IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2HL 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.
Conjugationz*: 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 algebraicallyz + w = (1 + 5) + (4 + 2)i = 6 + 6iStep 2: Sketch all three vectors from the originz + 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 subtractionz − w = (3 − 4) + (5 − (−2))i= −1 + 7iStep 2: Geometric interpretationsubtracting 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 = 6arg(z₁z₂) = π/4 + π/6 = 3π/12 + 2π/12 = 5π/12|z₁z₂| = 6, arg(z₁z₂) = 5π/12no 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 iziz = i(4 + 3i) = 4i + 3i²= 4i − 3 = −3 + 4iStep 2: Compare positionsz 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 originiz = −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 − 15iStep 2: Break into “scale” and “flip” partsmultiply 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 22z = 2(3 + i) = 6 + 2i (length doubled, same direction)Step 2: iz — rotate 90° counter-clockwiseiz = i(3 + i) = 3i + i² = −1 + 3i (turned left a quarter turn)Step 3: z* — reflect in real axisz* = 3 − i (flipped across the horizontal axis)Step 4: z·z* = |z|² — lands on the positive real axisz·z* = (3+i)(3−i) = 9 − i² = 9 + 1 = 102z = 6+2i, iz = −1+3i, z* = 3−i, z·z* = 10notice 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.
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