IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 HL only ~9 min read

Proof of De Moivre’s Theorem

De Moivre’s theorem is one of the most useful results in HL maths, and the IB loves to ask you to prove it. The good news? You only need to prove it for positive integer powers, which means you can use proof by induction — the same four-step technique you used for sums and divisibility results in Section 5. The structure is always identical: prove the base case, assume true for some k, show it follows for k+1, and write a clean conclusion. The clever bit comes in the inductive step, where you need to multiply two complex numbers in polar form using the compound-angle identities for cos and sin. If you’ve memorised those identities, the proof falls into place beautifully.

📘 What you need to know

What you’re proving: [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) for all n ∈ ℤ⁺.
The IB only requires the proof for positive integer n (even though the theorem holds for all real n).
The proof uses mathematical induction — four standard steps: base case, hypothesis, inductive step, conclusion.
The inductive step requires the compound-angle identities: cos(A + B) = cos A cos B − sin A sin B and sin(A + B) = sin A cos B + cos A sin B.
You’ll also need i² = −1 when expanding the brackets.
The inductive step uses the addition law for indices: a^k · a¹ = a^k+1. This is what lets you split [r(cos θ + i sin θ)]^k+1 into the assumed result times one more copy.
Conclude with the standard induction sentence: “true for n = 1, true for k implies true for k+1, therefore true for all n ∈ ℤ⁺.”

Why prove it by induction?

De Moivre’s theorem is a statement about every positive integer n — the modulus and argument behave a particular way no matter how big or small n is. That’s exactly the type of “for all natural numbers” claim that proof by induction is designed to handle.

🤔 Why doesn’t a single calculation work?

You could check the theorem for any specific n by direct computation. For example, you can verify it for n = 2 by squaring out (cos θ + i sin θ)² using the compound-angle identities and seeing it gives cos 2θ + i sin 2θ. But that only proves the case n = 2. Induction is what gets you all infinitely many cases at once.

The four-step structure

The four steps of the De Moivre proof by induction

The first step is usually a simple substitution. The second step costs you nothing — you just write the assumption. The fourth step is a memorised sentence. Step 3 is where all the actual work happens. For De Moivre, this means multiplying two complex numbers in polar form and recognising compound-angle identities in the result.

The two identities you’ll need in Step 3

In the inductive step you’ll multiply [r(cos θ + i sin θ)]^k by [r(cos θ + i sin θ)]¹. After expanding the brackets and using i² = −1, you’ll be left with two real expressions. To finish, you need to recognise these as cos((k+1)θ) and sin((k+1)θ). That recognition uses the compound-angle identities:

Compound angle identities cos(A + B) = cos A cos B − sin A sin B
sin(A + B) = sin A cos B + cos A sin B ✓ in the formula booklet

If you can’t recognise the compound-angle pattern, the inductive step won’t simplify and the proof gets stuck. Drill these two identities until they’re second nature — they’re in the booklet, but you need to spot them in your working without flipping pages.

The recipe for writing the proof

🧭 Recipe — Proof of De Moivre by induction

State what you’re proving clearly: [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) for n ∈ ℤ⁺.
Step 1 — Base case (n = 1): both sides equal r(cos θ + i sin θ). Done.
Step 2 — Inductive hypothesis: assume the result holds for some integer n = k.
Step 3 — Inductive step: consider [r(cos θ + i sin θ)]^k+1. Split it as […]^k · […]¹. Apply the hypothesis to the first factor. Multiply the two complex numbers, expand, use i² = −1, then group into real and imaginary parts.
Step 4: recognise the real part as cos((k+1)θ) and the imaginary part as sin((k+1)θ). State the result is now in the De Moivre form for n = k+1.
Conclusion: “Since the result is true for n = 1, and assuming true for n = k implies true for n = k+1, by the principle of mathematical induction the result holds for all n ∈ ℤ⁺.”

Worked examples

WE 1

Prove De Moivre’s theorem for positive integer n

Let z = r(cos θ + i sin θ). Use mathematical induction to prove that
zⁿ = rⁿ(cos nθ + i sin nθ) for all n ∈ ℤ⁺.

Step 1: Base case — show the result holds for n = 1 LHS: [r(cos θ + i sin θ)]¹ = r(cos θ + i sin θ) RHS: r¹(cos(1·θ) + i sin(1·θ)) = r(cos θ + i sin θ) LHS = RHS, so the result holds for n = 1 ✓Step 2: Inductive hypothesis — assume true for n = k [r(cos θ + i sin θ)]ᵏ = rᵏ(cos kθ + i sin kθ)Step 3: Inductive step — show true for n = k + 1 [r(cos θ + i sin θ)]ᵏ⁺¹ = [r(cos θ + i sin θ)]ᵏ · [r(cos θ + i sin θ)]¹ By the hypothesis: = rᵏ(cos kθ + i sin kθ) · r(cos θ + i sin θ) = rᵏ⁺¹ · (cos kθ + i sin kθ)(cos θ + i sin θ) Expand the bracket: = rᵏ⁺¹[cos kθ cos θ + i cos kθ sin θ + i sin kθ cos θ + i² sin kθ sin θ] Use i² = −1 and group real + imaginary parts: = rᵏ⁺¹[(cos kθ cos θ − sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ)] Recognise compound-angle identities: cos(kθ + θ) = cos kθ cos θ − sin kθ sin θ sin(kθ + θ) = sin kθ cos θ + cos kθ sin θ So: [r(cos θ + i sin θ)]ᵏ⁺¹ = rᵏ⁺¹[cos(kθ + θ) + i sin(kθ + θ)] = rᵏ⁺¹[cos((k+1)θ) + i sin((k+1)θ)] This is the De Moivre form with n = k + 1, so the result holds for n = k + 1 ✓Step 4: Conclusion The result is true for n = 1. If the result is true for n = k, it is also true for n = k + 1. Therefore, by the principle of mathematical induction, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) for all n ∈ ℤ⁺ ✓ memorise this layout — the structure scores marks even if you slip on a single algebraic line

WE 2

Verify the base case clearly

For z = 2(cos(π/3) + i sin(π/3)), verify that the De Moivre formula holds for n = 1 by computing both sides separately.

LHS: z¹ = z [2(cos(π/3) + i sin(π/3))]¹ = 2(cos(π/3) + i sin(π/3)) RHS: substitute n = 1 into the De Moivre formula 2¹(cos(1 · π/3) + i sin(1 · π/3)) = 2(cos(π/3) + i sin(π/3)) LHS = RHS ✓ Base case holds. the base case is usually trivial — but you still must write it down explicitly to score the mark

WE 3

Verify the inductive step in the case k = 1, going to k + 1 = 2

Assuming the De Moivre formula holds for n = 1 (i.e., z¹ = r(cos θ + i sin θ)), use the inductive-step argument to show it holds for n = 2.

Step 1: Split the (k+1) power as a product z² = z¹ · z¹ = r(cos θ + i sin θ) · r(cos θ + i sin θ) Step 2: Multiply r’s, expand the brackets = r²[cos θ cos θ + i cos θ sin θ + i sin θ cos θ + i² sin θ sin θ] = r²[cos²θ − sin²θ + i(2 sin θ cos θ)] (using i² = −1) Step 3: Apply double-angle identities cos²θ − sin²θ = cos(2θ) 2 sin θ cos θ = sin(2θ) z² = r²[cos(2θ) + i sin(2θ)] z² = r²(cos 2θ + i sin 2θ) ✓ De Moivre holds for n = 2. this is exactly the inductive step with k = 1 — the same argument works for any k

WE 4

Spot the error in this attempted inductive step

A student writes the following inductive step. State the line where the error occurs and explain how to fix it.
  Line A: z^k+1 = z^k · z¹
  Line B: = r^k(cos kθ + i sin kθ) · r(cos θ + i sin θ)
  Line C: = r^k+1(cos kθ cos θ + i sin kθ sin θ)
  Line D: = r^k+1(cos((k+1)θ) + i sin((k+1)θ))

Identify the error Lines A and B are correct. Line C is WRONG — the bracket expansion is incomplete. What’s missing (cos kθ + i sin kθ)(cos θ + i sin θ) has FOUR terms, not two: cos kθ cos θ + i cos kθ sin θ + i sin kθ cos θ + i² sin kθ sin θ Correct line C = rᵏ⁺¹[(cos kθ cos θ − sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ)] (using i² = −1) Line C: only two of four expansion terms shown — must include all four, then group into real and imaginary parts. expanding the product of two complex binomials is a “show all four terms” job — skipping any two costs marks

💡 Top tips

Memorise the four-step skeleton. Base case → hypothesis → inductive step → conclusion. The structure scores marks before any algebra.
State what you’re proving at the start. Examiners want to see the formula clearly written before you start the proof.
Don’t skip the base case. Even though n = 1 is trivial, you must write LHS = RHS = r(cos θ + i sin θ) explicitly.
The hypothesis line is free marks. “Assume true for n = k” then write the formula with k in it. No work, full credit.
Memorise the compound-angle identities. They’re in the booklet but you need to spot them in your working without searching.
Show all four terms when multiplying out. (cos kθ + i sin kθ)(cos θ + i sin θ) gives four terms before simplification — write them all.
End with the standard conclusion sentence. “Since true for n = 1 and true for k implies true for k+1, by the principle of mathematical induction the result holds for all n ∈ ℤ⁺.” Memorise it word for word.

⚠ Common mistakes

Skipping the base case or writing it without showing LHS = RHS explicitly. Examiners want both sides computed separately.
Writing the hypothesis in n instead of k. The whole point of the hypothesis is to introduce a new variable — write “assume true for n = k” and then the formula with k everywhere.
Forgetting to use i² = −1 when expanding the bracket. Without this substitution, the imaginary parts won’t combine cleanly.
Skipping the four-term expansion by writing only two terms (a common shortcut that loses marks). All four terms must appear before grouping.
Failing to recognise the compound-angle identities. If you stop at “r^k+1[(cos kθ cos θ − sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ)]” without simplifying further, the proof is incomplete.
Mixing up the signs in the compound-angle formulas. cos(A+B) has a minus; sin(A+B) has a plus. Don’t swap them.
Writing a vague conclusion. A line like “so it works for all n” doesn’t score. You need the full sentence with “principle of mathematical induction”.

The proof of De Moivre’s theorem is a classic IB Paper 2 question — a great way to score a full 7 or 8 marks if you’ve practised the layout. The mark scheme rewards structure at least as much as algebra: even if you slip on one expansion line, a clean four-step layout with a clear conclusion still picks up most of the marks. The next note brings De Moivre to its most beautiful application: finding the nth roots of complex numbers, where roots of unity sit on a regular polygon centred at the origin. That’s the topic that ties this entire section together.

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