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2015 Paper 3 Q4
D: 1700.0 B: 1516.0
complex numbers roots of polynomials symmetric functions Newton's sums cubic equation complex conjugate roots trigonometric proof

  1. If \(a\), \(b\) and \(c\) are all real, show that the equation \[ z^3+az^2+bz+c=0 \tag{\(*\)} \] has at least one real root.
  2. Let \[ S_1= z_1+z_2+z_3, \ \ \ \ S_2= z_1^2 + z_2^2 + z_3^2, \ \ \ \ S_3= z_1^3 + z_2^3 + z_3^3\,, \] where \(z_1\), \(z_2\) and \(z_3\) are the roots of the equation \((*)\). Express \(a\) and \(b\) in terms of \(S_1\) and \(S_2\), and show that \[ 6c =- S_1^3 + 3 S_1S_2 - 2S_3\,. \]
  3. The six real numbers \(r_k\) and \(\theta_k\) (\(k=1, \ 2, \ 3\)), where \(r_k>0\) and \(-\pi < \theta_k <\pi\), satisfy \[ \textstyle \sum\limits _{k=1}^3 r_k \sin (\theta_k) = 0\,, \ \ \ \ \textstyle \sum\limits _{k=1}^3 r_k^2 \sin (2\theta_k) = 0\,, \ \ \ \ \ \textstyle \sum\limits _{k=1}^3 r_k^3 \sin (3\theta_k) = 0\, . \] Show that \(\theta_k=0\) for at least one value of \(k\). Show further that if \(\theta_1=0\) then \(\theta_2 = - \theta_3\,\).

Solution:

  1. Let \(z \in \mathbb{R}\) and let \(z \to \pm \infty\) then \(z^3 + az^2 + bz + c\) changes sign, therefore somewhere it must have a real root.
  2. \begin{align*} &&z^3 + az^2 + bz + c &= (z-z_1)(z-z_2)(z-z_3) \\ && &= z^3 - (z_1+z_2+z_3)z^2 + (z_1z_2 + z_2z_3+z_3z_1)z - (z_1z_2z_3) \\ \\ \Rightarrow && S_1 &= z_1+z_2+z_3 \\ &&&= -a \\ \\ \Rightarrow && S_2 &= z_1^2+z_2^2+z_3^2 \\ &&&= (z_1+z_2+z_3)^2 - 2(z_1z_2 + z_2z_3+z_3z_1) \\ &&&= a^2 - 2b \\ \Rightarrow && a &= -S_1 \\ && b &= \frac12 \l S_1^2 - S_2\r \\ \\ && 0 &= z_i^3 + az_i^2+bz_i+c \\ \Rightarrow && 0 &= S_3 + aS_2+bS_1+3c \\ &&&= S_3 -S_1S_2 + \frac12 \l S_1^2 - S_2\r S_1 + 3c \\ \Rightarrow && 0 &= 2S_3 - 3S_1S_2 + S_1^3 + 6c \end{align*}
  3. Let \(z_k= r_ke^{i \theta_k}\), then we have \(\textrm{Im}(S_k) = 0\) and so the polynomial with roots \(z_k\) has real coefficients, and therefore at least one root is real. This root will have \(\theta_k = 0\). Moreover, since if \(w\) is a root of a real polynomial \(\overbar{w}\) is also a root, and therefore if \(\theta_1 = 0\), we must have that \(z_2\) and \(z_3\) are complex conjugate, ie \(\theta_2 = - \theta_3\)

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