IB Maths AI HLFurther Complex NumbersPaper 1 & 2Argand transformations~8 min read
Geometry of Complex Numbers
Every operation on a complex number has a picture on the Argand diagram. Adding is a translation, multiplying is a rotation paired with a scaling, and conjugating is a reflection. Once you see the geometry, the algebra makes intuitive sense.
📘 What you need to know
Additionz + w is the resultant of the two vectors — the diagonal of the parallelogram O, z, w, z+w.
Subtractionz − w uses the reverse vector −w. Order matters: z−w ≠ w−z.
Multiplyingz1 by z2scales by |z2| and rotates by arg z2.
Dividing by z2 scales by 1/|z2| and rotates by −arg z2.
Special cases: × a real k is pure scaling; × i is a pure 90° rotation (counter-clockwise).
Conjugationz → z* is a reflection in the real axis.
Addition and subtraction as vectors
Treat each complex number as a vector from O. Addition uses the head-to-tail rule: travel along z then w, and the resultant is the diagonal of the parallelogram O, z, w, z+w. Order doesn’t matter: z + w = w + z.
For subtraction reverse the second vector first — z − w is z followed by −w. Swapping the order flips the result: w − z = −(z − w).
Translation interpretation: adding w = a + bi to z is the same as translating z by the column vector (a, b); subtracting w translates by (−a, −b).
Multiplication and division as scale and rotation
Multiplication has the most striking geometry. To form z1z2, take the arrow for z1, rotate it about the origin by arg z2, then stretch (or shrink) its length by the factor |z2|. Division undoes both: rotate by −arg z2 and stretch by 1/|z2|.
Multiplying z = 3 + i by 2i: first i rotates 90° counter-clockwise to give iz (same length), then the factor 2 stretches to twice the length to land at 2iz.
Geometry of × and ÷
× z2: enlarge by |z2|, rotate by arg z2÷ z2: enlarge by 1/|z2|, rotate by −arg z2the geometric reading of |z1z2| = |z1||z2| and arg(z1z2) = arg z1 + arg z2
Special cases: real, imaginary, and the conjugate
Three multipliers come up so often that their pictures are worth memorising. A real multiplier k has zero argument, so it scales but doesn’t rotate. A purely imaginary multiplier ki has argument ±π/2, so it rotates 90° and scales by k. Conjugation swaps y for −y — geometrically a reflection across the real axis.
Quick reference: × k (k real, positive) — pure scaling, no rotation; × i — rotate 90° counter-clockwise; × (−i) — rotate 90° clockwise; × (−1) — rotate 180°; z → z* — reflect in the real axis.
🧠Recipe — sketching z₁z₂ geometrically
Plot z1 and z2 as vectors from the origin on an Argand diagram.
Read off |z2| and arg z2 — the scale factor and rotation angle.
Rotate the vector z1 about the origin by arg z2 (counter-clockwise if positive).
Scale the rotated vector by the factor |z2| — stretch if > 1, shrink if < 1.
Mark and label the endpoint as z1z2; check by expanding (z1)(z2) algebraically.
Worked examples
WE 1
Addition on an Argand diagram
Given z = 1 + 4i and w = 5 + 2i, find z + w and describe its position on the Argand diagram.
add real parts, then imaginary parts(1 + 5) + (4 + 2)i = 6 + 6igeometric meaningz + w = 6 + 6iit sits at the fourth corner of the parallelogram with corners O, z, w — the diagonal from O.
WE 2
Subtraction is not commutative
Given z = 4 + 3i and w = 2 − i, find (a) z − w and (b) w − z. Compare them geometrically.
(a) z − w(4 − 2) + (3 − (−1))i = 2 + 4i(b) w − z(2 − 4) + (−1 − 3)i = −2 − 4i(a) 2 + 4i · (b) −2 − 4iopposite vectors: same length, opposite direction — order of subtraction flips the arrow.
WE 3
Multiplication as scale and rotation
Let z = 3 + i and w = 1 + i. (a) Find |w| and arg w. (b) State the transformation z undergoes when multiplied by w. (c) Find zw.
(a) |w| = √(1 + 1), arg w from quadrant 1|w| = √2, arg w = π/4(b) z is rotated by π/4 and scaled by √2(c) expand zw = (3 + i)(1 + i)3 + 3i + i + i² = 2 + 4i(a) √2, π/4 · (b) rotate by π/4, scale by √2 · (c) 2 + 4icheck: |2 + 4i| = √20 = √10 × √2 = |z||w| — the rule lands true.
WE 4
Division by an imaginary number
Find (8 + 6i) ÷ (2i) in Cartesian form and describe what the transformation does geometrically.
multiply top and bottom by −i (the conjugate-trick for 2i)(8 + 6i)(−i) ÷ (2i)(−i) = (−8i − 6i²) ÷ 2= (6 − 8i) ÷ 2geometric meaning: |2i| = 2, arg(2i) = π/23 − 4i — scale by ½ and rotate by −π/2starts in quadrant 1, ends in quadrant 4 — a clockwise quarter-turn plus a shrink to half-length.
WE 5
The conjugate as a reflection
Let z = 5 + 2i. Write down z* and describe the geometric relation between z and z*. Then evaluate zz*.
flip the sign of the imaginary partz* = 5 − 2igeometrically: reflection in the real axiszz* = (5 + 2i)(5 − 2i) = 25 − 4i²= 25 + 4z* = 5 − 2i · zz* = 29zz* is the product of mirror-image vectors — always a real number, equal to |z|².
WE 6
Full question: three transformations
Let z = 2 + 2i. (a) Find |z| and arg z. (b) Find iz in Cartesian form and describe how it relates to z geometrically. (c) Find zz*.
(a) |z| = √(4 + 4), arg z in quadrant 1|z| = 2√2, arg z = π/4(b) iz = i(2 + 2i) = 2i + 2i²= −2 + 2i× i ⇒ rotate z by 90° counter-clockwise; same modulus(c) zz* = (2 + 2i)(2 − 2i) = 4 − 4i²= 8(a) 2√2, π/4 · (b) −2 + 2i, rotation +π/2 · (c) 8notice 8 = (2√2)² = |z|², confirming zz* = |z|².
💡 Top tips
For addition use the parallelogram rule; for subtraction draw −w first, then add.
Multiplying by a positive real changes length only — angles stay the same.
Multiplying by i is a 90° counter-clockwise rotation; multiplying by −i is 90° clockwise.
Multiplying by −1 is a 180° rotation — the same as taking −z.
A point lies on the real axis iff z = z* — the reflection coincides with the original.
âš Common mistakes
Swapping the order of subtraction — z − w and w − z are opposite vectors, not the same.
Forgetting the rotation in multiplication — only scaling the length gives a wrong answer.
Wrong rotation direction — arg z2 > 0 is counter-clockwise, arg z2 < 0 is clockwise.
Confusing z* with −z — conjugate reflects in the real axis; −z is a 180° rotation.
Not adjusting the new argument back into −π < arg z ≤ π after a rotation pushes it past π.
Next up: Modulus-Argument (Polar) Form — rewriting z using its modulus and argument as z = r(cos θ + isin θ). The geometric rules from this note become an algebraic shortcut.
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