Why is "Finding only one $n$th root instead of all $n$ of them" a common mistake in de Moivre’s theorem?

Why it happens: Forgetting to add 2 k to the argument before dividing by n; treating the equation like a real-number root. How to avoid it: Always write the argument as + 2 k at the start, then list k = 0, 1, , n-1. The number of roots equals the degree n. Source: 9231 Paper 2 Examiner Reports — recurring comment

Why is "Omitting the conclusion line of the induction proof" a common mistake in de Moivre’s theorem?

Why it happens: Stopping once (k+1) + i(k+1) is reached, treating the algebra as the whole proof. How to avoid it: End every induction with: 'true for n=1, and P(k) P(k+1), so by induction true for all positive integers n.' Source: 9231 Paper 2 mark schemes

Why is "Raising only the argument to the power and leaving the modulus unchanged" a common mistake in de Moivre’s theorem?

Why it happens: Memorising de Moivre for unit-modulus numbers and applying it blindly to r( + i). How to avoid it: For [r( + i)]^n, the modulus becomes r^n AND the argument becomes n. Both transform. Source: 9231 Paper 2 Examiner Reports

Why is "Giving decimal approximations where exact surd/$\pi$ answers are required" a common mistake in de Moivre’s theorem?

Why it happens: Reaching for a calculator instead of recalling exact values such as /6 = √3/2. How to avoid it: Keep arguments as exact multiples of and use exact trig values. Decimals lose accuracy marks. Source: 9231 Paper 2 Examiner Reports

Why is "Wrong argument because the quadrant of the complex number was ignored" a common mistake in de Moivre’s theorem?

Why it happens: Using ^-1y/x blindly without checking the signs of the real and imaginary parts. How to avoid it: Sketch the point first; adjust the principal-value ^-1 by when the point is in the 2nd or 3rd quadrant. Source: 9231 Paper 2 Examiner Reports

Why is "Skipping or fudging the base case in the induction" a common mistake in de Moivre’s theorem?

Why it happens: Assuming n = 1 is 'obvious' and not writing it down explicitly. How to avoid it: Always verify P(1) in writing: show both sides equal + i. No base case means no valid induction. Source: 9231 Paper 2 mark schemes

Cambridge International A Levels Further Mathematics (9231)

de Moivre’s theorem

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Detailed Study Notes

Detailed notes on Complex numbers for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

De Moivre's Theorem — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (Further Pure Mathematics 2, 2026-2027 syllabus)

De Moivre's theorem turns powers and roots of complex numbers into simple arithmetic on the modulus and argument: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ . Master the proof by induction, then use it to lay the $n$ th roots of unity out as a perfect regular polygon.

At a glance

De Moivre's theorem: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ — raise to a power by multiplying the argument by $n$ .
Geometric meaning: multiplying complex numbers multiplies moduli and adds arguments; powers just repeat this.
Proof by induction for positive integer $n$ is examinable — and is NOT in MF19.
$n$ th roots of unity sit at the vertices of a regular $n$ -gon on the unit circle, spaced $\frac{2\pi}{n}$ apart and summing to $0$ .
$n$ th roots of any $z$ : take the real $n$ th root of the modulus, divide the argument by $n$ , then add $\frac{2\pi k}{n}$ for $k = 0, 1, \ldots, n-1$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

2.5a — Understand de Moivre's theorem, for a positive or negative rational index, in the form $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ .
2.5b — Prove de Moivre's theorem for a positive integer exponent by mathematical induction.
2.5c — Use de Moivre's theorem to find the $n$ th roots of unity and the $n$ th roots of a general complex number, and represent them on an Argand diagram.

The theorem and what it means geometrically

Powers of a complex number multiply the modulus and rotate the argument.

The statement. For any real $\theta$ and any integer (in fact rational) $n$ , $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.$

Where it comes from — the rule for multiplying. If $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ , then expanding and using the compound-angle formulae gives $z_1 z_2 = r_1 r_2\bigl[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\bigr].$

So multiplying multiplies the moduli and adds the arguments. Raising $\cos\theta + i\sin\theta$ (which has modulus $1$ ) to the power $n$ therefore keeps the modulus at $1$ and adds $\theta$ to itself $n$ times — i.e. the argument becomes $n\theta$ . That is exactly de Moivre's theorem.

The exponential bridge. Writing $e^{i\theta} = \cos\theta + i\sin\theta$ (Euler's relation), de Moivre is just the index law $\left(e^{i\theta}\right)^n = e^{in\theta}$ . This view is enormously useful for the applications topic, but the inductive proof below does not assume it.

Squaring keeps the point on the unit circle but doubles the argument from θ to 2θ. The $n$th power rotates it to $n\theta$.

Cambridge tip. The theorem holds for negative indices too: $(\cos\theta + i\sin\theta)^{-1} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$ , which is why $\frac{1}{z}$ tricks (Topic 23.2) work so neatly.

Multiplying complex numbers: multiply moduli, add arguments.
$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for integer/rational $n$ .
Equivalent to the index law $\left(e^{i\theta}\right)^n = e^{in\theta}$ .

See the full worked example for de moivre’s theorem →

Proving de Moivre's theorem by induction

Base case n = 1, assume for n = k, multiply by one more factor to reach n = k+1.

For a positive integer $n$ , de Moivre is proved by induction. This proof is examinable and not in MF19, so learn its shape.

Statement. Let $P(n)$ be the proposition $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ .

Base case ( $n = 1$ ). $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta = \cos(1\cdot\theta) + i\sin(1\cdot\theta).$ So $P(1)$ is true.

Inductive step. Assume $P(k)$ holds, i.e. $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$ . Then $(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^{k}(\cos\theta + i\sin\theta) = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta).$

Multiply out (remember $i^2 = -1$ ): $= \cos k\theta\cos\theta - \sin k\theta\sin\theta + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta).$

Apply the compound-angle formulae $\cos(A+B) = \cos A\cos B - \sin A\sin B$ and $\sin(A+B) = \sin A\cos B + \cos A\sin B$ with $A = k\theta$ , $B = \theta$ : $= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos(k+1)\theta + i\sin(k+1)\theta.$

So $P(k+1)$ holds. Conclusion: since $P(1)$ is true and $P(k) \Rightarrow P(k+1)$ , by mathematical induction $P(n)$ is true for all positive integers $n$ .

Extending to negative integers. For $n = -m$ with $m > 0$ , use $(\cos\theta + i\sin\theta)^{-m} = \dfrac{1}{(\cos\theta + i\sin\theta)^{m}} = \dfrac{1}{\cos m\theta + i\sin m\theta}$ , then multiply top and bottom by $\cos m\theta - i\sin m\theta$ to get $\cos(-m)\theta + i\sin(-m)\theta$ .

Cambridge tip. Examiners insist on a complete induction: a clear base case, the assumption written out, the algebra using the compound-angle formulae explicitly, and a one-line conclusion naming the induction principle. Skipping the conclusion line loses the final mark.

Base case $n=1$ is immediate.
Inductive step multiplies $(\cos\theta + i\sin\theta)^k$ by one more factor.
Compound-angle formulae collapse the product to $\cos(k+1)\theta + i\sin(k+1)\theta$ .

See the full worked example for de moivre’s theorem →

The nth roots of unity

The n solutions of zⁿ = 1 form a regular n-gon on the unit circle and sum to zero.

Solving $z^n = 1$ means finding every complex number whose $n$ th power is $1$ .

Set up in modulus-argument form. Write $1 = \cos(2\pi k) + i\sin(2\pi k)$ for any integer $k$ — this is the crucial step, because $1$ has argument $0$ plus any whole number of full turns. Let $z = \cos\phi + i\sin\phi$ (its modulus must be $1$ since $|z|^n = 1$ ). By de Moivre: $z^n = \cos n\phi + i\sin n\phi = \cos 2\pi k + i\sin 2\pi k \;\Rightarrow\; n\phi = 2\pi k \;\Rightarrow\; \phi = \frac{2\pi k}{n}.$

Taking $k = 0, 1, 2, \ldots, n-1$ gives the $n$ distinct roots: $z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \qquad k = 0, 1, \ldots, n-1.$ (Beyond $k = n-1$ the values repeat.) Writing $\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}$ , the roots are simply $1, \omega, \omega^2, \ldots, \omega^{n-1}$ .

Geometry. All roots lie on the unit circle, one of them at $1$ , and they are equally spaced $\frac{2\pi}{n}$ apart — the vertices of a regular $n$ -gon.

The fifth-roots of unity: 1, ω, ω², ω³, ω⁴ spaced 72° apart. For any n ≥ 2 the roots sum to 0, since they are symmetric about the origin.

The roots sum to zero. For $n \ge 2$ , $1 + \omega + \omega^2 + \cdots + \omega^{n-1} = \dfrac{\omega^n - 1}{\omega - 1} = 0$ because $\omega^n = 1$ . Geometrically, the vectors to the vertices of a regular polygon cancel.

Cambridge tip. Always quote the roots exactly ( $\cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}$ ), and when sketching, mark one root at $1$ and show the equal spacing — examiners want the regular-polygon symmetry visible.

$z^n = 1$ has $n$ roots $z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}$ , $k = 0,\ldots,n-1$ .
They are the vertices of a regular $n$ -gon on the unit circle, one at $1$ .
For $n \ge 2$ the roots sum to $0$ .

See the full worked example for de moivre’s theorem →

The nth roots of a general complex number

Take the real nth root of the modulus, divide the argument by n, then step round by 2πk/n.

The same idea finds the $n$ th roots of any complex number $w = R(\cos\alpha + i\sin\alpha)$ (modulus $R$ , argument $\alpha$ ).

Method. To solve $z^n = w$ :

Write $w$ in modulus-argument form, including the $+2\pi k$ on the argument: $w = R\bigl[\cos(\alpha + 2\pi k) + i\sin(\alpha + 2\pi k)\bigr]$ .
The $n$ roots are $z_k = R^{1/n}\left[\cos\frac{\alpha + 2\pi k}{n} + i\sin\frac{\alpha + 2\pi k}{n}\right], \qquad k = 0, 1, \ldots, n-1.$

Here $R^{1/n}$ is the ordinary positive real $n$ th root of the modulus.

Why it works. Let $z = \rho(\cos\phi + i\sin\phi)$ . Then $z^n = \rho^n(\cos n\phi + i\sin n\phi)$ . Matching moduli gives $\rho^n = R$ , so $\rho = R^{1/n}$ ; matching arguments gives $n\phi = \alpha + 2\pi k$ , so $\phi = \frac{\alpha + 2\pi k}{n}$ .

Geometry. The roots lie on a circle of radius $R^{1/n}$ , again equally spaced $\frac{2\pi}{n}$ apart — a regular $n$ -gon, but rotated so the first vertex sits at argument $\frac{\alpha}{n}$ .

Worked sketch. To find the cube roots of $8i$ : here $R = 8$ , $\alpha = \frac{\pi}{2}$ , $n = 3$ , so $R^{1/3} = 2$ and the arguments are $\frac{\pi/2 + 2\pi k}{3} = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ (taking $k = 0, 1, 2$ ). The roots are $2(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6})$ , etc. — i.e. $\sqrt3 + i$ , $-\sqrt3 + i$ and $-2i$ .

Cambridge tip. The single biggest error is forgetting the $+2\pi k$ and so producing only one root. Always write the argument as $\alpha + 2\pi k$ at the start, then list $k = 0, 1, \ldots, n-1$ to guarantee all $n$ roots.

$z^n = w$ : $z_k = R^{1/n}\left[\cos\frac{\alpha + 2\pi k}{n} + i\sin\frac{\alpha + 2\pi k}{n}\right]$ .
Real $n$ th root of the modulus; argument divided by $n$ .
Add $2\pi k$ before dividing to get all $n$ roots — never just one.

See the full worked example for de moivre’s theorem →

Quick recap

$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for integer and rational $n$ .
Geometrically: powers multiply the modulus and multiply the argument by $n$ .
Prove for positive integer $n$ by induction — base case, assumption, compound-angle step, conclusion.
$n$ th roots of unity: $\cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}$ , a regular $n$ -gon summing to $0$ .
$n$ th roots of $w = R(\cos\alpha + i\sin\alpha)$ : $R^{1/n}\left[\cos\frac{\alpha + 2\pi k}{n} + i\sin\frac{\alpha + 2\pi k}{n}\right]$ .
De Moivre and the induction proof are NOT in MF19 — memorise both.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

De Moivre's theorem $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ .
The induction proof structure (base case $\to$ assume $P(k)$ $\to$ compound-angle step $\to$ conclusion).
$n$ th roots of unity $z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}$ , $k = 0,\ldots,n-1$ , and that they sum to $0$ .
$n$ th roots of a general $w$ : $R^{1/n}\left[\cos\frac{\alpha + 2\pi k}{n} + i\sin\frac{\alpha + 2\pi k}{n}\right]$ — remember the $+2\pi k$ .
Multiplication rule: multiply moduli, add arguments.

How it’s examined

De Moivre features in almost every Paper 2 (FPM2), typically 6-10 marks. A 'prove by induction' part is a recurring 4-5 mark request and must be fully written out — base case, assumption, compound-angle algebra and a named conclusion. Roots questions ask for the $n$ th roots of unity or of a given complex number in exact modulus-argument form plus an Argand sketch showing the regular-polygon symmetry. Marks are most often lost by (i) producing only one root (forgetting $+2\pi k$ ), (ii) omitting the conclusion line of the induction, and (iii) inexact or decimal answers where surds are expected.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 2, Further Pure Mathematics 2); 9231/21, 9231/23 Further Pure Mathematics 2 — past papers and mark schemes. Last reviewed 2026-06-23.

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Step-by-step worked examples — de Moivre’s theorem

Step-by-step solutions to past-paper-style questions on de moivre’s theorem, written exactly the way a tutor would explain them at the board.

Question type:

1A power via modulus-argument form
Getting startedDirect calculation• power, modulus-argument
Question
Use de Moivre's theorem to evaluate $\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)^4$ , giving your answer in the form $a + bi$ .
Step-by-step solution
1. Step 1
  Apply de Moivre — multiply the argument by the index $n = 4$ .
  $\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right)^4 = \cos\tfrac{4\pi}{6} + i\sin\tfrac{4\pi}{6} = \cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}$
2. Step 2
  Evaluate the exact trig values: $\cos\frac{2\pi}{3} = -\frac{1}{2}$ , $\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}$ .
  $= -\frac{1}{2} + \frac{\sqrt3}{2}\,i$
Answer
$\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)^4 = -\frac{1}{2} + \frac{\sqrt3}{2}\,i$ .
Examiner tip
The whole point is to avoid expanding a 4th power directly. Multiply the argument by 4, then read off exact values — keep the surd, never a decimal.
2A power of a complex number not on the unit circle
Getting startedMulti-step problem• power, modulus-argument
Question
Find $(1 + i)^8$ using de Moivre's theorem.
Step-by-step solution
1. Step 1
  Convert to modulus-argument form. $|1+i| = \sqrt{1^2 + 1^2} = \sqrt2$ and $\arg(1+i) = \frac{\pi}{4}$ .
  $1 + i = \sqrt2\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)$
2. Step 2
  Raise to the 8th power — the modulus is raised to the power 8, the argument multiplied by 8.
  $(1+i)^8 = (\sqrt2)^8\left(\cos\tfrac{8\pi}{4} + i\sin\tfrac{8\pi}{4}\right) = 16\left(\cos 2\pi + i\sin 2\pi\right)$
3. Step 3
  Evaluate. $\cos 2\pi = 1$ , $\sin 2\pi = 0$ .
  $(1+i)^8 = 16(1 + 0i) = 16$
Answer
$(1+i)^8 = 16$ .
Examiner tip
Don't forget to raise the modulus $\sqrt2$ to the 8th power as well — a common slip is to handle only the argument and leave the modulus at $1$ .
3Proving de Moivre's theorem by induction
Building confidenceShow that / prove• induction, proof
Question
Prove by mathematical induction that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for all positive integers $n$ .
Step-by-step solution
1. Step 1
  Base case ( $n = 1$ ). Substitute $n = 1$ : the left side is $\cos\theta + i\sin\theta$ and the right side is $\cos\theta + i\sin\theta$ . They agree, so $P(1)$ is true.
2. Step 2
  Inductive hypothesis. Assume the result holds for $n = k$ .
  $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$
3. Step 3
  Inductive step. Multiply both sides by one more factor of $(\cos\theta + i\sin\theta)$ .
  $(\cos\theta + i\sin\theta)^{k+1} = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$
4. Step 4
  Expand, using $i^2 = -1$ , and collect real and imaginary parts.
  $= (\cos k\theta\cos\theta - \sin k\theta\sin\theta) + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta)$
5. Step 5
  Apply the compound-angle formulae $\cos(A+B)$ and $\sin(A+B)$ with $A = k\theta$ , $B = \theta$ .
  $= \cos(k+1)\theta + i\sin(k+1)\theta$
6. Step 6
  Conclusion. $P(k+1)$ follows from $P(k)$ , and $P(1)$ is true, so by the principle of mathematical induction $P(n)$ holds for all positive integers $n$ .
Answer
Proved: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for all positive integers $n$ .
Examiner tip
Every component is a mark: base case, written-out assumption, the compound-angle collapse, and the final conclusion naming induction. Omitting the conclusion line is the single most common cause of a lost mark.
4The cube roots of unity
Building confidenceMulti-step problem• roots-of-unity, argand
Question
Find the three cube roots of unity in exact form, and state their sum.
Step-by-step solution
1. Step 1
  Write $1$ in modulus-argument form with the $+2\pi k$ term: $1 = \cos 2\pi k + i\sin 2\pi k$ . Solve $z^3 = 1$ with $z = \cos\phi + i\sin\phi$ .
  $\cos 3\phi + i\sin 3\phi = \cos 2\pi k + i\sin 2\pi k \;\Rightarrow\; \phi = \frac{2\pi k}{3}$
2. Step 2
  Take $k = 0, 1, 2$ to list the three distinct roots.
  $z_0 = \cos 0 + i\sin 0,\quad z_1 = \cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3},\quad z_2 = \cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}$
3. Step 3
  Evaluate to Cartesian form.
  $z_0 = 1,\quad z_1 = -\tfrac{1}{2} + \tfrac{\sqrt3}{2}i,\quad z_2 = -\tfrac{1}{2} - \tfrac{\sqrt3}{2}i$
4. Step 4
  Sum the roots.
  $1 + \left(-\tfrac{1}{2} + \tfrac{\sqrt3}{2}i\right) + \left(-\tfrac{1}{2} - \tfrac{\sqrt3}{2}i\right) = 0$
Answer
The cube roots of unity are $1$ , $-\frac{1}{2} + \frac{\sqrt3}{2}i$ and $-\frac{1}{2} - \frac{\sqrt3}{2}i$ ; they sum to $0$ .
Examiner tip
They form an equilateral triangle on the unit circle, one vertex at $1$ . The pair of complex roots are conjugates — a useful check. Their sum being $0$ is worth quoting.
5The cube roots of a general complex number
StretchMulti-step problem• roots, general-complex, argand
Question
Find the three values of $z$ satisfying $z^3 = 8i$ , giving each in exact Cartesian form.
Step-by-step solution
1. Step 1
  Modulus-argument form of $8i$ , with $+2\pi k$ : $|8i| = 8$ , $\arg(8i) = \frac{\pi}{2}$ .
  $8i = 8\left[\cos\left(\tfrac{\pi}{2} + 2\pi k\right) + i\sin\left(\tfrac{\pi}{2} + 2\pi k\right)\right]$
2. Step 2
  Take cube roots: real cube root of the modulus is $8^{1/3} = 2$ ; divide the argument by $3$ .
  $z_k = 2\left[\cos\frac{\tfrac{\pi}{2} + 2\pi k}{3} + i\sin\frac{\tfrac{\pi}{2} + 2\pi k}{3}\right]$
3. Step 3
  Substitute $k = 0, 1, 2$ to get arguments $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ .
  $z_0 = 2\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right),\; z_1 = 2\left(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\right),\; z_2 = 2\left(\cos\tfrac{3\pi}{2} + i\sin\tfrac{3\pi}{2}\right)$
4. Step 4
  Evaluate each to Cartesian form.
  $z_0 = \sqrt3 + i,\quad z_1 = -\sqrt3 + i,\quad z_2 = -2i$
Answer
$z = \sqrt3 + i,\;\; -\sqrt3 + i,\;\; -2i$ .
Examiner tip
Check: the three roots lie on a circle of radius $2$ , spaced $120°$ apart. Forgetting the $+2\pi k$ would give only $z_0$ — list $k = 0, 1, 2$ to secure all three.
6Using roots of unity to factorise an equation
StretchMulti-step problem• roots-of-unity, factorise
Question
Solve $z^4 = -16$ , giving each root in the form $a + bi$ , and hence write $z^4 + 16$ as a product of real quadratic factors.
Step-by-step solution
1. Step 1
  Modulus-argument form of $-16$ : $|-16| = 16$ , $\arg(-16) = \pi$ , so include $+2\pi k$ .
  $-16 = 16\left[\cos(\pi + 2\pi k) + i\sin(\pi + 2\pi k)\right]$
2. Step 2
  Fourth roots: modulus $16^{1/4} = 2$ ; arguments $\frac{\pi + 2\pi k}{4}$ for $k = 0,1,2,3$ give $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ .
  $z_k = 2\left[\cos\tfrac{\pi + 2\pi k}{4} + i\sin\tfrac{\pi + 2\pi k}{4}\right]$
3. Step 3
  Evaluate to Cartesian form (using $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2}$ , so $2 \times \frac{1}{\sqrt2} = \sqrt2$ ).
  $z = \sqrt2 + \sqrt2\,i,\;\; -\sqrt2 + \sqrt2\,i,\;\; -\sqrt2 - \sqrt2\,i,\;\; \sqrt2 - \sqrt2\,i$
4. Step 4
  Pair conjugate roots into real quadratics. Conjugate pair $\sqrt2 \pm \sqrt2\,i$ : sum $= 2\sqrt2$ , product $= 2 + 2 = 4$ , giving $z^2 - 2\sqrt2\,z + 4$ . Pair $-\sqrt2 \pm \sqrt2\,i$ gives $z^2 + 2\sqrt2\,z + 4$ .
  $z^4 + 16 = \left(z^2 - 2\sqrt2\,z + 4\right)\left(z^2 + 2\sqrt2\,z + 4\right)$
Answer
Roots $\sqrt2 \pm \sqrt2\,i$ and $-\sqrt2 \pm \sqrt2\,i$ ; $z^4 + 16 = (z^2 - 2\sqrt2\,z + 4)(z^2 + 2\sqrt2\,z + 4)$ .
Examiner tip
Conjugate roots always pair into a real quadratic $z^2 - (\text{sum})z + (\text{product})$ . This 'roots $\to$ real factors' route is a classic Paper 2 extension of a roots question.

Model Answers — de Moivre’s theorem

High-scoring sample answers for de moivre’s theorem on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Use de Moivre's theorem to express $\left(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}\right)^6$ in the form $a + bi$ , giving $a$ and $b$ exactly.
Model answer
Apply de Moivre's theorem — multiply the argument by $6$ : $\left(\cos\tfrac{\pi}{8} + i\sin\tfrac{\pi}{8}\right)^6 = \cos\tfrac{6\pi}{8} + i\sin\tfrac{6\pi}{8} = \cos\tfrac{3\pi}{4} + i\sin\tfrac{3\pi}{4}.$

Evaluate the exact values: $= -\frac{1}{\sqrt2} + \frac{1}{\sqrt2}\,i = -\frac{\sqrt2}{2} + \frac{\sqrt2}{2}\,i.$
Why this scores
M1 for applying de Moivre (argument $\times 6$ ), A1 for $\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}$ , A1 for the exact Cartesian form. A decimal answer loses the final A1.
Question 2
4 marks
Find $(\sqrt3 - i)^6$ using de Moivre's theorem.
Model answer
Convert to modulus-argument form. $|\sqrt3 - i| = \sqrt{3 + 1} = 2$ . The point $(\sqrt3, -1)$ is in the fourth quadrant, so $\arg(\sqrt3 - i) = -\frac{\pi}{6}$ . Thus $\sqrt3 - i = 2\left[\cos\left(-\tfrac{\pi}{6}\right) + i\sin\left(-\tfrac{\pi}{6}\right)\right].$

Raise to the 6th power (modulus to the power $6$ , argument $\times 6$ ): $(\sqrt3 - i)^6 = 2^6\left[\cos\left(-\pi\right) + i\sin\left(-\pi\right)\right] = 64(-1 + 0i) = -64.$
Why this scores
M1 for correct modulus and argument (sign of argument matters), M1 for applying de Moivre to both modulus and argument, A1 for $64(\cos(-\pi) + i\sin(-\pi))$ , A1 for $-64$ .
Question 3
6 marks
Prove by induction that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for all positive integers $n$ .
Model answer
Base case ( $n = 1$ ). The left-hand side is $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta$ , and the right-hand side is $\cos(1\cdot\theta) + i\sin(1\cdot\theta) = \cos\theta + i\sin\theta$ . They are equal, so the statement is true for $n = 1$ .

Inductive hypothesis. Assume the statement is true for $n = k$ : $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta.$

Inductive step. Consider $n = k + 1$ : $(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k(\cos\theta + i\sin\theta) = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta).$

Expanding and using $i^2 = -1$ : $= (\cos k\theta\cos\theta - \sin k\theta\sin\theta) + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta).$

Applying the compound-angle formulae: $= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos(k+1)\theta + i\sin(k+1)\theta.$

Conclusion. The statement is true for $n = 1$ , and its truth for $n = k$ implies its truth for $n = k + 1$ . Hence, by the principle of mathematical induction, it is true for all positive integers $n$ .
Why this scores
B1 base case; M1 writing $(\cdots)^{k+1} = (\cdots)^k(\cdots)$ and substituting the hypothesis; M1 expanding with $i^2 = -1$ ; A1 applying compound-angle formulae to reach $\cos(k+1)\theta + i\sin(k+1)\theta$ ; A1 a correct full conclusion; B1 overall rigour/clarity. The conclusion line is essential.
Question 4
5 marks
Find, in exact modulus-argument form, the four fourth-roots of unity. Show them on an Argand diagram and state their sum.
Model answer
Solve $z^4 = 1$ . Write $1 = \cos 2\pi k + i\sin 2\pi k$ and let $z = \cos\phi + i\sin\phi$ . By de Moivre, $\cos 4\phi + i\sin 4\phi = \cos 2\pi k + i\sin 2\pi k$ , so $4\phi = 2\pi k \;\Rightarrow\; \phi = \frac{\pi k}{2}, \qquad k = 0, 1, 2, 3.$

The four roots are $z_0 = \cos 0 + i\sin 0 = 1, \quad z_1 = \cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} = i,$ $z_2 = \cos\pi + i\sin\pi = -1, \quad z_3 = \cos\tfrac{3\pi}{2} + i\sin\tfrac{3\pi}{2} = -i.$

Argand diagram: the four points $1, i, -1, -i$ lie on the unit circle at the vertices of a square, equally spaced $90°$ apart.

Sum: $1 + i + (-1) + (-i) = 0.$
Why this scores
M1 for setting up $\phi = \frac{2\pi k}{4}$ with $+2\pi k$ ; A1 for all four exact roots; B1 for a correct Argand sketch showing the square symmetry; A1 for sum $= 0$ ; B1 for exact (not decimal) form. Listing $k = 0,1,2,3$ guarantees all four roots.
Question 5
7 marks
Find the five solutions of $z^5 = 32$ , giving each in the modulus-argument form $r(\cos\theta + i\sin\theta)$ with $-\pi < \theta \le \pi$ , and describe their geometric arrangement.
Model answer
Modulus-argument form of $32$ (a positive real, argument $0$ , plus full turns): $32 = 32\left[\cos(0 + 2\pi k) + i\sin(0 + 2\pi k)\right].$

Fifth roots. The real fifth root of the modulus is $32^{1/5} = 2$ , and the arguments are $\frac{2\pi k}{5}$ for $k = 0, 1, 2, 3, 4$ : $z_k = 2\left[\cos\frac{2\pi k}{5} + i\sin\frac{2\pi k}{5}\right].$

Listing, and adjusting arguments to lie in $-\pi < \theta \le \pi$ (so $k = 3, 4$ become negative angles): $z_0 = 2(\cos 0 + i\sin 0),\qquad z_1 = 2\left(\cos\tfrac{2\pi}{5} + i\sin\tfrac{2\pi}{5}\right),$ $z_2 = 2\left(\cos\tfrac{4\pi}{5} + i\sin\tfrac{4\pi}{5}\right),\qquad z_3 = 2\left(\cos\left(-\tfrac{4\pi}{5}\right) + i\sin\left(-\tfrac{4\pi}{5}\right)\right),$ $z_4 = 2\left(\cos\left(-\tfrac{2\pi}{5}\right) + i\sin\left(-\tfrac{2\pi}{5}\right)\right).$

Geometric arrangement. The five roots lie on a circle of radius $2$ centred at the origin, equally spaced $\frac{2\pi}{5} = 72°$ apart — the vertices of a regular pentagon, one vertex on the positive real axis at $z_0 = 2$ .
Why this scores
M1 setting up $32 = 32[\cos 2\pi k + i\sin 2\pi k]$ ; M1 modulus $32^{1/5} = 2$ and arguments $\frac{2\pi k}{5}$ ; A1 A1 for all five correct roots; M1 for reducing arguments into $(-\pi, \pi]$ ; A1 for the regular-pentagon description, radius $2$ . Forgetting the $+2\pi k$ is the classic error that yields only one root.
Question 6
8 marks
(a) Solve $z^3 = -27$ , giving each root in exact Cartesian form. (b) Hence express $z^3 + 27$ as the product of a real linear factor and a real quadratic factor.
Model answer
(a) Write $-27 = 27[\cos(\pi + 2\pi k) + i\sin(\pi + 2\pi k)]$ . The real cube root of $27$ is $3$ , and the arguments are $\frac{\pi + 2\pi k}{3}$ for $k = 0, 1, 2$ , namely $\frac{\pi}{3}, \pi, \frac{5\pi}{3}$ : $z_0 = 3\left(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\right) = 3\left(\tfrac{1}{2} + \tfrac{\sqrt3}{2}i\right) = \tfrac{3}{2} + \tfrac{3\sqrt3}{2}i,$ $z_1 = 3(\cos\pi + i\sin\pi) = -3,$ $z_2 = 3\left(\cos\tfrac{5\pi}{3} + i\sin\tfrac{5\pi}{3}\right) = \tfrac{3}{2} - \tfrac{3\sqrt3}{2}i.$

(b) The real root is $z = -3$ , giving the linear factor $(z + 3)$ . The two complex roots are conjugates $\frac{3}{2} \pm \frac{3\sqrt3}{2}i$ : their sum is $3$ and their product is $\left(\frac{3}{2}\right)^2 + \left(\frac{3\sqrt3}{2}\right)^2 = \frac{9}{4} + \frac{27}{4} = 9$ . Hence the real quadratic factor is $z^2 - 3z + 9$ , and $z^3 + 27 = (z + 3)(z^2 - 3z + 9).$
Why this scores
(a) M1 for $-27 = 27[\cos(\pi + 2\pi k) + \ldots]$ , A1 A1 for the three correct roots. (b) B1 for the linear factor from the real root, M1 for forming sum and product of the conjugate pair, A1 for $z^2 - 3z + 9$ , A1 for the complete factorisation. The result agrees with the standard sum-of-cubes identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ — a built-in check.

Key Formulae — de Moivre’s theorem

The formulae you need to memorise for de moivre’s theorem on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

De Moivre's theorem
$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
$\theta$
argument of the complex number (radians)
$n$
the (integer or rational) index
When to use
Raising a complex number in modulus-argument form to a power — multiply the argument by $n$ . NOT in MF19.
Product of two complex numbers
$r_1(\cos\theta_1 + i\sin\theta_1)\,r_2(\cos\theta_2 + i\sin\theta_2) = r_1 r_2\left[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right]$
$r_1, r_2$
moduli of the two numbers
$\theta_1, \theta_2$
their arguments
When to use
The geometric basis of de Moivre: multiplying multiplies moduli and adds arguments.
nth roots of unity
$z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \quad k = 0, 1, \ldots, n-1$
$n$
the degree — number of roots
$k$
the root index, $0$ to $n-1$
When to use
Solving $z^n = 1$ . The roots form a regular $n$ -gon on the unit circle and sum to $0$ (for $n \ge 2$ ).
nth roots of a general complex number
$z_k = R^{1/n}\left[\cos\frac{\alpha + 2\pi k}{n} + i\sin\frac{\alpha + 2\pi k}{n}\right], \quad k = 0, \ldots, n-1$
$R$
modulus of the number $w$ whose roots are sought
$\alpha$
argument of $w$
When to use
Solving $z^n = w$ . Real $n$ th root of the modulus; add $2\pi k$ to the argument BEFORE dividing by $n$ .
Euler's relation
$e^{i\theta} = \cos\theta + i\sin\theta$
$\theta$
argument (radians)
When to use
The exponential form of a unit complex number; makes de Moivre an index law $\left(e^{i\theta}\right)^n = e^{in\theta}$ .

Key Definitions and Keywords — de Moivre’s theorem

Definitions to memorise and the exact keywords mark schemes credit for de moivre’s theorem answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

De Moivre's theorem
Examiner keyword
The result $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ , valid for integer and rational $n$ . NOT given in MF19.
Modulus-argument (polar) form
Examiner keyword
Writing a complex number as $r(\cos\theta + i\sin\theta)$ , where $r = |z|$ is the modulus and $\theta = \arg z$ is the argument.
nth roots of unity
Examiner keyword
The $n$ solutions of $z^n = 1$ , given by $\cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}$ for $k = 0, \ldots, n-1$ ; they are the vertices of a regular $n$ -gon on the unit circle.
Principal argument
Examiner keyword
The unique value of the argument lying in the range $-\pi < \theta \le \pi$ . Roots are often required with arguments reduced into this range.
Argand diagram
The plane in which a complex number $x + iy$ is plotted as the point $(x, y)$ ; the modulus is the distance from the origin and the argument the angle from the positive real axis.
Exponential (Euler) form
$z = re^{i\theta}$ , equivalent to $r(\cos\theta + i\sin\theta)$ via Euler's relation $e^{i\theta} = \cos\theta + i\sin\theta$ .

Common Mistakes and Misconceptions — de Moivre’s theorem

The traps other students keep falling into on de moivre’s theorem questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Finding only one $n$ th root instead of all $n$ of them
9231 Paper 2 Examiner Reports — recurring comment
Why it happens
Forgetting to add $2\pi k$ to the argument before dividing by $n$ ; treating the equation like a real-number root.
How to avoid it
Always write the argument as $\alpha + 2\pi k$ at the start, then list $k = 0, 1, \ldots, n-1$ . The number of roots equals the degree $n$ .
✕Omitting the conclusion line of the induction proof
9231 Paper 2 mark schemes
Why it happens
Stopping once $\cos(k+1)\theta + i\sin(k+1)\theta$ is reached, treating the algebra as the whole proof.
How to avoid it
End every induction with: 'true for $n=1$ , and $P(k) \Rightarrow P(k+1)$ , so by induction true for all positive integers $n$ .'
✕Raising only the argument to the power and leaving the modulus unchanged
9231 Paper 2 Examiner Reports
Why it happens
Memorising de Moivre for unit-modulus numbers and applying it blindly to $r(\cos\theta + i\sin\theta)$ .
How to avoid it
For $[r(\cos\theta + i\sin\theta)]^n$ , the modulus becomes $r^n$ AND the argument becomes $n\theta$ . Both transform.
✕Giving decimal approximations where exact surd/ $\pi$ answers are required
9231 Paper 2 Examiner Reports
Why it happens
Reaching for a calculator instead of recalling exact values such as $\cos\frac{\pi}{6} = \frac{\sqrt3}{2}$ .
How to avoid it
Keep arguments as exact multiples of $\pi$ and use exact trig values. Decimals lose accuracy marks.
✕Wrong argument because the quadrant of the complex number was ignored
9231 Paper 2 Examiner Reports
Why it happens
Using $\tan^{-1}\frac{y}{x}$ blindly without checking the signs of the real and imaginary parts.
How to avoid it
Sketch the point first; adjust the principal-value $\tan^{-1}$ by $\pm\pi$ when the point is in the 2nd or 3rd quadrant.
✕Skipping or fudging the base case in the induction
9231 Paper 2 mark schemes
Why it happens
Assuming $n = 1$ is 'obvious' and not writing it down explicitly.
How to avoid it
Always verify $P(1)$ in writing: show both sides equal $\cos\theta + i\sin\theta$ . No base case means no valid induction.

de Moivre’s theorem

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1A power via modulus-argument form

2A power of a complex number not on the unit circle

3Proving de Moivre's theorem by induction

4The cube roots of unity

5The cube roots of a general complex number

6Using roots of unity to factorise an equation

De Moivre's theorem

Product of two complex numbers

nth roots of unity

nth roots of a general complex number

Euler's relation

De Moivre's theorem

Modulus-argument (polar) form

nth roots of unity

Principal argument

Argand diagram

Exponential (Euler) form

✕Finding only one nnnth root instead of all nnn of them

✕Omitting the conclusion line of the induction proof

✕Raising only the argument to the power and leaving the modulus unchanged

✕Giving decimal approximations where exact surd/π\piπ answers are required

✕Wrong argument because the quadrant of the complex number was ignored

✕Skipping or fudging the base case in the induction

✕Finding only one $n$ th root instead of all $n$ of them

✕Giving decimal approximations where exact surd/ $\pi$ answers are required