Home / A-Level Further Maths 9231 / Complex numbers / nth roots and roots of unity
Cambridge International AS & A Level · Topic 2.5
A-Level Further Maths 9231 / Topic 2.5 / AS & A Level

n roots, one regular polygon.

De Moivre's theorem with a fractional power finds every root of \(z^{n} = w\) at once: \(n\) points equally spaced on a circle. Use the nth roots of unity and \(\omega\) to simplify expressions, factorise \(z^{n} - 1\) into real quadratics, prove exact trigonometric results and solve equations such as \((z + 1)^{5} = z^{5}\).

Topic 2.5All complex numbers lessons →
The key idea

Write \(1 = e^{2\pi k i}\) for any integer \(k\). Then \(z^{n} = 1\) gives the nth roots of unity \(z = e^{2\pi k i/n}\), \(k = 0, 1, \ldots, n - 1\): exactly \(n\) of them, the vertices of a regular \(n\)-gon on the unit circle, one vertex at 1. Their sum is 0. More generally \(z^{n} = re^{i\theta}\) has the \(n\) roots \(z = r^{1/n}e^{i(\theta + 2\pi k)/n}\): modulus \(r^{1/n}\), arguments spaced \(\dfrac{2\pi}{n}\) apart. For cube roots, \(\omega = e^{2\pi i/3}\) satisfies \(\omega^{3} = 1\) and \(1 + \omega + \omega^{2} = 0\).

Re Im 2π/5 1 1 ω ω² ω³ ω⁴ ω = e2πi/5 roots of z⁵ = 1: 1, ω, ω², ω³, ω⁴ regular pentagon unit circle |z| = 1 angle 2π/5 apart sum of the five roots = 0 roots off the real axis come in conjugate pairs: ω⁴ = ω* and ω³ = (ω²)*
Fig. 1: the five roots of z^5 = 1 are 1, ω, ω², ω³ and ω⁴ with ω = e^(2πi/5). They sit on the unit circle 2π/5 apart and form a regular pentagon with one vertex at 1. The pentagon is symmetric about the real axis, so the roots pair off as complex conjugates, and by symmetry their sum is 0
Section 01

The nth roots of unity.

To solve \(z^{n} = 1\), write \(z = e^{i\theta}\) (every root has modulus 1, since \(|z|^{n} = 1\)) and \(1 = e^{2\pi k i}\). De Moivre's theorem gives \(e^{in\theta} = e^{2\pi k i}\), so \(\theta = \dfrac{2\pi k}{n}\). The values \(k = 0, 1, \ldots, n - 1\) give \(n\) different roots; \(k = n\) brings you back to \(k = 0\), so there are no more.

FactWhy
The roots are \(1, \omega, \omega^{2}, \ldots, \omega^{n-1}\), where \(\omega = e^{2\pi i/n}\)\(e^{2\pi k i/n} = \big(e^{2\pi i/n}\big)^{k}\)
They are the vertices of a regular \(n\)-gon on the unit circle, one vertex at 1modulus 1, arguments \(\dfrac{2\pi}{n}\) apart
Non-real roots come in conjugate pairs: \(\omega^{n-k} = \big(\omega^{k}\big)^{*}\)\(z^{n} - 1\) has real coefficients
Sum: \(1 + \omega + \omega^{2} + \cdots + \omega^{n-1} = 0\) for \(n \ge 2\)geometric series: \(\dfrac{\omega^{n} - 1}{\omega - 1} = 0\) as \(\omega^{n} = 1\), \(\omega \ne 1\)
Product of all \(n\) roots \(= (-1)^{n+1}\)product of the roots of \(z^{n} - 1 = 0\) is \((-1)^{n} \times (-1)\)

For \(n = 3\), the letter \(\omega\) means \(e^{2\pi i/3} = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\). The three cube roots of unity are \(1\), \(\omega\) and \(\omega^{2} = \omega^{*}\). Two facts do almost all the work: \(\omega^{3} = 1\) (so reduce every power modulo 3) and \(1 + \omega + \omega^{2} = 0\) (so \(1 + \omega = -\omega^{2}\), \(\omega + \omega^{2} = -1\) and so on).

Worked example The fifth roots of unity
Solve \(z^{5} = 1\), giving the roots in the form \(e^{i\theta}\) with \(-\pi \lt \theta \le \pi\). Describe their positions on an Argand diagram and verify that their sum is zero.
  1. \(z^{5} = 1 = e^{2\pi k i}\), so \(z = e^{2\pi k i/5}\) for \(k = 0, 1, 2, 3, 4\).
  2. Bring \(k = 3\) and \(k = 4\) into the range: \(\tfrac{6\pi}{5} - 2\pi = -\tfrac{4\pi}{5}\) and \(\tfrac{8\pi}{5} - 2\pi = -\tfrac{2\pi}{5}\).
  3. The roots are \(1\), \(e^{\pm 2\pi i/5}\) and \(e^{\pm 4\pi i/5}\): five points on the unit circle \(\tfrac{2\pi}{5}\) apart, the vertices of a regular pentagon with one vertex at 1 (Fig. 1).
  4. Sum: with \(\omega = e^{2\pi i/5}\), \[1 + \omega + \omega^{2} + \omega^{3} + \omega^{4} = \dfrac{\omega^{5} - 1}{\omega - 1} = \dfrac{1 - 1}{\omega - 1} = 0.\]
Answer: \(z = 1, e^{\pm 2\pi i/5}, e^{\pm 4\pi i/5}\), a regular pentagon on the unit circle; the sum is 0
Worked example Simplifying with omega
\(\omega\) is a complex cube root of unity. Simplify (a) \(\omega^{100}\) (b) \((1 + \omega)^{3}\) (c) \((2 + \omega)(2 + \omega^{2})\).
  1. (a) \(100 = 3 \times 33 + 1\), so \(\omega^{100} = (\omega^{3})^{33}\omega = \omega\).
  2. (b) \(1 + \omega + \omega^{2} = 0\) gives \(1 + \omega = -\omega^{2}\), so \((1 + \omega)^{3} = -\omega^{6} = -(\omega^{3})^{2} = -1\).
  3. (c) Expand: \(4 + 2\omega^{2} + 2\omega + \omega^{3} = 4 + 2(\omega + \omega^{2}) + 1\).
  4. \(\omega + \omega^{2} = -1\), so this is \(4 - 2 + 1 = 3\).
Answer: (a) \(\omega\)   (b) \(-1\)   (c) \(3\)
Your turn
Write the sixth roots of unity in the form \(a + bi\), and show that their product is \(-1\).
Examiner trap

Stop at \(k = n - 1\): \(z^{n} = 1\) has exactly \(n\) roots, and \(k = n\) repeats \(k = 0\). Do not forget the root \(z = 1\) itself (\(k = 0\)). With \(\omega\), the identity is \(1 + \omega + \omega^{2} = 0\), not \(\omega^{3} = 0\): \(\omega^{3} = 1\), so \(\omega^{4} = \omega\), never 0.

Section 02

The nth roots of any complex number.

To solve \(z^{n} = w\), first write \(w = re^{i\theta}\) in exponential form. Then \(w = re^{i(\theta + 2\pi k)}\) for every integer \(k\), and taking the \(n\)th root (de Moivre with exponent \(\tfrac{1}{n}\)) gives \[z = r^{1/n}e^{i(\theta + 2\pi k)/n}, \quad k = 0, 1, \ldots, n - 1.\]

All \(n\) roots have the same modulus \(r^{1/n}\), and their arguments are \(\dfrac{\theta}{n}\) plus multiples of \(\dfrac{2\pi}{n}\). So the roots are the vertices of a regular \(n\)-gon centred at the origin, with radius \(r^{1/n}\). Another way to say it: if \(z_{0}\) is any one root, the full set is \(z_{0}, z_{0}\omega, z_{0}\omega^{2}, \ldots, z_{0}\omega^{n-1}\) with \(\omega = e^{2\pi i/n}\), because multiplying by \(\omega\) rotates by \(\dfrac{2\pi}{n}\) without changing the modulus.

StepFor \(z^{3} = 8i\)
1. Exponential form of \(w\)\(8i = 8e^{i\pi/2}\)
2. Add \(2\pi k\) to the argument\(8e^{i(\pi/2 + 2\pi k)}\)
3. Root of the modulus, divide the argument by \(n\)\(z = 2e^{i(\pi/6 + 2\pi k/3)}\)
4. List \(k = 0, \ldots, n - 1\), adjust into range\(\arg z = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, -\tfrac{\pi}{2}\)
Worked example Cube roots of 8i
Solve \(z^{3} = 8i\), giving the roots in the form \(a + bi\). Describe the shape they form.
  1. \(8i = 8e^{i(\pi/2 + 2\pi k)}\), so \(z = 8^{1/3}e^{i(\pi/2 + 2\pi k)/3} = 2e^{i(\pi/6 + 2\pi k/3)}\).
  2. \(k = 0\): \(2e^{i\pi/6} = 2\big(\tfrac{\sqrt{3}}{2} + \tfrac{1}{2}i\big) = \sqrt{3} + i\).
  3. \(k = 1\): \(2e^{5\pi i/6} = -\sqrt{3} + i\).
  4. \(k = 2\): \(2e^{3\pi i/2} = -2i\).
  5. The three roots have modulus 2 and arguments \(\tfrac{2\pi}{3}\) apart: an equilateral triangle on the circle \(|z| = 2\).
Answer: \(z = \sqrt{3} + i\), \(-\sqrt{3} + i\) or \(-2i\), the vertices of an equilateral triangle
Worked example Fourth roots in exact form
Solve \(z^{4} = -8 + 8\sqrt{3}\,i\), giving your answers in the form \(re^{i\theta}\) with \(-\pi \lt \theta \le \pi\), and then in the form \(a + bi\).
  1. \(|w| = \sqrt{64 + 192} = 16\); \(w\) is in the second quadrant with \(\tan^{-1}\sqrt{3} = \tfrac{\pi}{3}\), so \(\arg w = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}\).
  2. \[z = 16^{1/4}e^{i(2\pi/3 + 2\pi k)/4} = 2e^{i(\pi/6 + \pi k/2)}.\]
  3. \(k = 0, 1, 2, 3\) give arguments \(\tfrac{\pi}{6}, \tfrac{2\pi}{3}, \tfrac{7\pi}{6}, \tfrac{5\pi}{3}\); the last two become \(-\tfrac{5\pi}{6}\) and \(-\tfrac{\pi}{3}\).
  4. In Cartesian form: \(2e^{i\pi/6} = \sqrt{3} + i\), \(2e^{2\pi i/3} = -1 + \sqrt{3}\,i\), \(2e^{-5\pi i/6} = -\sqrt{3} - i\), \(2e^{-i\pi/3} = 1 - \sqrt{3}\,i\).
  5. Check one: \((\sqrt{3} + i)^{2} = 2 + 2\sqrt{3}\,i\), and \((2 + 2\sqrt{3}\,i)^{2} = 4 - 12 + 8\sqrt{3}\,i = -8 + 8\sqrt{3}\,i\).
Answer: \(z = 2e^{i\pi/6}, 2e^{2\pi i/3}, 2e^{-5\pi i/6}, 2e^{-i\pi/3}\), that is \(\pm(\sqrt{3} + i)\) and \(\pm(1 - \sqrt{3}\,i)\)
Your turn
Solve \(z^{6} = -64\), giving your answers in the form \(a + bi\).
Examiner trap

Three slips cost most marks. The modulus is \(r^{1/n}\), not \(\dfrac{r}{n}\) (\(16^{1/4} = 2\), not 4). Add \(2\pi k\) BEFORE dividing by \(n\): writing \(\dfrac{\theta}{n} + 2\pi k\) gives the same root \(n\) times. And find \(\arg w\) from the quadrant: \(\tan^{-1}\!\big(\tfrac{8\sqrt{3}}{-8}\big)\) on a calculator gives \(-\tfrac{\pi}{3}\), the wrong quadrant.

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Which list gives all the roots of \(z^{3} = 1\)?

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\(\omega\) is a complex cube root of unity. What is \(1 + \omega + \omega^{2}\)?

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What is the sum of the seven 7th roots of unity?

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Which of these is a root of \(z^{4} = 16i\)?

Section 03

Using roots to factorise and solve.

Pairing each non-real root with its conjugate turns complex linear factors into real quadratic ones: \[(z - e^{i\alpha})(z - e^{-i\alpha}) = z^{2} - 2\cos\alpha\,z + 1.\] This factorises \(z^{n} - 1\) over the reals, and comparing coefficients then gives exact trigonometric values. The same roots of unity also solve equations like \((z + 1)^{5} = z^{5}\).

For \(n = 5\) the roots are \(1\), \(e^{\pm 2\pi i/5}\) and \(e^{\pm 4\pi i/5}\), so \[z^{5} - 1 = (z - 1)\Big(z^{2} - 2\cos\tfrac{2\pi}{5}\,z + 1\Big)\Big(z^{2} - 2\cos\tfrac{4\pi}{5}\,z + 1\Big).\] Dividing by \(z - 1\) gives \(z^{4} + z^{3} + z^{2} + z + 1\) as a product of two real quadratics.

For an equation of the form \((z + a)^{n} = z^{n}\), divide by \(z^{n}\) (note \(z = 0\) is not a root) and put \(w = \dfrac{z + a}{z}\). Then \(w^{n} = 1\), so \(w\) is an \(n\)th root of unity, but \(w = 1\) is impossible (it would need \(a = 0\)). That leaves \(n - 1\) roots, which matches the degree: the \(z^{n}\) terms cancel. A useful step is \(e^{i\phi} - 1 = e^{i\phi/2}\big(e^{i\phi/2} - e^{-i\phi/2}\big) = 2i\sin\tfrac{\phi}{2}\,e^{i\phi/2}\).

Worked example Factorise, then deduce a cosine result
Express \(z^{5} - 1\) as the product of one linear and two quadratic factors with real coefficients. Hence show that \(\cos\tfrac{2\pi}{5} + \cos\tfrac{4\pi}{5} = -\tfrac{1}{2}\).
  1. The roots are \(1\), \(e^{\pm 2\pi i/5}\), \(e^{\pm 4\pi i/5}\). Pair each conjugate pair: \((z - e^{i\alpha})(z - e^{-i\alpha}) = z^{2} - (e^{i\alpha} + e^{-i\alpha})z + 1 = z^{2} - 2\cos\alpha\,z + 1\).
  2. \[z^{5} - 1 = (z - 1)\Big(z^{2} - 2\cos\tfrac{2\pi}{5}\,z + 1\Big)\Big(z^{2} - 2\cos\tfrac{4\pi}{5}\,z + 1\Big).\]
  3. The sum of the five roots is 0 (the \(z^{4}\) coefficient of \(z^{5} - 1\) is 0). The conjugate pairs add to \(2\cos\alpha\), so \[1 + 2\cos\tfrac{2\pi}{5} + 2\cos\tfrac{4\pi}{5} = 0.\]
  4. Rearrange: \(\cos\tfrac{2\pi}{5} + \cos\tfrac{4\pi}{5} = -\tfrac{1}{2}\).
Answer: \(z^{5} - 1 = (z - 1)\big(z^{2} - 2\cos\tfrac{2\pi}{5}z + 1\big)\big(z^{2} - 2\cos\tfrac{4\pi}{5}z + 1\big)\), and \(\cos\tfrac{2\pi}{5} + \cos\tfrac{4\pi}{5} = -\tfrac{1}{2}\)
Worked example Solving (z + 1)^5 = z^5
Show that the roots of \((z + 1)^{5} = z^{5}\) are \(z = -\tfrac{1}{2} - \tfrac{1}{2}i\cot\tfrac{k\pi}{5}\) for \(k = 1, 2, 3, 4\).
  1. \(z = 0\) is not a root, so divide by \(z^{5}\): \(w^{5} = 1\) with \(w = \dfrac{z + 1}{z}\), so \(w = e^{i\phi}\), \(\phi = \tfrac{2k\pi}{5}\).
  2. \(w = 1\) would give \(z + 1 = z\), impossible, so \(k = 1, 2, 3, 4\): four roots, matching the degree (the \(z^{5}\) terms cancel).
  3. \(wz = z + 1\) gives \(z = \dfrac{1}{w - 1}\), and \(e^{i\phi} - 1 = 2i\sin\tfrac{\phi}{2}\,e^{i\phi/2}\), so \[z = \dfrac{e^{-i\phi/2}}{2i\sin\frac{\phi}{2}} = \dfrac{\cos\frac{\phi}{2} - i\sin\frac{\phi}{2}}{2i\sin\frac{\phi}{2}}.\]
  4. \(\dfrac{\cos\frac{\phi}{2}}{2i\sin\frac{\phi}{2}} = -\tfrac{1}{2}i\cot\tfrac{\phi}{2}\) and \(\dfrac{-i\sin\frac{\phi}{2}}{2i\sin\frac{\phi}{2}} = -\tfrac{1}{2}\).
  5. With \(\tfrac{\phi}{2} = \tfrac{k\pi}{5}\): \(z = -\tfrac{1}{2} - \tfrac{1}{2}i\cot\tfrac{k\pi}{5}\). Check: every root has real part \(-\tfrac{1}{2}\), as it must, since \(|z + 1| = |z|\) puts \(z\) on the perpendicular bisector of \(0\) and \(-1\).
Answer: \(z = -\tfrac{1}{2} - \tfrac{1}{2}i\cot\tfrac{k\pi}{5}\), \(k = 1, 2, 3, 4\)
Your turn
Use the sixth roots of unity to express \(z^{6} - 1\) as a product of two linear and two quadratic factors with real coefficients.
Examiner trap

In \((z + 1)^{5} = z^{5}\) the root of unity \(w = 1\) gives NO solution, so there are 4 roots, not 5. When pairing roots, the quadratic is \(z^{2} - 2\cos\alpha\,z + 1\): the middle term has a minus sign and a factor 2, and the constant is \(+1\) because \(e^{i\alpha}e^{-i\alpha} = 1\).

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\(\omega\) is a complex cube root of unity. Find the value of \((1 + \omega)(1 + \omega^{2})\).

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Which of these is a factor of \(z^{5} - 1\) with real coefficients?

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Solve \(z^{3} = -8\).

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The equation \((z + 1)^{5} = z^{5}\) has four roots. What is the real part of each root?

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Six original Cambridge-style questions
The sixth roots of unity, two \(\omega\) simplifications, cube roots of \(4\sqrt{2}(1 + i)\), an equation \((z + 1)^{4} = 16z^{4}\), the exact value of \(\cos\tfrac{2\pi}{5}\) from a factorisation of \(z^{5} - 1\), and a proof that the nth roots of unity sum to zero.