Year: 2025
Paper: 3
Question Number: 8
Course: UFM Pure
Section: Complex numbers 2
The majority of candidates focused solely on the pure questions, with questions 1, 2 and 8 the most popular. The statistics questions were more popular than the mechanics questions in this exam series. Candidates who did well on this paper generally: were careful to explain and justify the steps in their arguments, explaining what they had done rather than expecting the examiner to infer what had been done from disjointed groups of calculations; paid close attention to what was required by the questions; made fewer unnecessary mistakes with calculations; thought carefully about how to present rigorous arguments involving trig functions and their inverse functions, especially in relation to domain considerations; understood that questions set on the STEP papers require sufficient justification to earn full credit; knew the difference between 'positive' and 'non-negative'; attempted all parts of a question, picking up marks for later parts even when they had not necessarily attempted or completed previous parts. Candidates who did less well on this paper generally: did not pay attention to 'Hence' instructions: this means that you must use the previous part; presented explanations that were not precise enough (e.g. in Question 3 describing the transformations but not in the context of the graphs or in Question 8 not explaining use of trigonometric relationships sufficiently well); made additional assumptions, e.g. that a function was differentiable when this had not been given; tried to present if and only if arguments in a single argument when dealing with each direction separately would have been more appropriate and safer (note that this is not always the case; in general candidates need to consider what is the most appropriate presentation of an if and only if argument); tried to carry out too many steps in one go, resulting in them not justifying the key steps sufficiently; did not take sufficient care with graphs/curve sketching.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Show that
$$z^{m+1} - \frac{1}{z^{m+1}} = \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)$$
Hence prove by induction that, for $n \geq 1$,
$$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$
Find similarly $z^{2n} - \frac{1}{z^{2n}}$ as a product of $(z + \frac{1}{z})$ and a sum.
\item \begin{enumerate}
\item By choosing $z = e^{i\theta}$, show that
$$\sin 2n\theta = 2\sin\theta \sum_{r=1}^n \cos(2r-1)\theta$$
\item Use this result, with $n = 2$, to show that $\cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - \frac{1}{2}$.
\item Use this result, with $n = 7$, to show that $\cos\frac{2\pi}{15} + \cos\frac{4\pi}{15} + \cos\frac{8\pi}{15} + \cos\frac{16\pi}{15} = \frac{1}{2}$.
\end{enumerate}
\item Show that $\sin\frac{\pi}{14} - \sin\frac{3\pi}{14} + \sin\frac{5\pi}{14} = \frac{1}{2}$.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
RHS &= \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right) \\
&= z^{m+1} + \frac{1}{z^{m-1}} - z^{m-1} - \frac{1}{z^{m+1}} + z^{m-1} - \frac{1}{z^{m-1}} \\
&= z^{m+1} - \frac{1}{z^{m+1}} \\
&= LHS
\end{align*}.
Claim: For $n \geq 1$,
$$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$
Proof: (By Induction)
Base Case: ($n=1$).
\begin{align*}
LHS &= z^2 - \frac{1}{z^2} \\
&= (z-\frac1z)(z + \frac{1}{z}) \\
&= (z - \frac1z) \sum_{r=1}^1 \left ( z + \frac{1}{z} \right) \\
&= (z - \frac1z) \sum_{r=1}^1 \left ( z^{2r-1} + \frac{1}{z^{2r-1}} \right) \\
&= RHS
\end{align*}
as required.
Inductive step: Suppose our result is true for some $n=k$, then consider $n = k+1$.
\begin{align*}
RHS &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k+1} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) \\
&= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\
&= z^{2k} - \frac{1}{z^{2k}} + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\
&= z^{2k+2} - \frac{1}{z^{2k+2}} \\
&= LHS
\end{align*}.
Therefore if our result is true for $n=k$ is true, it is true for $n=k+1$. Since it is also true for $n=1$ it is true for all $n \geq 1$ but the principle of mathematical induction.
Since $\displaystyle z^{m+1} - \frac{1}{z^{m+1}} = \left(z + \frac{1}{z}\right)\left(z^m - \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)$, we must have $\displaystyle z^{2n}-\frac{1}{z^{2n}} = \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right)$
\item \begin{enumerate}
\item Since $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$
we have
\begin{align*}
&& e^{2n\theta i} - e^{-2n\theta i} &= \left(e^{\theta i} - e^{-\theta i}\right)\sum_{r=1}^n \left(e^{(2r-1)\theta i} + e^{-(2r-1)\theta i}\right) \\
\Rightarrow && 2i \sin 2n \theta &= 2i \sin \theta \sum_{r=1}^n 2 \cos (2r-1) \theta \\
\Rightarrow && \sin 2n \theta &= 2\sin \theta \sum_{r=1}^n \cos (2r-1) \theta
\end{align*}
\item When $n = 2, \theta = \frac{\pi}{5}$ we have:
\begin{align*}
&&\sin \frac{4\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}) \\
&&\sin \frac{\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} - \cos \frac{2\pi}{5}) \\
&&\frac12 &= \cos \frac{\pi}{5} - \cos \frac{2 \pi}{5} \\
\Rightarrow && \cos \frac{2\pi}{5} &= \cos \frac{\pi}{5} - \frac12
\end{align*}
\item When $n = 7, \theta = \frac{\pi}{15}$ we have:
\begin{align*}
&& \sin \frac{14 \pi}{15} &= 2 \sin \frac{\pi}{15} \sum_{r=1}^7 \cos (2r-1) \frac{\pi}{15} \\
\Rightarrow && \frac12 &= \cos \frac{\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}+ \cos \frac{7 \pi}{15}+ \cos \frac{9 \pi}{15}+ \cos \frac{11 \pi}{15}+ \cos \frac{13 \pi}{15} \\
&&&= -\cos \frac{16\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}- \cos \frac{8 \pi}{15}+ \cos \frac{9 \pi}{15}- \cos \frac{4 \pi}{15}- \cos \frac{2\pi}{15} \\
&&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \cos \frac{\pi}{5} + \cos \frac{\pi}{3} + \cos \frac{3 \pi}{5} \\
&&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \frac12 + \frac12 \\
\Rightarrow && \frac12 &= cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}
\end{align*}
\end{enumerate}
\item By using $z = e^{i \theta}$ we have that:
\begin{align*}
&& z^{2n}-\frac{1}{z^{2n}} &= \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right ) \\
\Rightarrow && e^{2n \theta i} - e^{-2n \theta i} &= (e^{\theta i} + e^{-\theta i}) \sum_{r=1}^n (e^{(2r-1)\theta i} - e^{(2r-1) \theta i}) \\
\Rightarrow && 2i \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n 2i \sin(2r-1) \theta \\
\Rightarrow && \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n \sin(2r-1) \theta
\end{align*}
When $n = 3, \theta = \frac{\pi}{14}$ we must have:
\begin{align*}
&&\sin \frac{3 \pi}{7} &= 2 \cos \frac{\pi}{14}( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\
&&&= 2 \sin \left (\frac{\pi}{2} - \frac{\pi}{14} \right)( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\
&&&= 2 \sin \frac{3\pi}{7} ( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\
\Rightarrow && \frac12 &= \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}
\end{align*} as required.
\end{questionparts}
In terms of the number of attempts, this was the second most popular question. The induction in part (i) was generally done very well with a clearly laid out proof. The fifth (method) mark in part (i) was often gained by finding a relevant identity. However, the final mark in part (i) was missed by a large majority of candidates due to not handling the alternating sign in the summation correctly. The summation is quite tricky in this respect, requiring the notation to be set up so that either the final term in the sum is positive, if terms are kept in the same order as for the previous part, or reversing the order of the sum (in which case the first term is positive). Candidates often missed out on one or both marks in part (ii) (a) due to forgetting factors of 2 and i. Most candidates that attempted parts (ii) (b) and (ii) (c) gained two method marks for successfully substituting a valid value of θ in both. However, a significant number of attempts at parts (ii) (b) and (ii) (c) did not gain full credit due to insufficient justification when manipulating trigonometric expressions. In general, candidates who stated the trigonometric identities they were using, or which specific terms were equivalent were successful here. Those that did multiple steps at once without justification often missed out on marks because it was not possible to pick out the results they had used. It is very significant here that the answer was given. Candidates should in general attempt to give more details when proving an answer given in the question to show they understand the intermediate steps between the starting point and given answer. Most candidates did not attempt the part (iii). The successful attempts were from candidates who had given very clear answers to previous parts. There were a few different choices of n and θ that led to the required result.