AS & A Level · Practice questions · nth roots and roots of unity

Every root, exactly.

Six original Cambridge-style questions on nth roots: roots of unity and \(\omega\), roots of a general complex number as a regular polygon, real quadratic factors of \(z^{n} - 1\), exact cosines and an equation solved with roots of unity, with full model solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Exponential form, add \(2\pi k\), then divide.

Write \(w = re^{i\theta}\), checking the quadrant for \(\theta\).
Add \(2\pi k\): \(z = r^{1/n}e^{i(\theta + 2\pi k)/n}\), \(k = 0, \ldots, n - 1\), then adjust arguments into the range asked for.
Picture the roots as a regular \(n\)-gon of radius \(r^{1/n}\); roots of unity sum to 0.
Simplify with \(\omega^{3} = 1\) and \(1 + \omega + \omega^{2} = 0\).
Factorise by pairing conjugates: \((z - e^{i\alpha})(z - e^{-i\alpha}) = z^{2} - 2\cos\alpha\,z + 1\).
Solve \((z + a)^{n} = cz^{n}\) with \(w = \dfrac{z + a}{z}\), discarding any \(w\) that gives no \(z\).

Recall

[3 marks]

Find the sixth roots of unity, giving each in the form \(a + bi\), and state the shape they form on an Argand diagram.

\(z^{6} = 1 = e^{2\pi k i}\), so \(z = e^{\pi k i/3}\), \(k = 0, 1, \ldots, 5\). ✓
\(k = 0\): 1; \(k = 1, 5\): \(\tfrac{1}{2} \pm \tfrac{\sqrt{3}}{2}i\); \(k = 2, 4\): \(-\tfrac{1}{2} \pm \tfrac{\sqrt{3}}{2}i\); \(k = 3\): \(-1\). ✓
They are the vertices of a regular hexagon inscribed in the unit circle, with a vertex at 1. ✓

Calculation

[4 marks]

\(\omega\) is a complex cube root of unity.

(a) Show that \((1 - \omega)(1 - \omega^{2}) = 3\).

(b) Find the value of \((2 + 5\omega + 2\omega^{2})^{3}\).

(a) Expand: \(1 - \omega^{2} - \omega + \omega^{3} = 1 - (\omega + \omega^{2}) + 1\). ✓
\(\omega + \omega^{2} = -1\), so this is \(1 + 1 + 1 = 3\). ✓
(b) \(2 + 5\omega + 2\omega^{2} = 2(1 + \omega + \omega^{2}) + 3\omega = 0 + 3\omega = 3\omega\). ✓
So \((2 + 5\omega + 2\omega^{2})^{3} = 27\omega^{3} = 27\). ✓

Calculation

[5 marks]

(a) Solve the equation \(z^{3} = 4\sqrt{2}(1 + i)\), giving your answers in the form \(re^{i\theta}\), where \(r \gt 0\) and \(-\pi \lt \theta \le \pi\).

(b) Write the root with the largest argument in the form \(a + bi\), and describe the shape formed by the three roots.

(a) \(|w| = 4\sqrt{2} \times \sqrt{2} = 8\) and \(\arg w = \tfrac{\pi}{4}\) (first quadrant), so \(w = 8e^{i(\pi/4 + 2\pi k)}\). ✓
\[z = 8^{1/3}e^{i(\pi/4 + 2\pi k)/3} = 2e^{i(\pi/12 + 2\pi k/3)}.\] ✓
\(k = 0, 1, 2\): arguments \(\tfrac{\pi}{12}, \tfrac{3\pi}{4}, \tfrac{17\pi}{12}\); the last becomes \(\tfrac{17\pi}{12} - 2\pi = -\tfrac{7\pi}{12}\). ✓
So \(z = 2e^{i\pi/12}\), \(2e^{3\pi i/4}\) or \(2e^{-7\pi i/12}\). ✓
(b) \(2e^{3\pi i/4} = 2\big(-\tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2}i\big) = -\sqrt{2} + \sqrt{2}\,i\). ✓
The roots have modulus 2 and are \(\tfrac{2\pi}{3}\) apart: an equilateral triangle inscribed in the circle \(|z| = 2\). ✓

Analysis

[6 marks]

By writing \(w = \dfrac{z + 1}{z}\), solve the equation \((z + 1)^{4} = 16z^{4}\), giving each root in the form \(a + bi\). Explain why the equation has exactly four roots.

\(z = 0\) is not a root (it gives \(1 = 0\)), so divide by \(z^{4}\): \(w^{4} = 16\). ✓
\(16 = 16e^{2\pi k i}\), so \(w = 2e^{\pi k i/2}\): \(w = 2, 2i, -2, -2i\). ✓
From \(wz = z + 1\), \(z = \dfrac{1}{w - 1}\). ✓
\(w = 2\): \(z = 1\). \(w = -2\): \(z = -\tfrac{1}{3}\). ✓
\(w = 2i\): \(z = \dfrac{1}{-1 + 2i} = \dfrac{-1 - 2i}{5}\). \(w = -2i\): \(z = \dfrac{1}{-1 - 2i} = \dfrac{-1 + 2i}{5}\). ✓
Expanding, \((z + 1)^{4} - 16z^{4} = -15z^{4} + 4z^{3} + \cdots\) is a quartic, so it has exactly four roots: \(1\), \(-\tfrac{1}{3}\), \(-\tfrac{1}{5} \pm \tfrac{2}{5}i\). ✓

Proof

[7 marks]

(a) Express \(z^{5} - 1\) as the product of one linear and two quadratic factors with real coefficients, in terms of \(\cos\tfrac{2\pi}{5}\) and \(\cos\tfrac{4\pi}{5}\).

(b) By comparing coefficients, show that \(\cos\tfrac{2\pi}{5} + \cos\tfrac{4\pi}{5} = -\tfrac{1}{2}\) and \(\cos\tfrac{2\pi}{5}\cos\tfrac{4\pi}{5} = -\tfrac{1}{4}\).

(a) The roots are \(1, e^{\pm 2\pi i/5}, e^{\pm 4\pi i/5}\), and \((z - e^{i\alpha})(z - e^{-i\alpha}) = z^{2} - 2\cos\alpha\,z + 1\), so \[z^{5} - 1 = (z - 1)\big(z^{2} - 2az + 1\big)\big(z^{2} - 2bz + 1\big),\] with \(a = \cos\tfrac{2\pi}{5}\), \(b = \cos\tfrac{4\pi}{5}\). ✓
(b) Dividing by \(z - 1\): \(z^{4} + z^{3} + z^{2} + z + 1 = (z^{2} - 2az + 1)(z^{2} - 2bz + 1)\). ✓
Expand the right side: \[z^{4} - 2(a + b)z^{3} + (2 + 4ab)z^{2} - 2(a + b)z + 1.\] ✓
\(z^{3}\): \(-2(a + b) = 1\), so \(a + b = -\tfrac{1}{2}\). \(z^{2}\): \(2 + 4ab = 1\), so \(ab = -\tfrac{1}{4}\). ✓
(c) \(a\) and \(b\) are the roots of \(t^{2} + \tfrac{1}{2}t - \tfrac{1}{4} = 0\), that is \(4t^{2} + 2t - 1 = 0\), so \(t = \dfrac{-2 \pm \sqrt{20}}{8} = \dfrac{-1 \pm \sqrt{5}}{4}\). ✓
\(\tfrac{2\pi}{5}\) is acute, so \(\cos\tfrac{2\pi}{5} \gt 0\) and takes the plus sign: \(\cos\tfrac{2\pi}{5} = \dfrac{\sqrt{5} - 1}{4}\) (about 0.309). The other root, \(\dfrac{-1 - \sqrt{5}}{4}\), is \(\cos\tfrac{4\pi}{5}\). ✓

Proof

[6 marks]

(a) Let \(\omega = e^{2\pi i/n}\), where \(n \ge 2\) is an integer. Prove that \(1 + \omega + \omega^{2} + \cdots + \omega^{n-1} = 0\).

(b) By taking \(n = 7\) and considering real parts, show that \(\cos\tfrac{2\pi}{7} + \cos\tfrac{4\pi}{7} + \cos\tfrac{6\pi}{7} = -\tfrac{1}{2}\).

(a) This is a geometric series with first term 1 and ratio \(\omega\). Since \(n \ge 2\), \(\omega \ne 1\), so the sum is \(\dfrac{\omega^{n} - 1}{\omega - 1}\). ✓
By de Moivre, \(\omega^{n} = e^{2\pi i} = 1\), so the sum is \(\dfrac{0}{\omega - 1} = 0\). ✓
(b) With \(n = 7\), the real part of \(\omega^{k}\) is \(\cos\tfrac{2\pi k}{7}\), so \(\displaystyle\sum_{k=0}^{6}\cos\tfrac{2\pi k}{7} = 0\). ✓
\(\cos\tfrac{2\pi(7 - k)}{7} = \cos\big(2\pi - \tfrac{2\pi k}{7}\big) = \cos\tfrac{2\pi k}{7}\), so the terms for \(k = 4, 5, 6\) repeat those for \(k = 3, 2, 1\). ✓
Hence \(1 + 2\Big(\cos\tfrac{2\pi}{7} + \cos\tfrac{4\pi}{7} + \cos\tfrac{6\pi}{7}\Big) = 0\). ✓
So \(\cos\tfrac{2\pi}{7} + \cos\tfrac{4\pi}{7} + \cos\tfrac{6\pi}{7} = -\tfrac{1}{2}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.