| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2025 |
| Session | February |
| Marks | 8 |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring recognition that C + iS forms a geometric series in complex form, then algebraic manipulation to extract the imaginary part. While it involves multiple techniques (complex exponentials, geometric series, De Moivre's theorem, and rationalizing complex denominators), the path is relatively standard for FM students who have practiced similar problems. The geometric series pattern is fairly evident once written in exponential form, and the algebraic steps, though lengthy, are mechanical. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.06a Summation formulae: sum of r, r^2, r^3 |
The infinite series $C$ and $S$ are defined by
$$C = \cos \theta + \frac{1}{2}\cos 5\theta + \frac{1}{4}\cos 9\theta + \frac{1}{8}\cos 13\theta + \ldots$$
$$S = \sin \theta + \frac{1}{2}\sin 5\theta + \frac{1}{4}\sin 9\theta + \frac{1}{8}\sin 13\theta + \ldots$$
Given that the series $C$ and $S$ are both convergent,
\begin{enumerate}[label=(\alph*)]
\item show that
$$C + iS = \frac{2e^{i\theta}}{2 - e^{4i\theta}}$$ [4]
\item Hence show that
$$S = \frac{4\sin \theta + 2\sin 3\theta}{5 - 4\cos 4\theta}$$ [4]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2025 Q11 [8]}}