Year: 2025
Paper: 2
Question Number: 2
Course: LFM Stats And Pure
Section: Complex Numbers (L8th)
As is commonly the case, the vast majority of candidates focused on the Pure questions in Section A of the paper, with a good number of attempts made on all of those questions. Candidates that attempted the Mechanics questions in Section B generally answered both questions. More candidates attempted Question 11 in Section C than either Mechanics question, but very few attempted Question 12 in that section. There were a large number of good responses seen for all the questions, but a significant number of responses lacked sufficient detail in the presentation, particularly when asked to prove a given result or provide an explanation. Candidates who did well on this paper generally: gave careful explanations of each step within their solutions; indicated all points of interest on graphs and other diagrams clearly; made clear comments about the approach that needed to be taken, particularly when having to explore a number of cases as part of the solution to a question; used mathematical terminology accurately within their solutions. Candidates who did less well on this paper generally: made errors with basic algebraic manipulation, such as incorrect processing of indices; produced sketches of graphs in which significant points were difficult to see clearly because of the chosen scale; skipped important lines within lengthy sections of algebraic reasoning.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item
\begin{enumerate}
\item Show that if the complex number $z$ satisfies the equation
\[z^2 + |z + b| = a,\]
where $a$ and $b$ are real numbers, then $z$ must be either purely real or purely imaginary.
\item Show that the equation
\[z^2 + \left|z + \frac{5}{2}\right| = \frac{7}{2}\]
has no purely imaginary roots.
\item Show that the equation
\[z^2 + \left|z + \frac{7}{2}\right| = \frac{5}{2}\]
has no purely real roots.
\item Show that, when $\frac{1}{2} < b < \frac{3}{4}$, the equation
\[z^2 + |z + b| = \frac{1}{2}\]
will have at least one purely imaginary root and at least one purely real root.
\end{enumerate}
\item Solve the equation
\[z^3 + |z + 2|^2 = 4.\]
\end{questionparts}\begin{questionparts}
\item
\begin{enumerate}
\item Suppose $z^2 + |z + b| = a$, then $z^2 = a- |z + b| \in \mathbb{R}$, since $a \in \mathbb{R}$. Since the square root of a real number is either purely real or purely imaginary, $z$ is purely real or purely imaginary.
\item Suppose $z = it$ for some $t \in \mathbb{R}$, then
\begin{align*}
&& \frac72 &= -t^2 +\sqrt{t^2 + \frac{25}{4}} \\
\Rightarrow && \left ( \frac72 + t^2\right)^2 &= t^2 + \frac{25}{4} \\
\Rightarrow && t^4 + 7t^2 + \frac{49}{4} &= t^2 + \frac{25}{4} \\
\Rightarrow && t^4 + 6t^2 + 6 &= 0
\end{align*}
but since $\Delta = 6^2 - 4 \cdot 1 \cdot 6 < 0$ there are no real solutions.
\item Suppose $z = t$ for some $t \in \mathbb{R}$, then either $t^2 + t + \frac72 = \frac52 \Rightarrow t^2 + t + 1 = 0$ (no solutions) or $t^2 - t - \frac72 = \frac52 \Rightarrow t^2 - t - 6 = (t-3)(t+2) = 0$. When $t = 3$ then we must take the positive part for $|z + \frac72|$ so this cannot work. When $t = -2$ we also have $\frac72-2 > 0$ so we are still taking the positive part. Hence no solutions
\item Suppose $\frac{1}{2} < b < \frac{3}{4}$, the equation then consider $z^2 + |z + b| = \frac{1}{2}$.
Case 1: $z = t \in \mathbb{R}$, then we have two cases:
Case 1a: $z+b > 0$.
$z^2 + z + b = \frac12 \Rightarrow z = \frac{-1 \pm \sqrt{1-4b+2}}{2}$ which clearly is a valid real number an $z + b > 0$.
Case 1b: $z+b < 0$
$z^2 - z - b = \frac12 \Rightarrow z = \frac{1 \pm \sqrt{1+4b-2}}{2}$
\end{enumerate}
\item Let $\omega$ be a (primitive) cube root of unity. \begin{align*}
&& z^3 &= 4 - |z+2|^2 \\
\Rightarrow && z &\in \mathbb{R} \cup \omega \mathbb{R} \cup \omega^2 \mathbb{R}
\end{align*}
Case 1:
\end{questionparts}
This was a popular question and there was a wide variety in the quality of responses seen, with a small proportion of candidates producing perfect, or close to perfect, solutions. Part (i) (a) was generally completed well, with most candidates explaining the reasoning carefully in their responses. Part (i) (b) was also completed well by many candidates, with numerical errors being the main area where marks were lost. In part (i) (c), however, a large number of candidates did not realise there were two cases to be considered and instead only analysed one case. In part (i) (d) there was a lot of variation in the quality of responses seen. Those who had successfully completed part (i) (b) were often able to demonstrate that there was a purely imaginary root. Those who had successfully completed part (i) (c) were often able to make good progress in showing that there must be a purely real root, although many stopped at the point of showing that the discriminant was positive and did not show that at least one of the roots of the quadratic equation lay within the appropriate range. As with part (i) (c), there were a large number of candidates who did not recognise that there were two cases to be explored in this part of the question. Part (ii) was completed well by many candidates, including those who had lost marks in previous parts of the question. Several candidates lost marks through numerical errors or by not explaining clearly enough that all of the solutions had been found in each case.