Year: 2015
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Complex numbers 2
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: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item 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.
\item 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
\[
6c =- S_1^3 + 3 S_1S_2 - 2S_3\,.
\]
\item 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 $\theta_2 = - \theta_3\,$.
\end{questionparts}\begin{questionparts}
\item 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.
\item
\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*}
\item 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 $\theta_2 = - \theta_3$
\end{questionparts}
Along with questions 5 and 7, attempted by just over three quarters, this was the third most popular question, though a little less successful than the most popular question 1. The first part was frequently not well attempted, but the second part was usually mastered. Attempts at the third part suffered from arguments with poor logical structure, though many did not get a start on this part.