iy| = |x + (y - 3)i| \;\Rightarrow\; (x - 1)^2 + y^2 = x^2 + (y - 3)^2. \quad (\text{M1 square both moduli})$∣(x−1)+iy∣=∣x+(y−3)i∣⇒(x−1)2+y2=x2+(y−3)2.(M1 square both moduli) $x^2 - 2x + 1 + y^2 = x^2 + y^2 - 6y + 9 \;\Rightarrow\; -2x + 1 = -6y + 9. \quad (\text{M1 expand and cancel})$x2−2x+1+y2=x2+y2−6y+9⇒−2x+1=−6y+9.(M1 expand and cancel) $\therefore\; 6y = 2x + 8 \;\Rightarrow\; y = \tfrac13 x + \tfrac43. \quad (\text{A1})$∴6y=2x+8⇒y=31​x+34​.(A1)

(M1 square; M1 expand/cancel the $x^2, y^2$x2,y2 terms; A1 line. The $x^2$x2 and $y^2$y2 terms always cancel for a "$=$=" of two moduli — the locus is a straight line, never a circle.)

Example 5 — using rotation to find a vertex

Squares $OABC$OABC (labelled anticlockwise) have $O$O at the origin and $A$A at $2 + i$2+i. Find the complex numbers representing $B$B and $C$C.

Going anticlockwise, each side is the previous one rotated $90^\circ$90∘, i.e. multiplied by $i$i. The vector $\vec{OA} = 2 + i$OA=2+i; rotating it gives $\vec{AB} = i(2 + i) = -1 + 2i$AB=i(2+i)=−1+2i. So

$C = i \cdot A = i(2 + i) = -1 + 2i, \qquad B = A + \vec{AB} = (2 + i) + (-1 + 2i) = 1 + 3i. \quad (\text{M1 rotate by } i;\ \text{A1 each})$C=i⋅A=i(2+i)=−1+2i,B=A+AB=(2+i)+(−1+2i)=1+3i.(M1 rotate by i; A1 each)

(M1 for using multiplication by $i$i as the $90^\circ$90∘ rotation; A1 for $C = -1 + 2i$C=−1+2i; A1 for $B = 1 + 3i$B=1+3i. Check: $|OA| = |AB| = \sqrt 5$∣OA∣=∣AB∣=5​ and $\vec{OA}\perp\vec{AB}$OA⊥AB since $\vec{AB} = i\,\vec{OA}$AB=iOA ✓.)

Example 6 — midpoint and centroid

The points $P$P, $Q$Q, $R$R represent $z_1 = 2 + 6i$z1​=2+6i, $z_2 = 8 - 2i$z2​=8−2i, $z_3 = -1 + 2i$z3​=−1+2i. Find the midpoint of $PQ$PQ and the centroid of triangle $PQR$PQR.

The midpoint of a segment is the average of its endpoints, just as for position vectors:

$M = \frac{z_1 + z_2}{2} = \frac{(2 + 6i) + (8 - 2i)}{2} = \frac{10 + 4i}{2} = 5 + 2i. \quad (\text{M1 A1})$M=2z1​+z2​​=2(2+6i)+(8−2i)​=210+4i​=5+2i.(M1 A1)

The centroid is the average of the three vertices:

$G = \frac{z_1 + z_2 + z_3}{3} = \frac{9 + 6i}{3} = 3 + 2i. \quad (\text{M1 A1})$G=3z1​+z2​+z3​​=39+6i​=3+2i.(M1 A1)

(M1 each for the averaging formula; A1 each value. These mirror exactly the vector formulae $\tfrac12(\mathbf{a} + \mathbf{b})$21​(a+b) and $\tfrac13(\mathbf{a} + \mathbf{b} + \mathbf{c})$31​(a+b+c).)

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Specimen-style exam question

(specimen-style — not from any past paper)

The points $A$A, $B$B, $C$C represent $z_1 = 1 + i$z1​=1+i, $z_2 = 5 + i$z2​=5+i, $z_3 = 5 + 4i$z3​=5+4i. (a) Find $|z_2 - z_1|$∣z2​−z1​∣ and $|z_3 - z_2|$∣z3​−z2​∣. (b) Find the complex number represented by the fourth vertex $D$D of the rectangle $ABCD$ABCD. (c) State $|z_3 - z_1|$∣z3​−z1​∣.

(a) $|z_2 - z_1| = |4| = 4$∣z2​−z1​∣=∣4∣=4 and $|z_3 - z_2| = |3i| = 3$∣z3​−z2​∣=∣3i∣=3.

(b) $ABCD$ABCD is a rectangle, so $\vec{AD} = \vec{BC} = z_3 - z_2 = 3i$AD=BC=z3​−z2​=3i. Thus $D = z_1 + 3i = 1 + 4i$D=z1​+3i=1+4i.