Complex numbers and polynomials - University College Cork

To utilize this, replace x by ρ cos(θ) and multiply through by 4 to getρ 3 cos(3θ) + ρ(3ρ 2 + 4q) cos θ + 4r = 0.Now choose ρ so that 3ρ 2 + 4q = 0. Thencos 3θ = −4rρ 3 .Now solve this for θ! This is the strategy. When does it work? For the lastequation to be solvable for real θ, we require ρ to be real and −1 ≤ −4r ≤ρ 31. Hence, we require q ≤ 0 and −1 ≤ −4r ≤ 1 to hold, i.e., we requireρ 327r 2 + 4q 3 ≤ 0, the same condition we encountered before that is necessaryand sufficient for the roots to be real. (NB. The method can also be used evenif these conditions aren’t fulfilled, by allowing θ to be complex and using thefact that cos z = (e iz + e −iz )/2, which is valid for complex z. In particular,cos(iθ) = (e θ + e −θ )/2 = cosh θ, the hyperbolic equivalent of cosine.)9 Outline solutions to some problems on polys1. (IMO 1973) Find the minimum value of a 2 + b 2 , where a, b are real numbersfor which the equationhas at least one real root.x 4 + ax 3 + bx 2 + ax + 1 = 0Solution. Denote by Γ the set of points (x, y) in the plane for which there isa real number w such thatSuppose (x, y) ∈ Γ. Lettingw 4 + xw 3 + yw 2 + xw + 1 = 0.t = w + 1 w , t2 = w 2 + 1 w 2 + 2 ≥ 4,we see that there is a real number t, with |t| ≥ 2, such thatThe reality of t requires thatt 2 + xt + y − 2 = 0.x 2 − 4(y − 2) ≥ 0, ∀(x, y) ∈ Γ.Thus,Γ ⊂ {(x, y) : 4(y − 2) ≤ x 2 }.29