nth Roots of Unity

If De Moivre's theorem tells you how to raise a complex number to a power, running it backwards tells you how to take roots — and the answer is beautifully geometric. Every non-zero complex number has exactly $n$n distinct $n$nth roots, and they sit at the vertices of a regular $n$n-gon on a circle. The simplest and most important case is $z^n = 1$zn=1: the $n$nth roots of unity, equally spaced points on the unit circle whose remarkable algebra (they multiply like a clock, and they sum to zero) underpins everything from factorising $z^n - 1$zn−1 to the discrete Fourier transform. This lesson is one of the most elegant in the whole A-Level.

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Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2, the direct sequel to De Moivre's theorem. It tests AO1 (write down the $n$n roots systematically), AO2 (justify why there are exactly $n$n, prove the sum-is-zero and the $1 + \omega + \cdots + \omega^{n-1} = 0$1+ω+⋯+ωn−1=0 relation, and reason about the regular-polygon geometry) and AO3 (use roots of unity to factorise, to solve $z^n = w$zn=w efficiently, or to handle a multi-step geometric problem). It connects backwards to De Moivre and conjugate roots, and forwards to the further-algebra strand's work on roots and coefficients.

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Core theory: roots come in a ring of $n$n, equally spaced

The $n$nth roots of a general complex number

To solve $z^n = w$zn=w where $w = R(\cos\phi + i\sin\phi)$w=R(cosϕ+isinϕ) (with $R = |w| > 0$R=∣w∣>0), write $z = r(\cos\theta + i\sin\theta)$z=r(cosθ+isinθ). By De Moivre, $z^n = r^n(\cos n\theta + i\sin n\theta)$zn=rn(cosnθ+isinnθ), so matching with $w$w:

• moduli: $r^n = R \Rightarrow r = R^{1/n}$rn=R⇒r=R1/n (the positive real $n$nth root — there is only one);
• arguments: $n\theta = \phi + 2k\pi$nθ=ϕ+2kπ for any integer $k$k, because the argument is only defined up to whole turns. Hence $\theta = \dfrac{\phi + 2k\pi}{n}$θ=nϕ+2kπ​.

The crucial point is the $+2k\pi$+2kπ: adding whole turns to $\phi$ϕ before dividing by $n$n gives genuinely different roots. Taking $k = 0, 1, 2, \ldots, n-1$k=0,1,2,…,n−1 yields exactly $n$n distinct values (then $k = n$k=n repeats $k = 0$k=0, since it adds $\frac{2n\pi}{n} = 2\pi$n2nπ​=2π to the argument):

$\boxed{\,z_k = R^{1/n}\left(\cos\frac{\phi + 2k\pi}{n} + i\sin\frac{\phi + 2k\pi}{n}\right), \qquad k = 0, 1, \ldots, n-1.\,}$zk​=R1/n(cosnϕ+2kπ​+isinnϕ+2kπ​),k=0,1,…,n−1.​

Consecutive roots differ in argument by $\frac{2\pi}{n}$n2π​ and share the modulus $R^{1/n}$R1/n, so they are equally spaced on a circle of radius $R^{1/n}$R1/n — the vertices of a regular $n$n-gon.

The $n$nth roots of unity

Taking $w = 1 = \cos 0 + i\sin 0$w=1=cos0+isin0 (so $R = 1$R=1, $\phi = 0$ϕ=0) gives the $n$nth roots of unity:

$z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \qquad k = 0, 1, \ldots, n-1.$zk​=cosn2kπ​+isinn2kπ​,k=0,1,…,n−1.

Write $\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n} = e^{2\pi i/n}$ω=cosn2π​+isinn2π​=e2πi/n, the primitive $n$nth root of unity (the first one anticlockwise from $1$1). Then $z_k = \omega^k$zk​=ωk, and the complete set of roots is

$1,\; \omega,\; \omega^2,\; \ldots,\; \omega^{n-1}.$1,ω,ω2,…,ωn−1.

They all lie on the unit circle, are equally spaced by $\frac{2\pi}{n}$n2π​, form a regular $n$n-gon, and always include $z = 1$z=1. Multiplying by $\omega$ω just rotates one root to the next — the roots behave like the hours on a clock face. Note too that $\omega^{n} = 1$ωn=1 and $\omega^{-1} = \omega^{n-1} = \overline{\omega}$ω−1=ωn−1=ω, so the powers of $\omega$ω cycle with period $n$n and reciprocals coincide with conjugates — both facts used constantly when manipulating roots.

The sum is zero

For $n \ge 2$n≥2, the $n$nth roots of unity sum to zero:

$1 + \omega + \omega^2 + \cdots + \omega^{n-1} = \sum_{k=0}^{n-1}\omega^k = \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0,$1+ω+ω2+⋯+ωn−1=∑k=0n−1​ωk=1−ω1−ωn​=1−ω1−1​=0,

a finite geometric series with $\omega^n = 1$ωn=1 and $\omega \neq 1$ω=1. Geometrically this says the position vectors to the vertices of a regular polygon centred at the origin cancel — a tidy AO2 result examiners like to ask you to prove.

Roots of $z^n = w$zn=w from one root plus the roots of unity

Once you have any single solution $z_0$z0​ of $z^n = w$zn=w, the full set is

$z_0,\; z_0\omega,\; z_0\omega^2,\; \ldots,\; z_0\omega^{n-1},$z0​,z0​ω,z0​ω2,…,z0​ωn−1,

because multiplying by $\omega^k$ωk keeps the modulus ($|\omega| = 1$∣ω∣=1) and rotates by $\frac{2k\pi}{n}$n2kπ​, stepping between the equally spaced roots. This is often the fastest route in the exam.

Factorising $z^n - 1$zn−1

Since the roots of unity are exactly the roots of $z^n - 1 = 0$zn−1=0,

$z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1}),$zn−1=(z−1)(z−ω)(z−ω2)⋯(z−ωn−1),

and dividing by $(z - 1)$(z−1) gives the cyclotomic-style factor

$z^{n-1} + z^{n-2} + \cdots + z + 1 = (z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1}).$zn−1+zn−2+⋯+z+1=(z−ω)(z−ω2)⋯(z−ωn−1).

Two consequences are worth noting. First, setting $z = 1$z=1 in the right-hand product gives $\prod_{k=1}^{n-1}(1 - \omega^k) = n$∏k=1n−1​(1−ωk)=n (the left side is $1 + 1 + \cdots + 1 = n$1+1+⋯+1=n) — a tidy identity that appears in harder problems. Second, when $n$n is even the real-coefficient factorisation pairs each non-real root $\omega^k$ωk with its conjugate $\omega^{n-k} = \overline{\omega^k}$ωn−k=ωk to give real quadratic factors $z^2 - 2\cos\frac{2k\pi}{n}\,z + 1$z2−2cosn2kπ​z+1, since each pair has sum $2\cos\frac{2k\pi}{n}$2cosn2kπ​ and product $1$1. This is exactly the route to factorising $z^n - 1$zn−1 (or $z^n + 1$zn+1) over the reals.

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Worked examples (with mark scheme)

Example 1 — cube roots of unity

Solve $z^3 = 1$z3=1.

With $R = 1$R=1, $\phi = 0$ϕ=0, $n = 3$n=3: $z_k = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}$zk​=cos32kπ​+isin32kπ​, $k = 0, 1, 2$k=0,1,2. (M1 set up the formula.)

$z_0 = 1; \quad z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt 3}{2}i; \quad z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt 3}{2}i. \quad (\text{A1 all three})$z0​=1;z1​=cos32π​+isin32π​=−21​+23​​i;z2​=cos34π​+isin34π​=−21​−23​​i.(A1 all three)

The roots form an equilateral triangle on the unit circle, and $z_2 = \bar z_1$z2​=zˉ1​ (the non-real roots are a conjugate pair). Writing $\omega = -\tfrac12 + \tfrac{\sqrt 3}{2}i$ω=−21​+23​​i, the roots are $1, \omega, \omega^2$1,ω,ω2, with $\omega^3 = 1$ω3=1, $1 + \omega + \omega^2 = 0$1+ω+ω2=0 and $\omega^2 = \bar\omega$ω2=ωˉ. (M1 formula; A1 three correct roots in exact form.)

Example 2 — fourth roots of unity

Solve $z^4 = 1$z4=1.

$z_k = \cos\frac{2k\pi}{4} + i\sin\frac{2k\pi}{4} = \cos\frac{k\pi}{2} + i\sin\frac{k\pi}{2}, \quad k = 0,1,2,3 \;\Longrightarrow\; z = 1,\, i,\, -1,\, -i. \quad (\text{M1 A1})$zk​=cos42kπ​+isin42kπ​=cos2kπ​+isin2kπ​,k=0,1,2,3⟹z=1,i,−1,−i.(M1 A1)

These are the vertices of a square on the unit circle, lying on the axes. (A simple sanity check: their sum $1 + i - 1 - i = 0$1+i−1−i=0 ✓.)

Example 3 — cube roots of a general number

Find all cube roots of $8i$8i.

Write $8i = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$8i=8(cos2π​+isin2π​), so $R = 8$R=8, $\phi = \frac{\pi}{2}$ϕ=2π​, and $R^{1/3} = 2$R1/3=2. (M1 polar form and $R^{1/3}$R1/3.)

$z_k = 2\left(\cos\frac{\frac{\pi}{2} + 2k\pi}{3} + i\sin\frac{\frac{\pi}{2} + 2k\pi}{3}\right), \quad k = 0,1,2. \quad (\text{M1 add } 2k\pi \text{ before dividing})$zk​=2(cos32π​+2kπ​+isin32π​+2kπ​),k=0,1,2.(M1 add 2kπ before dividing)

$z_0 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \sqrt 3 + i;$z0​=2(cos6π​+isin6π​)=3​+i; $z_1 = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = -\sqrt 3 + i;$z1​=2(cos65π​+isin65π​)=−3​+i; $z_2 = 2\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = -2i. \quad (\text{A1 all three})$z2​=2(cos23π​+isin23π​)=−2i.(A1 all three)

Check: $(\sqrt 3 + i)$(3​+i) has modulus $2$2 and argument $\frac{\pi}{6}$6π​, so by De Moivre $(\sqrt 3 + i)^3 = 2^3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 8i$(3​+i)3=23(cos2π​+isin2π​)=8i ✓. (M1 polar form; M1 the $+2k\pi$+2kπ step; A1 three roots — with a verification that one cubes to $8i$8i.)

Example 4 — fourth roots, leading to a real-coefficient factorisation

Find the fourth roots of $-16$−16, and hence factorise $z^4 + 16$z4+16 into real quadratics.

$-16 = 16(\cos\pi + i\sin\pi)$−16=16(cosπ+isinπ), $R^{1/4} = 2$R1/4=2:

$z_k = 2\left(\cos\frac{\pi + 2k\pi}{4} + i\sin\frac{\pi + 2k\pi}{4}\right), \;k=0,1,2,3 \;\Rightarrow\; \sqrt 2(1+i),\, \sqrt 2(-1+i),\, \sqrt 2(-1-i),\, \sqrt 2(1-i).$zk​=2(cos4π+2kπ​+isin4π+2kπ​),k=0,1,2,3⇒2​(1+i),2​(−1+i),2​(−1−i),2​(1−i).

The roots pair into conjugates $\sqrt 2(\pm 1 + i)$2​(±1+i) and $\sqrt 2(\pm 1 - i)$2​(±1−i). Each conjugate pair gives a real quadratic; e.g. $\sqrt 2(1 \pm i)$2​(1±i) has sum $2\sqrt 2$22​ and product $2(1 + 1) = 4$2(1+1)=4, giving $z^2 - 2\sqrt 2\,z + 4$z2−22​z+4; the other pair gives $z^2 + 2\sqrt 2\,z + 4$z2+22​z+4. Hence

$z^4 + 16 = (z^2 - 2\sqrt 2\,z + 4)(z^2 + 2\sqrt 2\,z + 4).$z4+16=(z2−22​z+4)(z2+22​z+4).

Example 5 — using one root and the roots of unity

Given that $z_0 = \sqrt 3 + i$z0​=3​+i is one cube root of $8i$8i, write down the other two without re-deriving from scratch.

The other roots are $z_0\omega$z0​ω and $z_0\omega^2$z0​ω2, where $\omega = e^{2\pi i/3} = -\tfrac12 + \tfrac{\sqrt 3}{2}i$ω=e2πi/3=−21​+23​​i is the primitive cube root of unity. (M1 multiply by $\omega, \omega^2$ω,ω2.) Geometrically each multiplication rotates by $\frac{2\pi}{3}$32π​ and keeps the modulus $2$2, so:

$z_0\omega = 2\,\operatorname{cis}\!\left(\frac{\pi}{6} + \frac{2\pi}{3}\right) = 2\,\operatorname{cis}\frac{5\pi}{6} = -\sqrt 3 + i; \quad z_0\omega^2 = 2\,\operatorname{cis}\!\left(\frac{\pi}{6} + \frac{4\pi}{3}\right) = 2\,\operatorname{cis}\frac{3\pi}{2} = -2i. \quad (\text{A1})$z0​ω=2cis(6π​+32π​)=2cis65π​=−3​+i;z0​ω2=2cis(6π​+34π​)=2cis23π​=−2i.(A1)

These match Example 3 exactly. (M1 the rotation idea; A1 the two roots. This "one root times the roots of unity" route is often the quickest in the exam once you have any single root.)

Example 6 — a sum of selected roots

If $\omega = e^{2\pi i/6}$ω=e2πi/6 is a primitive sixth root of unity, evaluate $1 + \omega^2 + \omega^4$1+ω2+ω4.

The powers $1, \omega^2, \omega^4$1,ω2,ω4 are themselves the three cube roots of unity (since $(\omega^2)^3 = \omega^6 = 1$(ω2)3=ω6=1, and they are distinct). (M1 recognise as cube roots of unity.) Therefore their sum is $0$0:

$1 + \omega^2 + \omega^4 = 0. \quad (\text{A1})$1+ω2+ω4=0.(A1)

(M1 spotting that $\{1, \omega^2, \omega^4\}${1,ω2,ω4} are the cube roots of unity; A1 for $0$0. Selecting every other root of unity reproduces a smaller set of roots of unity — a recurring exam idea, and the seed of the "roots-of-unity filter" in the stretch section.)

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Specimen-style exam question

(specimen-style — not from any past paper)

(a) Solve $z^4 = -8 + 8\sqrt 3\, i$z4=−8+83​i, giving the roots in the form $r(\cos\theta + i\sin\theta)$r(cosθ+isinθ) with $-\pi < \theta \le \pi$−π<θ≤π. (b) State the geometric figure formed by the four roots on an Argand diagram.

(a) Modulus: $|-8 + 8\sqrt 3\, i| = \sqrt{64 + 192} = \sqrt{256} = 16$∣−8+83​i∣=64+192​=256​=16. The point is in the second quadrant ($\operatorname{Re} < 0$Re<0, $\operatorname{Im} > 0$Im>0), so $\arg = \pi - \arctan\frac{8\sqrt 3}{8} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$arg=π−arctan883​​=π−3π​=32π​. With $R^{1/4} = 16^{1/4} = 2$R1/4=161/4=2:

$z_k = 2\left(\cos\frac{\frac{2\pi}{3} + 2k\pi}{4} + i\sin\frac{\frac{2\pi}{3} + 2k\pi}{4}\right) = 2\left(\cos\frac{\pi + 3k\pi}{6} + i\sin\frac{\pi + 3k\pi}{6}\right), \quad k = 0,1,2,3.$zk​=2(cos432π​+2kπ​+isin432π​+2kπ​)=2(cos6π+3kπ​+isin6π+3kπ​),k=0,1,2,3.

Evaluating the arguments $\frac{\pi}{6}, \frac{4\pi}{6}=\frac{2\pi}{3}, \frac{7\pi}{6}, \frac{10\pi}{6}=\frac{5\pi}{3}$6π​,64π​=32π​,67π​,610π​=35π​ and reducing the last two into $(-\pi, \pi]$(−π,π] (subtract $2\pi$2π): $\frac{7\pi}{6} \to -\frac{5\pi}{6}$67π​→−65π​, $\frac{5\pi}{3} \to -\frac{\pi}{3}$35π​→−3π​. So

$z = 2\,\operatorname{cis}\frac{\pi}{6},\; 2\,\operatorname{cis}\frac{2\pi}{3},\; 2\,\operatorname{cis}\!\left(-\frac{5\pi}{6}\right),\; 2\,\operatorname{cis}\!\left(-\frac{\pi}{3}\right).$z=2cis6π​,2cis32π​,2cis(−65π​),2cis(−3π​).

(b) The four roots have equal modulus $2$2 and arguments spaced by $\frac{\pi}{2}$2π​, so they are the vertices of a square inscribed in the circle $|z| = 2$∣z∣=2.

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