Conjugate roots and x⁴ + 4
Tonight's idea
The conjugate of \(z = a + bi\) is \(\bar z = a - bi\), its mirror image in the real axis. Two facts make it useful. Conjugation respects addition and multiplication: \(\overline{z + w} = \bar z + \bar w\) and \(\overline{zw} = \bar z\,\bar w\) (multiply out to check). And a real number is its own conjugate.
Take a polynomial \(p(x) = a_n x^n + \dots + a_1 x + a_0\) with real coefficients and conjugate \(p(z)\). Each power \(z^k\) becomes \(\bar z^k\), and each coefficient stays put. So \(\overline{p(z)} = p(\bar z)\). If \(p(z) = 0\), then \(p(\bar z) = \bar 0 = 0\).
Non-real roots of a real polynomial come in conjugate pairs, and each pair gives a real quadratic factor: \((x - z)(x - \bar z) = x^2 - 2\,\mathrm{Re}(z)\,x + \lvert z\rvert^2\).
The quadratic is real because \(z + \bar z = 2a\) and \(z\bar z = a^2 + b^2\) are real.
The fundamental theorem of algebra says every polynomial of degree \(n \geq 1\) has exactly \(n\) complex roots, counted with multiplicity. You met it for \(x^n = 1\) in grade 10; the general argument is on sheet b17. Together with the pairing: every real polynomial factors over the reals into linear factors (one per real root) and quadratics (one per conjugate pair). In particular, a real polynomial of odd degree has at least one real root.
Watch one: \(x^4 + 4\)
The roots solve \(x^4 = -4\). In polar form \(-4\) has modulus 4 and angle \(180°\), so a fourth root has modulus \(\sqrt[4]{4} = \sqrt2\) and angle \(45°\), or \(45°\) plus a multiple of \(90°\). The four roots are \(1 + i\), \(-1 + i\), \(-1 - i\), \(1 - i\) (figure). Pair each with its conjugate:
\[(x - (1+i))(x - (1-i)) = x^2 - 2x + 2, \qquad (x - (-1+i))(x - (-1-i)) = x^2 + 2x + 2.\]
So \(x^4 + 4 = (x^2 - 2x + 2)(x^2 + 2x + 2)\). Check by writing \(x^4 + 4 = (x^2 + 2)^2 - (2x)^2\), a difference of squares: that is the same factorisation, reached without seeing a single root.
Drag \(z\) and watch \(iz\), \(z^2\) and \(-z\) move: mathdetours.com/demos/complex.html.
Your turn
- Solve \(x^2 - 6x + 13 = 0\). Check the roots are conjugates, and that their sum and product are the \(6\) and \(13\) Vieta predicts.
- (a) Find a monic real quadratic with \(1 - 3i\) as a root. (b) Find a monic real cubic with roots \(1 - 3i\) and \(2\).
- \(2 + i\) is a root of \(x^3 - x^2 - 7x + 15\). Find the other two roots. (Build the quadratic factor first, then divide.)
- Plot the six roots of \(x^6 + 1\) on the unit circle. Pair them off and write \(x^6 + 1\) as a product of three real quadratics. Multiply back to check.
Stretch: a prime that never is
Show that \(n^4 + 4\) is never prime for an integer \(n > 1\). (Tonight's factorisation works for integers too.) Why does the argument fail at \(n = 1\)?
Answers
- \(x = 3 \pm 2i\), conjugates. Sum \(6\), product \(9 - (2i)^2 = 9 + 4 = 13\).
- (a) The partner root is \(1 + 3i\): \(x^2 - 2x + 10\). (b) \((x^2 - 2x + 10)(x - 2) = x^3 - 4x^2 + 14x - 20\).
- \(2 - i\) is also a root, so \((x - 2 - i)(x - 2 + i) = x^2 - 4x + 5\) is a factor. Dividing, \(x^3 - x^2 - 7x + 15 = (x^2 - 4x + 5)(x + 3)\). The roots are \(2 + i\), \(2 - i\), \(-3\).
- \(x^6 = -1\) has roots of modulus 1 at \(30°, 90°, 150°, 210°, 270°, 330°\). The pair at \(\pm 90°\) is \(\pm i\), giving \(x^2 + 1\). The pair at \(\pm 30°\) has real part \(\cos 30° = \tfrac{\sqrt3}{2}\) and modulus 1, giving \(x^2 - \sqrt3\,x + 1\); the pair at \(\pm 150°\) gives \(x^2 + \sqrt3\,x + 1\). So \(x^6 + 1 = (x^2 + 1)(x^2 - \sqrt3\,x + 1)(x^2 + \sqrt3\,x + 1)\). The last two multiply to \((x^2 + 1)^2 - 3x^2 = x^4 - x^2 + 1\), and \((x^2 + 1)(x^4 - x^2 + 1) = x^6 + 1\).
Stretch. \(n^4 + 4 = (n^2 - 2n + 2)(n^2 + 2n + 2) = \big((n-1)^2 + 1\big)\big((n+1)^2 + 1\big)\). For \(n \geq 2\) the smaller factor is at least \(2\) and the larger is bigger still, so \(n^4 + 4\) is a product of two integers greater than 1: composite. For example \(3^4 + 4 = 85 = 5 \times 17\) and \(5^4 + 4 = 629 = 17 \times 37\). At \(n = 1\) the smaller factor is \((1-1)^2 + 1 = 1\), which proves nothing, and indeed \(1 + 4 = 5\) is prime.