A-Level Complex numbers in cartesian form and the Argand diagram: worked solution

Question

(a) The equation -i x^3 + 5i x^2 + a x + b = 0, with a and b purely imaginary, has roots 2 + i, 2 - i and 1. Explain why the complex roots form a conjugate pair. (b) Using part (a) and a suitable substitution, find the roots of i x^3 + 5i x^2 + a* x + b = 0, where a* is the conjugate of a.

(a) The conjugate root theorem applies only to polynomials whose coefficients are all real, so the aim is to show this equation is equivalent to one with real coefficients. Write the purely imaginary constants explicitly: a = i(alpha) and b = i(beta), with alpha and beta real. Multiply the whole equation by -i (this is allowed and changes no roots, since -i is not zero): (-i)(-i x³) = (-i)² x³ = (i²) x³ = -x³, a real coefficient (-i)(5i x²) = -5 i² x² = 5x², a real coefficient (-i)(a x) = (-i)(i alpha) x = (-i²) alpha x = alpha x, a real coefficient (-i)(b) = (-i)(i beta) = beta, real So the equation is equivalent to -x³ + 5x² + alpha x + beta = 0, i.e. x³ - 5x² - alpha x - beta = 0, whose coefficients are all real. For a polynomial with real coefficients, taking the conjugate of the whole equation leaves the coefficients unchanged, so if z is a root then conj(z) is also a root. Hence the two non-real roots must occur as a conjugate pair, which is why 2 + i and 2 - i appear together (the remaining root, 1, is then forced to be real). (b) The new equation is i x³ + 5i x² + a* x + b = 0. Since a is purely imaginary, a = i(alpha) gives a* = -i(alpha) = -a. Note the constant term b is unchanged. Compare the two cubics: the x³ coefficient has changed sign and the x coefficient has changed sign, while the x² and constant terms have not. Changing the sign of the odd-power terms is exactly what the substitution x = -w does, so try it. Put x = -w: i(-w)³ + 5i(-w)² + a*(-w) + b = 0 Use (-w)³ = -w³ and (-w)² = w²: -i w³ + 5i w² - a* w + b = 0 and -a* = a (from above), so this is -i w³ + 5i w² + a w + b = 0, which is precisely the equation of part (a). Therefore w takes the roots found in part (a): w = 2 + i, w = 2 - i, w = 1. Since x = -w, the roots of the new equation are x = -(2 + i) = -2 - i, x = -(2 - i) = -2 + i, x = -1.