Roots of Unity and Trigonometric Deductions

This problem combines the algebra of complex roots of unity with trigonometric identities, a classic Extension 2 topic. By solving $z^5 + 1 = 0$ and factorising the polynomial, you’ll practise writing complex roots in polar form, pairing conjugates to obtain real quadratic factors, and comparing coefficients to deduce exact values for $\cos\frac{\pi}{5}$ and $\cos\frac{2\pi}{5}$.

Problem Statement

a) Solve the equation $z^5 + 1 = 0$, expressing the roots in mod-arg form. b) Hence show that $z^5 + 1 = (z + 1)\left(z^2 - \left(2\cos\frac{\pi}{5}\right)z + 1\right)\left(z^2 + \left(2\cos\frac{2\pi}{5}\right)z + 1\right)$. c) Deduce that $2\cos\frac{2\pi}{5} - 2\cos\frac{\pi}{5} + 1 = 0$, and hence find the exact values of $\cos\frac{\pi}{5}$ and $\cos\frac{2\pi}{5}$.

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Hints

• For part (a), rewrite the equation as $z^5 = -1$ and express $-1$ in polar form as $\text{cis}(\pi)$.
• For part (b), group the conjugate pairs of roots from part (a) to form real quadratic factors using the property $(z - z_k)(z - \overline{z_k}) = z^2 - 2\text{Re}(z_k)z + |z_k|^2$.
• For part (c), equate coefficients by comparing the coefficient of $z^3$ after expanding the identity from part (b). Then use the double angle identity $\cos\frac{2\pi}{5} = 2\cos^2\frac{\pi}{5} - 1$ to form a quadratic equation.

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Solutions

• Part (a): We need to solve $z^5 = -1$. In polar form, $-1 = \text{cis}(\pi)$. Using the general formula for $n$th roots, the five roots are obtained by letting $k$ range over five consecutive integers, such as $k = -2, -1, 0, 1, 2$. The roots are given by $z_k = \text{cis}\left(\frac{\pi + 2k\pi}{5}\right)$ for $k = -2, -1, 0, 1, 2$. The five roots are:

  $$z_0 = \text{cis}\left(\frac{\pi}{5}\right)$$

  $$z_1 = \text{cis}\left(\frac{3\pi}{5}\right)$$

  $$z_2 = \text{cis}(\pi) = -1$$

  $$z_{-1} = \text{cis}\left(-\frac{\pi}{5}\right)$$

  $$z_{-2} = \text{cis}\left(-\frac{3\pi}{5}\right)$$

• Part (b): By the Factor Theorem, we can write $z^5 + 1$ as the product of linear factors corresponding to its roots:

  $$z^5 + 1 = (z - z_2)(z - z_0)(z - z_{-1})(z - z_1)(z - z_{-2})$$

  Since the polynomial has real coefficients, its non-real roots occur in conjugate pairs. Pairing each conjugate pair and multiplying the corresponding linear factors yields a quadratic with real coefficients:

  $$(z - z_0)(z - \overline{z_0}) = z^2 - (z_0 + \overline{z_0})z + z_0\overline{z_0} = z^2 - 2\cos\left(\frac{\pi}{5}\right)z + 1$$

  $$(z - z_1)(z - \overline{z_1}) = z^2 - (z_1 + \overline{z_1})z + z_1\overline{z_1} = z^2 - 2\cos\left(\frac{3\pi}{5}\right)z + 1$$

  Using the identity $\cos\left(\frac{3\pi}{5}\right) = -\cos\left(\frac{2\pi}{5}\right)$ (because $\frac{3\pi}{5} = \pi - \frac{2\pi}{5}$), this second quadratic factor can be written as:

  $$z^2 + 2\cos\left(\frac{2\pi}{5}\right)z + 1$$

  Therefore, multiplying these factors together with the real root factor $(z+1)$:

  $$z^5 + 1 = (z + 1)\left(z^2 - \left(2\cos\frac{\pi}{5}\right)z + 1\right)\left(z^2 + \left(2\cos\frac{2\pi}{5}\right)z + 1\right)$$

• Part (c): We know the algebraic identity for the sum of fifth powers:

  $$z^5 + 1 = (z + 1)(z^4 - z^3 + z^2 - z + 1)$$

  Since both expressions represent the same complete factorisation of $z^5+1$, the quartic factor from this identity must equal the product of the two quadratics from part (b):

  $$z^4 - z^3 + z^2 - z + 1 = \left(z^2 - \left(2\cos\frac{\pi}{5}\right)z + 1\right)\left(z^2 + \left(2\cos\frac{2\pi}{5}\right)z + 1\right)$$

  Expanding the right side and comparing the coefficients of $z^3$ gives us:

  $$-1 = 2\cos\frac{2\pi}{5} - 2\cos\frac{\pi}{5}$$

  $$\therefore 2\cos\frac{2\pi}{5} - 2\cos\frac{\pi}{5} + 1 = 0$$

  Now, let $x = \cos\frac{\pi}{5}$. Using the double angle formula, $\cos\frac{2\pi}{5} = 2x^2 - 1$. Substitute this into the deduced equation:

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

  $$4x^2 - 2x - 1 = 0$$

  Solving this quadratic for $x$:

  $$x = \frac{2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{2 \pm \sqrt{20}}{8} = \frac{1 \pm \sqrt{5}}{4}$$

  Since $\frac{\pi}{5}$ is in the first quadrant, $\cos\frac{\pi}{5} > 0$. Thus, $\cos\frac{\pi}{5} = \frac{1 + \sqrt{5}}{4}$. Finally, we can find $\cos\frac{2\pi}{5}$:

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

  $$\therefore \cos\frac{2\pi}{5} = \frac{\sqrt{5} - 1}{4}$$

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Takeaways

• Roots of unity are powerfully connected to the exact values of trigonometric functions for fractional multiples of $\pi$.
• Forming real quadratic factors by grouping conjugate pairs is a standard technique for breaking down polynomials with real coefficients.
• Comparing coefficients between two different factorised forms of the same polynomial is a reliable way to deduce relationships between their terms.

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Further Readings

HSC Complex Numbers, HSC Polynomials