IB Maths AI HLComplex NumbersPaper 1 & 2Argand diagram~8 min read
Modulus & Argument
Plot a complex number on an Argand diagram and it becomes an arrow from the origin. That arrow has two measurements: its length, the modulus, and the angle it turns from the positive real axis, the argument.
📘 What you need to know
The modulus |z| is the distance from the origin: |z| = √(x² + y²).
A modulus is never negative, and |zz*| relates to it by zz* = |z|².
The argument arg z is the angle to the positive real axis, in radians, with −π < arg z ≤ π.
The argument’s sign and size depend on the quadrant — always sketch first.
Multiplying numbers multiplies moduli and adds arguments.
Dividing numbers divides moduli and subtracts arguments.
The modulus of a complex number
On an Argand diagram, the complex number z = x + yi is a point at (x, y). The modulus, written |z| or r, is its straight-line distance from the origin. By Pythagoras on the right triangle with legs x and y, |z| = √(x² + y²).
The modulus |z| is the hypotenuse of the right triangle with legs x and y; the argument θ is the angle the arrow makes with the positive real axis.
A modulus is never negative — it is a distance. Note too that adding complex numbers does not add their moduli: |z1 + z2| ≠ |z1| + |z2| in general. The modulus also links to the conjugate: since zz* = (x + yi)(x − yi) = x² + y², it follows that zz* = |z|².
Modulus & the conjugate
|z| = √(x² + y²)
zz* = z*z = |z|²multiplying by the conjugate produces a real number — the modulus squared
The argument of a complex number
The argument, arg z or θ, is the angle the arrow makes with the positive real axis, measured counter-clockwise, in radians. The standard (principal) range is −π < arg z ≤ π, so anticlockwise angles are positive and clockwise ones negative.
Use right-angled trigonometry: the reference angle is α = tan−1(|y| / |x|). The final argument then depends on the quadrant, so a quick sketch is essential.
Quadrant rule for the argument (reference angle α): quadrant 1 → θ = α (positive acute); quadrant 2 → θ = π − α (positive obtuse); quadrant 3 → θ = −(π − α) (negative obtuse); quadrant 4 → θ = −α (negative acute). The argument of 0 is undefined.
Modulus and argument under multiplication and division
Modulus and argument behave very neatly when complex numbers are multiplied or divided. The moduli combine by multiplication or division, while the arguments combine by addition or subtraction.
Multiplication & division rules: |z1z2| = |z1||z2| and arg(z1z2) = arg z1 + arg z2; |z1/z2| = |z1| / |z2| and arg(z1/z2) = arg z1 − arg z2. In short — moduli multiply or divide; arguments add or subtract.
🧠Recipe — finding the modulus and argument
Sketch z = x + yi on an Argand diagram to see which quadrant it lies in.
Modulus: compute |z| = √(x² + y²).
Reference angle: find α = tan−1(|y| / |x|).
Adjust for the quadrant: 1 → α, 2 → π−α, 3 → −(π−α), 4 → −α.
State |z| and arg z, keeping arg z in −π < arg z ≤ π.
Worked examples
WE 1
Finding the modulus
Find the modulus of (a) z = 5 − 12i and (b) w = −8 + 6i.
Find the argument of z = 4 + 4i, giving an exact answer.
sketch: x > 0, y > 0 ⇒ first quadrantargument is positive and acute: θ = tan⁻¹(y / x)θ = tan⁻¹(4 / 4) = tan⁻¹(1)arg z = π/4π/4 ≈ 0.785 — no quadrant adjustment is needed in quadrant 1.
WE 3
Argument in the second quadrant
Find the argument of z = −3 + 3i, giving an exact answer.
sketch: x < 0, y > 0 ⇒ second quadrantreference angle α = tan⁻¹(|y| / |x|)α = tan⁻¹(3 / 3) = π/4quadrant 2: θ = π − αθ = π − π/4arg z = 3π/4positive and obtuse — exactly what quadrant 2 should give.
WE 4
Modulus from the conjugate
For z = 6 − 8i, find zz* and hence write down |z|.
z* = 6 + 8i; multiply zz*(6 − 8i)(6 + 8i) = 36 − 64i² = 36 + 64zz* = 100use zz* = |z|², so |z| = √(zz*)zz* = 100 · |z| = 10zz* lands on a real number — its square root is the modulus.
WE 5
Multiplying and dividing
Two complex numbers have |z1| = 6, arg z1 = π/3 and |z2| = 2, arg z2 = π/12. Find the modulus and argument of (a) z1z2 and (b) z1/z2.
(a) product: multiply moduli, add arguments|z₁z₂| = 6 × 2 = 12arg = π/3 + π/12 = 4π/12 + π/12 = 5π/12(b) quotient: divide moduli, subtract arguments|z₁/z₂| = 6 ÷ 2 = 3arg = π/3 − π/12 = 3π/12 = π/4(a) 12, 5π/12 · (b) 3, π/4no need to convert to Cartesian form — work directly with modulus and argument.
WE 6
Full question: fourth quadrant and a product
Let z = √3 − i. (a) Find |z|. (b) Find arg z. (c) Given w has |w| = 5 and arg w = π/2, find the modulus and argument of zw.
(a) |z| = √((√3)² + (−1)²)= √(3 + 1) = √4 = 2(b) x > 0, y < 0 ⇒ quadrant 4, negative acuteα = tan⁻¹(1 / √3) = π/6 ⇒ arg z = −π/6(c) zw: multiply moduli, add arguments|zw| = 2 × 5 = 10; arg = −π/6 + π/2 = π/3(a) 2 · (b) −π/6 · (c) |zw| = 10, arg(zw) = π/3a quadrant-4 number has a negative argument — the sketch confirms it.
💡 Top tips
Always sketch first — the quadrant decides the sign and size of the argument.
Work the reference angle from positive side lengths, then adjust for the quadrant.
Keep arg z in the principal range −π < arg z ≤ π unless told otherwise.
For products and quotients, use modulus–argument rules — far quicker than expanding.
Use zz* = |z|² as a shortcut to a modulus without a square root step.
âš Common mistakes
A negative modulus — a distance is never negative; check your arithmetic.
Skipping the quadrant adjustment — tan−1 alone only gives the reference angle.
Working in degrees — arguments are given in radians.
Adding moduli — |z1 + z2| is not |z1| + |z2|.
An argument outside the range — add or subtract 2π to land inside −π < arg z ≤ π.
Next up: Introduction to Argand Diagrams — the complex plane in full. You will plot complex numbers as points and vectors, and read numbers straight off a diagram.
Need help with AI HL Complex Numbers?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
