Complex numbers and de Moivre’s theorem
Use polar form and de Moivre’s theorem for powers, roots, multiple-angle identities and trigonometric sums.
Section 2.5 · Complex numbers
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Interpret multiplication and division geometrically in polar form
- Use de Moivre’s theorem for integer and rational powers
- Find roots of unity and express powers or sums of trigonometric ratios
Polar form makes powers a rotation and scaling
Write z=r(cos θ+i sin θ). Multiplying complex numbers multiplies moduli and adds arguments; division divides moduli and subtracts arguments.
De Moivre gives zⁿ=rⁿ(cos nθ+i sin nθ) for integer n. Equating real and imaginary parts produces multiple-angle identities; negative powers reverse the rotation and scale. The positive-integer result can be proved by induction using the angle-addition formulae.
Rational powers create several roots
For z^q=w, write w in polar form and divide each argument by q, adding 2πk/q for k=0,…,q−1. The roots are equally spaced around a circle.
For trigonometric sums, use e^{iθ}=cos θ+i sin θ and the finite geometric sum. Take the real or imaginary part only after summing; this is the C+iS method.
Pause and explain
Which habit best checks your work on complex numbers and de moivre’s theorem?
Worked example
Find (1+i)⁶ exactly.
Show the worked solution
- Write 1+i=√2(cos π/4+i sin π/4).
- Raise the modulus and argument: (√2)⁶=8 and 6π/4=3π/2.
- So the value is 8(cos 3π/2+i sin 3π/2).
Answer −8i.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find all cube roots of 8i.
Give me a hint
Write 8i with modulus 8 and principal argument π/2, then divide each argument by 3.
Compare my reasoning
- The modulus of each root is 2.
- Arguments are (π/2+2πk)/3 for k=0,1,2.
- These are π/6, 5π/6 and 3π/2.
2cis(π/6), 2cis(5π/6), and 2cis(3π/2).
Look for these in your work
- Root modulus correct
- Three argument branches included
- Roots equally spaced
Use the C+iS method to find Σ from r=0 to 5 of cos(rπ/3).
Give me a hint
Sum the sixth roots of unity, then take the real part.
Compare my reasoning
- Σe^{irπ/3}=(1−e^{i2π})/(1−e^{iπ/3})=0.
- The required cosine sum is the real part of this complex sum.
- Therefore the cosine sum is zero.
0.
Look for these in your work
- Geometric sum used
- Full cycle recognised
- Real part selected
Explain why zⁿ=1 has n equally spaced roots on the unit circle for positive integer n.
Give me a hint
Write 1 in polar form and include the full set of arguments modulo 2π.
Compare my reasoning
- A root must have modulus 1 because |z|ⁿ=1.
- Its argument θ must satisfy nθ=2πk.
- Thus θ=2πk/n for k=0,…,n−1, equally spaced by 2π/n.
The roots are cis(2πk/n), k=0,…,n−1; they lie equally spaced on the unit circle.
Look for these in your work
- Modulus condition derived
- Argument equation solved
- All distinct roots counted
Common mistakes
- Forgetting to include all argument branches when finding roots
- Raising the modulus to n but not multiplying the argument by n
- Taking real parts term-by-term before applying the finite geometric sum
What can you explain now?
What happens to modulus and argument when two complex numbers are multiplied?
Compare with the explanation
Moduli multiply and arguments add
This is why polar form is useful for powers.
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Make it stick
Why does polar form turn powers and roots of complex numbers into modulus and angle calculations?
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