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Prove the conjugate-power identity for positive integers n.

Read the idea, work independently, then explain what changed.

TOPIC 01

2022 JM02

The negative power is a reciprocal; it is a conjugate because the modulus is one.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Prove the conjugate-power identity for positive integers n.

ωn+ω−n=2cos⁡2nπ7,ω=cis⁡(2π/7)\omega^n+\omega^{-n}=2\cos\frac{2n\pi}7,\quad\omega=\operatorname{cis}(2\pi/7)

Official paper · jm02-2022 · 4(b)(ii) · PDF 6

Official original and suggested answers ↗ · Suggested answer PDF page 11

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use |ω|=1 to interpret the negative power.
Hint 2
The two imaginary parts cancel.
Worked solution
  1. Apply De Moivre to both signs.

    ω±n=cos⁡(2nπ/7)±isin⁡(2nπ/7)\omega^{\pm n}=\cos(2n\pi/7)\pm i\sin(2n\pi/7)
  2. Add the conjugate expressions.

    ωn+ω−n=2cos⁡(2nπ/7)\omega^n+\omega^{-n}=2\cos(2n\pi/7)

The identity holds for every positive integer n.

Checks and common pitfalls: The negative power is a reciprocal; it is a conjugate because the modulus is one.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Apply De Moivre to both signs.
    ω±n=cos⁡(2nπ/7)±isin⁡(2nπ/7)\omega^{\pm n}=\cos(2n\pi/7)\pm i\sin(2n\pi/7)
  • Add the conjugate expressions.
    ωn+ω−n=2cos⁡(2nπ/7)\omega^n+\omega^{-n}=2\cos(2n\pi/7)

Think first. Reveal a hint when the class is ready.

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Curriculum and source notes ↗