Paper 2 · Further Pure 2 ✦

De Moivre's Theorem

The engine behind multiple-angle identities, trig series, and the roots of any complex number — and one of the biggest sources of lost A* marks in FP2.

Syllabus 2.5Complex numbersAssumes: mod–arg & exponential form (9709 P3)

What the examiner expects you to do

Understand & prove de Moivre's theorem for a positive integer exponent, and use it for negative/rational exponents.
Express trig ratios of multiple angles (e.g. $\cos 5 θ$ ) in terms of powers of $\cos θ, \sin θ$ .
Express powers of $\sin θ, \cos θ$ in terms of multiple angles (for integration & proof).
Use the C + iS method to sum trigonometric series.
Find and use the n th roots of unity and the n th roots of any complex number.

§1 The statement

Recall the modulus–argument and exponential forms $z = r (\cos θ + i \sin θ) = r e^{i θ}$ . De Moivre's theorem says what happens to the argument when you take a power.

◆ Theorem — De Moivre

For any integer $n$ ,

(\cos θ + i \sin θ)^{n} = \cos n θ + i \sin n θ .

Geometrically: a power multiplies the argument by $n$ and takes the modulus to the power $n$ : $(r e^{i θ})^{n} = r^{n} e^{i n θ}$ .

For a rational exponent $p / q$ it gives one value of a multivalued expression — exactly what powers the n th roots later.

§2 Proof (the part that scores)

A* answers prove the positive-integer case by induction, then extend to negative integers separately. Don't just quote it.

Proof by induction — positive integer

n

Let $P (n)$ be the statement $(\cos θ + i \sin θ)^{n} = \cos n θ + i \sin n θ$ .

Base case $n = 1$ : LHS $= \cos θ + i \sin θ =$ RHS. So $P (1)$ is true.
Inductive hypothesis: assume $P (k)$ : $(\cos θ + i \sin θ)^{k} = \cos k θ + i \sin k θ$ .
Multiply both sides by $(\cos θ + i \sin θ)$ : $(\cos θ + i \sin θ)^{k + 1} = (\cos k θ + i \sin k θ) (\cos θ + i \sin θ) .$
Expand and collect real/imaginary parts (using $i^{2} = - 1$ ): $= (\cos k θ \cos θ - \sin k θ \sin θ) + i (\sin k θ \cos θ + \cos k θ \sin θ) .$
Apply the compound-angle formulae: $= \cos (k θ + θ) + i \sin (k θ + θ) = \cos (k + 1) θ + i \sin (k + 1) θ .$ So $P (k + 1)$ holds.
Conclusion: $P (1)$ true and $P (k) \Rightarrow P (k + 1)$ , so by induction $P (n)$ holds for all integers $n \geq 1$ . $■$

Extension to negative integers

Let $n = - m$ with $m \in ℤ^{+}$ . Then $(\cos θ + i \sin θ)^{- m} = \frac{1}{\cos m θ + i \sin m θ}$ . Multiply by the conjugate and use $\cos^{2} m θ + \sin^{2} m θ = 1$ :

= \cos m θ - i \sin m θ = \cos (- m θ) + i \sin (- m θ) = \cos n θ + i \sin n θ . ■

§3 Multiple angles: $\cos n θ, \sin n θ$

Expand $(\cos θ + i \sin θ)^{n}$ with the binomial theorem, then equate real parts for $\cos n θ$ , imaginary parts for $\sin n θ$ .

✎ Worked example —

\cos 3 θ

in terms of

\cos θ

By de Moivre $\cos 3 θ + i \sin 3 θ = (\cos θ + i \sin θ)^{3}$ . With $c = \cos θ, s = \sin θ$ :

(c + i s)^{3} = (c^{3} - 3 c s^{2}) + i (3 c^{2} s - s^{3}) .

Real part: $\cos 3 θ = c^{3} - 3 c s^{2}$ ; sub $s^{2} = 1 - c^{2}$ :

\cos 3 θ = 4 \cos^{3} θ - 3 \cos θ .

(The imaginary part gives $\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ$ for free.)

★ A* tip —

\tan n θ

Form $\frac{\sin n θ}{\cos n θ}$ from the expansion, then divide top and bottom by $\cos^{n} θ$ to turn everything into $\tan θ$ .

§4 Powers of $\sin θ, \cos θ$ as multiple angles

The reverse direction — essential for integrating $\sin^{6} θ$ and friends. Let $z = \cos θ + i \sin θ = e^{i θ}$ .

◇ The

z + \frac{1}{z}

results

z^{k} + \frac{1}{z^{k}} = 2 \cos k θ, z^{k} - \frac{1}{z^{k}} = 2 i \sin k θ .

✎ Worked example —

\sin^{4} θ

in terms of multiple angles

Since $2 i \sin θ = z - \frac{1}{z}$ , raise to the 4th power and group conjugate pairs:

(2 i \sin θ)^{4} = (z - \frac{1}{z})^{4} = 2 \cos 4 θ - 8 \cos 2 θ + 6 .

As $(2 i)^{4} = 16$ : $\sin^{4} θ = \frac{1}{8} (\cos 4 θ - 4 \cos 2 θ + 3) .$

§5 Summing series: the C + iS method

To sum $C = \sum \cos r θ$ and $S = \sum \sin r θ$ , combine them as $C + i S = \sum z^{r}$ — a geometric series in $z = e^{i θ}$ .

✎ Worked example —

\sum_{r = 0}^{n - 1} \cos r θ

$C + i S = \frac{z^{n} - 1}{z - 1}$ . Factor $e^{i n θ / 2}$ from the top and $e^{i θ / 2}$ from the bottom:

= e^{i (n - 1) θ / 2} \cdot \frac{\sin \frac{n θ}{2}}{\sin \frac{θ}{2}} \Rightarrow C = \frac{\sin \frac{n θ}{2}}{\sin \frac{θ}{2}} \cos \frac{(n - 1) θ}{2} .

⚠ Trap

The $e^{- i θ / 2}$ factoring is the step examiners reward. Don't leave it as $\frac{z^{n} - 1}{z - 1}$ — that scores no method marks for the split into sines/cosines.

§6 The n th roots of a complex number

To solve $w^{n} = z$ , write $z = r (\cos (θ + 2 π k) + i \sin (θ + 2 π k))$ — adding $2 π k$ so no roots are lost — then apply de Moivre with power $\frac{1}{n}$ :

◆ n th roots

w_{k} = r^{1 / n} (\cos \frac{θ + 2 π k}{n} + i \sin \frac{θ + 2 π k}{n}), k = 0,1, \dots, n - 1 .

Exactly $n$ roots, all of modulus $r^{1 / n}$ , equally spaced by $\frac{2 π}{n}$ — the vertices of a regular $n$ -gon about the origin.

✎ Worked example — cube roots of

8 i

$8 i = 8 cis \frac{π}{2}$ , so modulus $2$ and arguments $\frac{\frac{π}{2} + 2 π k}{3}$ :

w_{0} = \sqrt{3} + i, w_{1} = - \sqrt{3} + i, w_{2} = - 2 i .

★ Roots of unity

With $z = 1$ : the n th roots of unity $1, ω, \dots, ω^{n - 1}$ , $ω = e^{2 π i / n}$ . They sum to $0$ (for $n \geq 2$ ) and their product is $(- 1)^{n + 1}$ .

§7 Interactive: roots on the Argand diagram

Set the modulus and argument of $z$ , pick $n$ , and watch the $n$ roots of $w^{n} = z$ form a regular polygon. Set $r = 1, θ = 0$ for the roots of unity.

Modulus

r

1.00

Argument

θ

0°

Root index

n

drag the sliders ✎

§8 Examiner traps & A* checklist

⚠ The five most common mark-losers

De Moivre applies to $(\cos θ + i \sin θ)^{n}$ only. Rewrite $\cos θ - i \sin θ$ as $\cos (- θ) + i \sin (- θ)$ first.
Finding only one root. $w^{n} = z$ has exactly $n$ — use $k = 0, \dots, n - 1$ , and add $2 π k$ before dividing by $n$ .
Arguments outside the required range — give them in $- π < θ \leq π$ and in exact form.
Proving only the positive-integer case when negatives are asked — state the separate argument.
In C + iS, forgetting to take the real/imaginary part, or dropping the $e^{i (n - 1) θ / 2}$ phase factor.

§9 Video explainers (curated)

Board-agnostic — the maths of de Moivre is identical across CAIE/Edexcel/OCR. The clearest walk-throughs:

▶De Moivre's theorem — ExamSolutions
The statement, proof idea and first examples.↗ ▶n th roots of a complex number — ExamSolutions
The full method, worked on the Argand diagram.↗ ▶Roots of unity & series — TLMaths
Playlist on C + iS summation and roots-of-unity geometry.↗

§10 Check yourself

Score 0 / 5

Q1 · warm-up

Using de Moivre, $(\cos θ + i \sin θ)^{7}$ equals:

Solution

A power multiplies the argument: $(\cos θ + i \sin θ)^{7} = \cos 7 θ + i \sin 7 θ$ .

Q2 · multiple angles

Which is the correct expansion of $\cos 3 θ$ ?

Solution

Real part of $(\cos θ + i \sin θ)^{3}$ is $\cos^{3} θ - 3 \cos θ \sin^{2} θ$ ; sub $\sin^{2} θ = 1 - \cos^{2} θ$ to get $4 \cos^{3} θ - 3 \cos θ$ .

Q3 · concept

For $n \geq 2$ , the sum of the $n$ th roots of unity is:

Solution

They are the roots of $z^{n} - 1 = 0$ ; the coefficient of $z^{n - 1}$ is $0$ , so the sum of roots is $0$ .

Q4 · powers → multiple angles

$\sin^{4} θ$ in terms of multiple angles is:

Solution

From $(z - \frac{1}{z})^{4} = 16 \sin^{4} θ = 2 \cos 4 θ - 8 \cos 2 θ + 6$ .

Q5 · full method — try it on paper first

Find the three cube roots of $- 8$ , each in the form $a + b i$ (exact).

Solution

$- 8 = 8 (\cos (π + 2 π k) + i \sin (π + 2 π k))$ . Modulus $8^{1 / 3} = 2$ , arguments $\frac{π + 2 π k}{3}$ , $k = 0,1,2$ :

1 + \sqrt{3} i, - 2, 1 - \sqrt{3} i .

Equally spaced by $120^{\circ}$ on a circle of radius $2$ . ✓

FP2 · Topic 12 · de Moivre's Theorem

De Moivre's Theorem

What the examiner expects you to do

§1 The statement

§2 Proof (the part that scores)

§3 Multiple angles: cos⁡nθ, sin⁡nθ

§4 Powers of sin⁡θ,cos⁡θ as multiple angles

§5 Summing series: the C + iS method

§6 The n th roots of a complex number

§7 Interactive: roots on the Argand diagram

§8 Examiner traps & A* checklist

§9 Video explainers (curated)

§10 Check yourself

§3 Multiple angles: $\cos n θ, \sin n θ$

§4 Powers of $\sin θ, \cos θ$ as multiple angles