IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2HL only~10 min read
De Moivre’s Theorem
If exponential form felt slick, brace yourself: De Moivre’s theorem turns “raise a complex number to any power” into a single line. Want to find z10? Just raise the modulus to the 10th and multiply the argument by 10. That’s it. No expanding (a + bi)10. No 11 terms in a binomial expansion. The same theorem works for negative powers, fractional powers, and even irrational ones. Geometrically, De Moivre says: raising to the nth power means rotating n times around the origin and scaling the length to the nth power. It’s the natural extension of “scale and rotate” — only now you’re rotating again and again. This single result powers nearly every advanced complex-number question on Paper 1.
📘 What you need to know
De Moivre’s theorem: [r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ).
In exponential form, this is just (reiθ)n = rneinθ — standard index laws.
In cis notation: (r cis θ)n = rn cis(nθ). Same theorem, three ways to write it.
The theorem is in the formula booklet, with all three forms listed.
It works for any real n — positive integer, negative, fractional, rational, irrational. IB exam questions usually use n = 3, 10, −3, ½, −¾, …
For negative powers, use cos(−θ) = cos θ and sin(−θ) = −sin θ, so 1/z = (1/r)(cos θ − i sin θ).
For numbers in Cartesian form, convert to polar/exponential form first, apply De Moivre, then convert back if a Cartesian answer is needed.
If r = 1, the powers z0, z1, z2, … may form a periodic sequence on the unit circle — useful for spotting patterns.
What is De Moivre’s theorem?
De Moivre’s theorem is a clean rule for raising a complex number to a power, when the number is in polar form:
De Moivre’s Theorem
[r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ)
✓ in the formula booklet
Two things happen to the complex number when you raise it to the nth power:
modulus: r → rn (raised to the nth power)
argument: θ → nθ (multiplied by n)
That’s the whole rule. The same idea, written in exponential form, is just standard index laws:
De Moivre in exponential form
(reiθ)n = rn einθ
🤔 Why does this work?
Because z2 = z · z means doubling the argument and squaring the modulus (from the multiplication rule). And z3 = z2 · z triples the argument and cubes the modulus. By induction, the same pattern works for every positive integer n — and with a bit of care, for negative and fractional n too. The proof is the topic of the next note.
The geometric picture — n-fold rotation
De Moivre’s theorem has a clean geometric meaning. Each time you multiply z by itself, you rotate by θ (the argument) and stretch by r. So raising to the nth power means:
Each successive power rotates by θ and stretches by r
So if z is on the unit circle (r = 1), every power stays on the unit circle — only the angle changes. If r > 1, powers spiral outwards; if 0 < r < 1, they spiral inwards toward the origin.
Positive integer powers
The recipe is the same for any positive integer power. The key is to start in polar form — if the number is in Cartesian form, convert first.
🧭 Recipe — finding zn (positive integer n)
If z is in Cartesian form, convert to polar using r = √(x² + y²) and θ = arg(z).
Apply De Moivre’s theorem: zn = rn(cos nθ + i sin nθ).
Reduce nθ modulo 2π if needed, to bring the argument into the standard range.
Convert back to Cartesian form if the question asks for an answer in a + bi form.
For exam efficiency, switch to exponential form wherever possible. The calculation (reiθ)n = rneinθ is one line. Polar form is good for showing working in writing; exponential form is faster for arithmetic.
Negative powers
De Moivre works with negative n too. Just take 1/z = z−1:
Negative power as De Moivre with n = −11z = z−1 = r−1(cos(−θ) + i sin(−θ)) = 1r(cos θ − i sin θ)
The last step uses the even/odd identities: cos(−θ) = cos θ and sin(−θ) = −sin θ. So flipping a complex number by 1/z inverts the modulus and negates the argument — geometrically, a reflection in the real axis combined with a reciprocal of the length.
Quick result: for any negative integer n = −m with m > 0, z−m = (1/rm)(cos(mθ) − i sin(mθ)) — same modulus rule, but with a sign flip on the imaginary part.
Periodic patterns when r = 1
If a complex number has modulus 1 — meaning it sits on the unit circle — then its powers all stay on the unit circle. The arguments march around in regular steps, and after enough powers, they come back to where they started.
So the powers cycle every 8. To find z101, just compute 101 mod 8 = 5, so z101 = z5 = cis(5π/4). Faster than a calculator and absolutely exact.
Pattern recognition saves time. Whenever a power-of-z question has |z| = 1 and a “round” argument like π/3, π/4, π/6, look for the period before applying De Moivre directly. The answer often falls out from one mod calculation.
Worked examples
WE 1
Compute a positive integer power
Given z = 2(cos(π/8) + i sin(π/8)), find z6, giving your answer in Cartesian form.
Step 1: Apply De Moivre’s theoremz⁶ = 2⁶(cos(6 · π/8) + i sin(6 · π/8))= 64(cos(3π/4) + i sin(3π/4))Step 2: Evaluate the trig functions (3π/4 is in Q2)cos(3π/4) = −√2/2sin(3π/4) = √2/2Step 3: Multiply through by 64z⁶ = 64(−√2/2 + (√2/2)i)= −32√2 + 32√2 iz⁶ = −32√2 + 32√2 iif a Cartesian answer is asked, evaluate the trig in exact form — IB markschemes prefer surds
WE 2
Power of a Cartesian-form complex number
Find (1 + i)10, giving your answer in Cartesian form.
Step 1: Convert 1 + i to polar form (Q1)|1 + i| = √(1 + 1) = √2arg(1 + i) = π/41 + i = √2 cis(π/4)Step 2: Apply De Moivre’s theorem(1 + i)¹⁰ = (√2)¹⁰ cis(10 · π/4)(√2)¹⁰ = 2⁵ = 3210 · π/4 = 10π/4 = 5π/2Step 3: Reduce the argument modulo 2π5π/2 − 2π = 5π/2 − 4π/2 = π/2so (1+i)¹⁰ = 32 cis(π/2) = 32i(1 + i)¹⁰ = 32iexpanding (1+i)¹⁰ in Cartesian form would mean an 11-term binomial expansion — De Moivre takes one line
WE 3
Negative integer power
Find z−4 where z = 3 cis(π/6).
Step 1: Apply De Moivre with n = −4z⁻⁴ = 3⁻⁴ · cis(−4 · π/6)3⁻⁴ = 1/3⁴ = 1/81−4 · π/6 = −2π/3Step 2: Check the range — −2π/3 is in (−π, π]no adjustment neededz⁻⁴ = (1/81) cis(−2π/3)negative powers shrink the modulus and reverse the rotation direction
WE 4
Fractional power
Given z = 16 ei·2π/3, find z1/2 in exponential form.
Step 1: Apply index laws (= De Moivre with n = 1/2)z^(1/2) = 16^(1/2) · e^(i · (1/2) · 2π/3)Step 2: Compute each piece16^(1/2) = 4(1/2) · 2π/3 = π/3z^(1/2) = 4 e^(iπ/3)this is one of the square roots of z — there’s another, with argument π/3 + π = 4π/3 (covered in the Roots note)
WE 5
Very high power using periodicity
Given z = cos(π/3) + i sin(π/3), find z50.
Step 1: Note that |z| = 1 → modulus stays 1 foreverarg(z) = π/3, so the powers cycle every 6 (since 6 · π/3 = 2π)Step 2: Reduce 50 mod 650 = 8 · 6 + 2so z⁵⁰ = z² (same position on unit circle)Step 3: Apply De Moivre to z² (or use 50 · π/3 directly)arg(z⁵⁰) = 50 · π/3 = 50π/350π/3 mod 2π: 50π/3 − 16π = 50π/3 − 48π/3 = 2π/3Step 4: Convert to Cartesianz⁵⁰ = cos(2π/3) + i sin(2π/3) = −1/2 + (√3/2)iz⁵⁰ = −1/2 + (√3/2)iwhen |z| = 1, the modulus never changes — only the angle marches around the unit circle
WE 6
Combined application — Cartesian form output
Find the value of (√3 − i)5, giving your answer in Cartesian form a + bi.
Step 1: Convert √3 − i to polar form|√3 − i| = √(3 + 1) = 2sketch: real > 0, imag < 0 → Q4 → arg negative acuteα = arctan(1/√3) = π/6arg = −π/6, so √3 − i = 2 cis(−π/6)Step 2: Apply De Moivre with n = 5(√3 − i)⁵ = 2⁵ cis(5 · −π/6)= 32 cis(−5π/6)Step 3: Convert back to Cartesiancos(−5π/6) = cos(5π/6) = −√3/2sin(−5π/6) = −sin(5π/6) = −1/2(√3 − i)⁵ = 32(−√3/2 + (−1/2)i)= −16√3 − 16i(√3 − i)⁵ = −16√3 − 16ithree-step pattern: Cartesian → polar → De Moivre → Cartesian. Always the same workflow.
💡 Top tips
Convert to polar or exponential form before applying De Moivre. The theorem doesn’t work directly on Cartesian form — convert first.
Use exponential form for the cleanest arithmetic. (reiθ)n = rneinθ is a one-line index-law calculation.
Reduce the new argument modulo 2π when it gets large. Subtract 2π until you’re back in the standard range.
For |z| = 1, look for periodicity first. If the period of the powers is small (4, 6, 8, …), you can find zn for huge n using a quick mod calculation.
Negative powers: use cos(−θ) = cos θ and sin(−θ) = −sin θ. You don’t need to memorise extra rules — these two identities handle everything.
Always check the range the question asks for. −π < θ ≤ π and 0 ≤ θ < 2π are both common; use whichever the question specifies.
For “find zn in Cartesian form”, the workflow is: convert to polar → apply De Moivre → reduce angle → convert back. Same four steps every time.
⚠ Common mistakes
Trying to apply De Moivre to Cartesian form. The theorem requires polar or exponential form — convert first.
Multiplying instead of raising the modulus. The modulus becomes rn, not n·r. Index law, not arithmetic.
Adding to the argument instead of multiplying. The argument becomes nθ, not n + θ.
Forgetting to reduce the argument modulo 2π when it gets bigger than π. Especially common for high powers like z10.
Mishandling negative powers. Don’t forget the sign flip on the sin term: 1/z = (1/r)(cos θ − i sin θ).
Sign errors when converting back to Cartesian. If the new argument is in Q3 or Q4, both cos and/or sin will be negative — work carefully with the exact values.
Using degrees by mistake. All HL complex-number arguments are in radians — always check your calculator.
De Moivre’s theorem turns one of the hardest calculations in algebra (raising to a high power) into one of the easiest in geometry (rotating n times). It’s the engine behind nearly every Paper 1 question on advanced complex numbers — you’ll see it in roots, in trigonometric identities, and in the proofs of identities like cos(2θ) = 2cos²θ − 1. The next note covers the proof of De Moivre’s theorem by induction, which is itself a regular Paper 2 question. After that, you’ll use De Moivre to find roots of complex numbers — where the theorem really shines.
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