Year: 2015
Paper: 3
Question Number: 6
Course: UFM Pure
Section: Roots of polynomials
A very similar number of candidates to 2014 once again ensured that all questions received a decent number of attempts, with seven questions being very popular rather than five being so in 2014, but the most popular questions were attempted by percentages in the 80s rather than 90s. All but one question was answered perfectly at least once, the one exception receiving a number of very close to perfect solutions. About 70% attempted at least six questions, and in those cases where more than six were attempted, the extra attempts were usually fairly superficial.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1553.5
Banger Comparisons: 6
\begin{questionparts}
\item Let $w$ and $z$ be complex numbers, and let $u= w+z$ and $v=w^2+z^2$. Prove that $w$ and $z$ are real if and only if $u$ and $v$ are real and $u^2\le2v$.
\item The complex numbers $u$, $w$ and $z$ satisfy the equations
\begin{align*}
w+z-u&=0 \\
w^2+z^2 -u^2 &= - \tfrac 23 \\
w^3+z^3 -\lambda u &= -\lambda\,
\end{align*}
where $\lambda $ is a positive real number. Show that for all values of $\lambda$ except one (which you should find) there are three possible values of $u$, all real.
Are $w$ and $z$ necessarily real? Give a proof or counterexample.
\end{questionparts}\begin{questionparts}
\item Notice that $u^2 = v+2wz$, so $w,z$ are roots of the quadratic $t^2 -ut+\frac{u^2-v}{2}$. Therefore they are both real if $u^2 \geq 2(u^2-v) \Rightarrow 2v \geq u^2$.
\item \begin{align*}
&& w+z &= u \\
&& w^2+z^2 &= u^2 - \tfrac23 \\
&& w^3+z^3 &= \lambda(u-1) \\
\\
&& wz &= \frac{u^2 - (u^2-\tfrac23)}{2} = \tfrac13\\
\\
&& (w+z)(w^2+z^2) &= w^3+z^3+wz(w+z) \\
&&u(u^2-\tfrac23)&= \lambda(u-1)+\frac13u \\
\Rightarrow && u^3-u&= \lambda (u-1) \\
\Rightarrow && 0 &= (u-1)(u(u+1) - \lambda) \\
\Rightarrow && 0 &= (u-1)(u^2+u - \lambda)
\end{align*}
Therefore there will be at most 3 values for $u$, unless $1$ is a root of $u^2+u-\lambda$, ie $\lambda = 2$.
Suppose $u = 1$, then we have:
$w+z = 1, wz = 1/3 \Rightarrow w,z = \frac{-1 \pm \sqrt{-1/3}}{2}$ which are clearly complex.
\end{questionparts}
About three fifths of the candidates attempted this question but without great success. The first part tripped up many through needing to prove 'if and only if'. The first part of (ii) yielded good scoring opportunities for those that did make progress on this question, though some fell by the wayside when it came to the situation that would not generate three possible values. Some attempts at the last result failed as the counter-example was not always shown to be a counter-example.