Fifth Roots of Unity and Quadratic Equations

This problem explores the elegant connection between complex roots of unity and quadratic equations, a classic topic in the HSC Mathematics Extension 2 syllabus. By pairing roots with their conjugates, we transform a fifth‑degree complex polynomial into a real quadratic, ultimately yielding an exact value for $\cos(2\pi/5)$. Working through this will strengthen your skills in de Moivre’s theorem, Vieta’s formulas, and the geometry of complex numbers.

Problem Statement

a) Find the fifth roots of unity in polar form. b) Let $\alpha$ be the complex fifth root of unity with the smallest positive argument, and suppose that $u = \alpha + \alpha^4$ and $v = \alpha^2 + \alpha^3$. c) Show that $u$ and $v$ are the roots of the quadratic equation $x^2 + x - 1 = 0$. d) Show that $u - v = \sqrt{5}$. e) Use the values of $u + v$ and $u - v$ to deduce that $\cos\frac{2\pi}{5} = \frac{1}{4}(\sqrt{5} - 1)$.

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Hints

• For part (a), solve the equation $z^5 = 1$ using de Moivre's theorem.
• For part (c), find the sum $u+v$ and product $uv$, keeping in mind that the sum of the roots of $z^5 - 1 = 0$ is $0$, meaning $1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 = 0$.
• For part (d), express $(u-v)^2$ in terms of $(u+v)$ and $uv$. Then determine the sign of $u-v$ by converting the roots back to trigonometric form.

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Solutions

• Part (a): We apply de Moivre’s theorem to find all solutions of $z^5 = 1$. The solutions are given by $z_k = \text{cis}\left(\frac{2k\pi}{5}\right)$ for $k = -2, -1, 0, 1, 2$. The fifth roots of unity are: $1$, $\text{cis}\left(\frac{2\pi}{5}\right)$, $\text{cis}\left(\frac{4\pi}{5}\right)$, $\text{cis}\left(-\frac{2\pi}{5}\right)$, and $\text{cis}\left(-\frac{4\pi}{5}\right)$.

• Part (b): The complex fifth root of unity with the smallest positive argument is $\alpha = \text{cis}\left(\frac{2\pi}{5}\right)$. Because $\alpha^5=1$, $\alpha^4 = \overline{\alpha}$, so $u$ and $v$ each pair a root with its conjugate, guaranteeing real values. We are given $u = \alpha + \alpha^4$ and $v = \alpha^2 + \alpha^3$. Notice that $\alpha^4 = \text{cis}\left(\frac{8\pi}{5}\right) = \text{cis}\left(-\frac{2\pi}{5}\right) = \bar{\alpha}$. Similarly, $\alpha^3 = \text{cis}\left(\frac{6\pi}{5}\right) = \text{cis}\left(-\frac{4\pi}{5}\right) = \bar{\alpha^2}$.

• Part (c): To show $u$ and $v$ are roots of $x^2 + x - 1 = 0$, we find their sum and product and then use Vieta’s formulas. Sum: $u + v = \alpha + \alpha^4 + \alpha^2 + \alpha^3$. Since $1, \alpha, \alpha^2, \alpha^3, \alpha^4$ are the roots of $z^5 - 1 = 0$, their sum is $0$.

  $$1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 = 0 \implies u + v = -1$$

  Product: $uv = (\alpha + \alpha^4)(\alpha^2 + \alpha^3) = \alpha^3 + \alpha^4 + \alpha^6 + \alpha^7$. Since $\alpha^5 = 1$, we can simplify $\alpha^6 = \alpha$ and $\alpha^7 = \alpha^2$.

  $$uv = \alpha^3 + \alpha^4 + \alpha + \alpha^2 = -1$$

  A quadratic equation with roots $u$ and $v$ is given by $x^2 - (u+v)x + uv = 0$.

  $$x^2 - (-1)x + (-1) = 0 \implies x^2 + x - 1 = 0$$

• Part (d): Rather than expanding directly, we note that $(u-v)^2$ can be expressed entirely in terms of the known sum and product. We can evaluate $(u - v)^2$ using the sum and product:

  $$(u - v)^2 = (u + v)^2 - 4uv = (-1)^2 - 4(-1) = 1 + 4 = 5$$

  Therefore, $u - v = \pm\sqrt{5}$. We must determine the correct sign by looking at the trigonometric forms.

  $$u = \alpha + \bar{\alpha} = 2\cos\left(\frac{2\pi}{5}\right)$$

  $$v = \alpha^2 + \bar{\alpha^2} = 2\cos\left(\frac{4\pi}{5}\right)$$

  Since $\frac{2\pi}{5} < \frac{\pi}{2}$, $\cos\left(\frac{2\pi}{5}\right) > 0$. Since $\frac{4\pi}{5} > \frac{\pi}{2}$, $\cos\left(\frac{4\pi}{5}\right) < 0$. This means $u > 0$ and $v < 0$, so $u - v > 0$. Thus, $u - v = \sqrt{5}$.

• Part (e): With both the sum and difference known, we solve a simple linear system for $u$ and then equate it to its trigonometric expression. We have a system of linear equations:

  1. $u + v = -1$
  2. $u - v = \sqrt{5}$ Adding these equations gives:

  $$2u = \sqrt{5} - 1 \implies u = \frac{\sqrt{5} - 1}{2}$$

  From part (d), we know that $u = 2\cos\left(\frac{2\pi}{5}\right)$.

  $$2\cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} - 1}{2} \implies \cos\left(\frac{2\pi}{5}\right) = \frac{1}{4}(\sqrt{5} - 1)$$

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Takeaways

• Grouping the roots of unity into sums like $\alpha + \alpha^{-1}$ efficiently transforms complex polynomials into real ones.
• The sum of all $n$-th roots of unity is always $0$. This property is incredibly useful for evaluating sums of complex numbers.
• By leveraging the sum and product of specifically chosen grouped roots, we can construct lower-degree polynomial equations to find exact trigonometric values.

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Further Readings

HSC Complex Numbers, HSC Polynomials