IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 HL only ~10 min read

Geometry of Complex Numbers

Every algebraic operation on complex numbers has a visual meaning on an Argand diagram. Adding two complex numbers? That’s tip-to-tail vector addition — the result is the diagonal of a parallelogram. Subtracting? Reverse the second arrow first. Multiplying? Scale and rotate. Conjugating? Flip across the horizontal axis. This note is about turning every operation you already know how to do algebraically into a picture you can read at a glance. Once you can see the geometry, exam questions about Argand diagrams stop being puzzles and start being obvious.

📘 What you need to know

Addition: z + w is the diagonal of the parallelogram formed by the vectors for z and w. Equivalent to a translation of z by the vector for w.
Subtraction: z − w is the diagonal of the parallelogram with vertices at z, −w and z − w. Always plot −w (the reverse of w) first.
Multiplication by z₂: scales by |z₂| and rotates by arg(z₂) counter-clockwise.
Division by z₂: scales by 1/|z₂| and rotates by −arg(z₂) (clockwise).
Multiplying by a real number k: pure scaling — the vector keeps its direction (or flips 180° if k is negative).
Multiplying by i: rotates 90° counter-clockwise. Multiplying by −i rotates 90° clockwise.
Conjugation z*: reflects z in the real axis. z · z* is always a positive real number equal to |z|².

Addition — vectors tip-to-tail

Adding two complex numbers z and w looks exactly like adding two vectors. Travel along z, then continue along w from where z ended. The arrow from the origin to where you stop represents z + w.

Addition: parallelogram law

Equivalently, you can think of adding w = a + bi to z as a translation — slide z across by a and up by b. Same result, different mental picture.

Addition is commutative: z + w = w + z. So you can travel along z first or w first — the destination is the same. That’s why the picture is a parallelogram, not just a triangle.

Subtraction — flip the second arrow

Subtracting w from z isn’t quite as simple as addition. The trick is: plot −w first (the reverse of w), then add it tip-to-tail to z.

Subtraction: travel along z, then along −w

Geometrically, subtracting w = a + bi from z is a translation by the vector with components −a and −b — exactly opposite to addition.

Order matters! z − w ≠ w − z. The two are negatives of each other, so they point in opposite directions on the diagram.

Multiplication — scale and rotate

This is where complex numbers get really beautiful. When you multiply z₁ by z₂, the result is found by:

Scale

enlarge by |z₂|

The length of z₁ is multiplied by |z₂|.

Rotate

turn by arg(z₂)

Rotate counter-clockwise by the argument of z₂.

Multiplying z₁ by z₂: scale by |z₂|, rotate by arg(z₂)

🤔 Why scale and rotate?

From the previous note, you already know |z₁·z₂| = |z₁|·|z₂| and arg(z₁·z₂) = arg(z₁) + arg(z₂). Geometrically, multiplying lengths is “scaling”, and adding arguments is “rotating”. So multiplication does both at once — the algebraic rule and the geometric picture are exactly the same fact.

Division reverses both moves: scale by 1/|z₂| (a shrink) and rotate by −arg(z₂) (clockwise).

Special cases — multiplying by real numbers and by i

Two special cases are worth committing to memory because they show up in nearly every Argand-diagram question.

Multiplying by a real number k

A real number has argument 0 (if positive) or π (if negative). So multiplying by k doesn’t rotate — it just scales:

Multiplying by a real number multiply z by k (real) → scales by |k|
same direction if k > 0, flipped 180° if k < 0

Multiplying by i (or −i)

The number i has modulus 1 and argument π/2. So multiplying by i doesn’t change the length — but rotates 90° counter-clockwise. Multiplying by −i rotates 90° clockwise.

Multiplying by i multiply by i → rotate 90° counter-clockwise
multiply by −i → rotate 90° clockwise

Quick check: if z = 3 + 2i, then iz = i(3 + 2i) = 3i − 2 = −2 + 3i. The point (3, 2) has rotated 90° counter-clockwise to (−2, 3) — exactly as predicted.

Conjugation — reflection in the real axis

You met this in the previous Argand-diagram note, but it’s worth repeating because it shows up everywhere. The conjugate z* of z = a + bi is a − bi — the same horizontal position, with the imaginary part flipped.

Conjugate: reflection in the real axis

Useful fact: z · z* is always a positive real number — it equals |z|². This sits on the positive real axis on the Argand diagram.

Worked examples

WE 1

Addition on an Argand diagram

Let z = 1 + 4i and w = 5 + 2i. On an Argand diagram, sketch z, w and z + w, and verify that the result is the diagonal of a parallelogram.

Step 1: Compute z + w algebraically z + w = (1 + 5) + (4 + 2)i = 6 + 6i Step 2: Sketch all three vectors from the origin z + w = 6 + 6i, the diagonal of the parallelogram ✓ the dashed lines complete the parallelogram with vertices O, z, w, and z+w

WE 2

Subtraction on an Argand diagram

Let z = 3 + 5i and w = 4 − 2i. Find z − w and describe it as a translation of z.

Step 1: Compute the subtraction z − w = (3 − 4) + (5 − (−2))i = −1 + 7i Step 2: Geometric interpretation subtracting w means translating z by the vector for −w −w = −4 + 2i, so translation is “4 left, 2 up” z − w = −1 + 7i (z translated 4 left and 2 up) starting at z = (3, 5), shifting by (−4, 2) lands at (−1, 7) — exactly the algebraic answer

WE 3

Multiplication as scale and rotate

The complex number z₁ has modulus 3 and argument π/4. The complex number z₂ has modulus 2 and argument π/6. Describe geometrically the effect of multiplying z₁ by z₂, and find the modulus and argument of z₁z₂.

Step 1: Multiplication = scale by |z₂|, rotate by arg(z₂) z₁ gets scaled by 2 (so length 3·2 = 6) z₁ gets rotated by π/6 (counter-clockwise) Step 2: Compute new modulus and argument |z₁z₂| = |z₁|·|z₂| = 3 · 2 = 6 arg(z₁z₂) = π/4 + π/6 = 3π/12 + 2π/12 = 5π/12 |z₁z₂| = 6, arg(z₁z₂) = 5π/12 no need to multiply (a+bi)(c+di) — the geometric rule does it in one line

WE 4

Multiplying by i — the 90° rotation

Let z = 4 + 3i. Compute iz, and describe the geometric relationship between z and iz.

Step 1: Compute iz iz = i(4 + 3i) = 4i + 3i² = 4i − 3 = −3 + 4i Step 2: Compare positions z is at point (4, 3) iz is at point (−3, 4) Step 3: Geometric description (4, 3) → (−3, 4) is a 90° counter-clockwise rotation about the origin iz = −3 + 4i (z rotated 90° counter-clockwise about O) multiplying by i is the cleanest rotation in mathematics — turn 90° to the left, no scale change

WE 5

Multiplying by a negative real number

Let z = 2 + 5i. Compute −3z and describe the geometric effect.

Step 1: Compute −3z −3z = −3(2 + 5i) = −6 − 15i Step 2: Break into “scale” and “flip” parts multiply by 3 → enlarge by factor 3 (length 3 times bigger) multiply by −1 → rotate 180° about the origin (flip direction) −3z = −6 − 15i (enlarged ×3 and rotated 180°) a negative real number = positive multiplier × (−1), so it’s “scale + 180° flip”

WE 6

All transformations at once

Let z = 3 + i. Find the values of 2z, iz, z* and z·z*, and describe each transformation geometrically.

Step 1: 2z — pure scaling by 2 2z = 2(3 + i) = 6 + 2i (length doubled, same direction) Step 2: iz — rotate 90° counter-clockwise iz = i(3 + i) = 3i + i² = −1 + 3i (turned left a quarter turn) Step 3: z* — reflect in real axis z* = 3 − i (flipped across the horizontal axis) Step 4: z·z* = |z|² — lands on the positive real axis z·z* = (3+i)(3−i) = 9 − i² = 9 + 1 = 10 2z = 6+2i, iz = −1+3i, z* = 3−i, z·z* = 10 notice z·z* = |z|² = 3² + 1² = 10, sitting on the real axis as a positive number

💡 Top tips

Sketch the parallelogram for addition. Drawing all four vertices (O, z, w, z+w) makes the diagonal fall into place automatically.
Always plot −w before drawing z−w. Forgetting this is the most common mistake on subtraction sketches.
Multiplication = scale and rotate, simultaneously. The two effects happen together; you don’t do one then the other in any specific order.
Multiplying by i = rotate 90° counter-clockwise. Drill this into memory — it’s a one-mark gift in many exam questions.
Multiplying by a negative real number = scale + 180° flip. The sign causes a half-turn; the magnitude does the scaling.
Conjugation is just a reflection. No length change, no rotation — the modulus is preserved.
z·z* always lands on the positive real axis at the value |z|². This is a standard exam answer worth memorising.

⚠ Common mistakes

Treating z−w as if it were w−z. They’re negatives of each other, pointing in opposite directions. Order matters for subtraction.
Forgetting to plot −w first when sketching z−w. Without −w, the parallelogram doesn’t close in the right place.
Adding moduli for sums. |z+w| ≠ |z|+|w| in general. Vectors don’t add lengths unless they’re parallel.
Multiplying lengths but forgetting to add arguments. Multiplication has both effects — you need both.
Multiplying arguments instead of adding. arg(z₁·z₂) = arg(z₁) + arg(z₂), not the product of the two arguments.
Confusing “multiply by i” with “multiply by −1”. The first is a 90° rotation; the second is a 180° rotation. Different turns entirely.
Confusing conjugation with negation. z* reflects across the real axis (flip the imaginary part). −z rotates 180° about the origin (flip both parts).

Once you can see the algebra in the picture, complex numbers feel a lot more natural. Every operation has a clean geometric meaning, and the rest of the Further Complex Numbers section — polar form, exponential form, De Moivre’s theorem — leans heavily on this geometric picture. The next note formalises the modulus-and-argument view into polar form, which is the standard way to write complex numbers when multiplication and powers are the main concern.

Need help with the Geometry of Complex Numbers?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →