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IB DP Maths AA · Topic 1 Number and algebra · AHL 1.12, 1.13, 1.14 · ~6 min read

Complex numbers

Cartesian form and arithmetic with complex numbers, the Argand diagram, modulus and argument, polar and Euler forms, conjugate roots of polynomials, De Moivre's theorem, nth roots and trig identities.

i² = −1Cartesian formArgand diagramModulus & argumentPolar & Euler formConjugate rootsDe Moivrenth roots
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No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • HL only. A complex number z = a + bi has real part a, imaginary part b, and i2 = −1. Its conjugate is z* = a − bi.
  • On an Argand diagram z is a point; its distance from 0 is the modulus |z| = r and its angle from the positive real axis is the argument θ.
  • Polar form z = r(cos θ + i sin θ) = r cis θ = reiθ makes multiplication easy: multiply the moduli, add the arguments.
  • De Moivre: (r cis θ)n = rn cis nθ. It gives powers, all n nth roots, and multiple-angle identities.
  • Real polynomials have complex roots in conjugate pairs.

By the end you can

  • Add, subtract, multiply and divide complex numbers in Cartesian form. HL
  • Represent complex numbers on an Argand diagram and find modulus and argument. HL
  • Convert between Cartesian, polar and Euler forms and multiply and divide in polar form. HL
  • Use the conjugate root theorem to solve polynomial equations. HL
  • Use De Moivre's theorem for powers, roots and trigonometric identities. HL

1Cartesian form HL

  • Arithmetic: add real and imaginary parts separately; multiply out brackets using i2 = −1; divide by multiplying top and bottom by the conjugate of the bottom.
Worked example

Write 3 + 4i1 − 2i in the form a + bi.

  1. × 1 + 2i1 + 2i: numerator (3 + 4i)(1 + 2i) = 3 + 10i + 8i2 = −5 + 10i M1A1
  2. denominator 12 + 22 = 5
  3. = −1 + 2i A1
Worked example

Solve z2 + 4z + 13 = 0.

  1. z = −4 ± √(16 − 52)2 = −4 ± 6i2 M1
  2. z = −2 ± 3i A1
  • Equal complex numbers have equal real parts AND equal imaginary parts: one complex equation gives two real equations.

2Modulus, argument and the Argand diagram HL

ReIm-3-2-1123-2-1120z = −1 + √3 iz* = −1 − √3 iarg z = 2π/3|z| = 2z = r(cos θ + i sin θ)= re^(iθ) = r cis θr = 2, θ = 2π/3
Modulus is length, argument is angle. The conjugate is the reflection in the real axis.
|z| = r = √(a2 + b2), arg z = θ with tan θ = ba
choose θ in the correct quadrant (usually −π < θ ≤ π)
z = r(cos θ + i sin θ) = reiθ = r cis θ FORMULA BOOKLET
polar and Euler forms
Worked example

Write z = −1 + √3 i in polar form.

  1. r = √(1 + 3) = 2 A1
  2. z is in the 2nd quadrant: θ = π − π/3 = 2π/3 M1A1
  3. z = 2 cis (2π/3) = 2e2πi/3 A1
Common mistakeTaking θ = arctan(b/a) without checking the quadrant. arctan(√3/−1) gives −π/3, which points into the 4th quadrant: wrong.
  • Multiplying: |z1z2| = |z1||z2| and arg(z1z2) = arg z1 + arg z2. Dividing: divide moduli, subtract arguments. Geometrically, multiplying by cis θ rotates a point by θ.

3Conjugate roots HL

  • Conjugate root theorem: if a polynomial has real coefficients and a + bi is a root, then a − bi is also a root. Their product (z − (a + bi))(z − (a − bi)) = z2 − 2az + (a2 + b2) is a real quadratic factor.
Worked example

Given that 3i is a root of z3 − 2z2 + 9z − 18 = 0, find the other roots.

  1. −3i is also a root, so (z2 + 9) is a factor M1
  2. z3 − 2z2 + 9z − 18 = (z2 + 9)(z − 2) A1
  3. roots 3i, −3i, 2 A1

4De Moivre's theorem HL

[r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ) = rneinθ FORMULA BOOKLET
true for all n ∈ ℤ (and proved by induction for n ∈ ℤ+)
Worked example

Find (1 + i)8.

  1. 1 + i = √2 cis (π/4) A1
  2. (1 + i)8 = (√2)8 cis 2π M1
  3. = 16 A1

nth roots

ReIm√3 + i−√3 + i−2iThe 3 cube roots of 8i: equal modulus 2, arguments 2π/3 apart: a regular triangle.
The n roots of any complex number are equally spaced around a circle.
Worked example

Solve z3 = 8i.

  1. 8i = 8 cis(π/2 + 2kπ) M1
  2. z = 2 cis(π6 + 2kπ3), k = 0, 1, 2 M1A1
  3. z = √3 + i, −√3 + i, −2i A1
  • Roots of unity: zⁿ = 1 has roots cis(2kπ/n), k = 0, 1, …, n − 1. They add up to zero.

Trig identities from De Moivre

Worked example

Show that cos 3θ = 4cos3 θ − 3cos θ.

  1. (cos θ + i sin θ)3 = cos 3θ + i sin 3θ (De Moivre) M1
  2. expand: real part = cos3 θ − 3cos θ sin2 θ A1
  3. = cos3 θ − 3cos θ(1 − cos2 θ) = 4cos3 θ − 3cos θ A1AG
Exam tipEuler's form makes powers and products quickest: (reiθ)n = rneinθ. Convert back to Cartesian only at the end if the question asks.
GDC (calculator) methodPaper 2: set the GDC to complex mode (a + bi or reiθ) to check arithmetic, moduli and arguments. Still write the method: polar form, then De Moivre.
Linking questionComplex numbers make rotations and oscillations simple: physics and engineering use eiωt for waves and AC circuits, and Euler's formula links e, i and π: eiπ + 1 = 0.
Idea for your IA / explorationExplore fractals built from complex iteration (the Mandelbrot set, z → z2 + c), or the geometry of the nth roots of unity and regular polygons.

Test yourself — cover the answers first

Quick check

  1. Simplify i3 and i4.
  2. Find |3 − 4i|.
  3. Find the conjugate of 2 − 5i.
  4. What happens to z on an Argand diagram when multiplied by i?
  5. Write 4eiπ/2 in Cartesian form.
  6. How many roots does z5 = 1 have?

Exam-style

  1. Given z = 2cis(π/3) and w = 3cis(π/6), find zw and z/w in polar form.
  2. Find all solutions of z4 = −16.
Show answers
  1. −i and 1
  2. 5
  3. 2 + 5i
  4. it rotates by π/2 anticlockwise
  5. 4i
  6. 5
  7. zw = 6 cis(π/2) = 6i
    z/w = 23 cis(π/6)
  8. −16 = 16 cis(π + 2kπ)
    z = 2 cis(π/4 + kπ/2)
    z = √2 ± √2 i, −√2 ± √2 i
Practise this chapter
IB Maths AA practice papers
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.