| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Locus with parameter variation |
| Difficulty | Standard +0.8 This FP3 question requires manipulating complex exponentials, using half-angle formulas to derive a non-obvious trigonometric identity, and understanding how transformations map loci in the complex plane. While the algebraic manipulation is systematic once you multiply by the conjugate and apply Euler's formula, recognizing the half-angle cotangent form and correctly sketching the transformed locus (unit circle → imaginary axis) requires solid understanding beyond routine application. The 6-mark allocation and multi-step reasoning place it moderately above average difficulty. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02d Exponential form: re^(i*theta)4.02k Argand diagrams: geometric interpretation4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER \(\frac{1+e^{i\theta}}{1-e^{i\theta}} = \frac{e^{-\frac{i\theta}{2}}+e^{\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}}\) | M1 | EITHER For changing LHS terms to \(e^{\pm i\theta}\) |
| \(= \frac{2\cos\frac{\theta}{2}}{-2i\sin\frac{\theta}{2}} = i\cot\frac{\theta}{2}\) | M1 | OR in reverse For using \(\cot\frac{\theta}{2} = \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\) |
| OR in reverse with similar working | ||
| \(\frac{e^{\frac{i\theta}{2}} \pm e^{-i\theta}}{(2)(i)}\) soi | M1 A1 3 | For either of \(\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = \frac{e^{\frac{i\theta}{2}} \pm e^{-i\theta}}{(2)(i)}\); For fully correct proof to AG; SR If factors of 2 or \(i\) are not clearly seen, award M1 M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER \(\frac{1+e^{i\theta}}{1-e^{i\theta}} = \frac{1-e^{-i\theta}}{1-e^{i\theta}} \cdot \frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}\) | M1 | For multiplying top and bottom by complex conjugate in exp or trig form |
| \(\text{OR} \frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta} \cdot \frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}\) | ||
| \(= \frac{2\sin\theta}{2-2\cos\theta} \cdot \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = i\cot\frac{\theta}{2}\) | M1 A1 | For using both double angle formulae correctly; For fully correct proof to AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta} = \frac{2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\) | M1 | For using both double angle formulae correctly |
| \(= \frac{2\cos\frac{\theta}{2}(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})}{2\sin\frac{\theta}{2}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})}\) | M1 | For appropriate factorisation |
| \(= i\cot\frac{\theta}{2} \cdot \frac{(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})}{(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})} = i\cot\frac{\theta}{2}\) | A1 | For fully correct proof to AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1+e^{i\theta}}{1-e^{i\theta}} = \frac{1+\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}}{1-\frac{1-t^2}{1+t^2}-i\frac{2t}{1+t^2}}\) | M1 | For substituting both \(t\) formulae correctly |
| \(= \frac{2+2it}{2t^2-2it} = \frac{1+1it}{t(t-i)} = i\cot\frac{\theta}{2}\) | M1 A1 | For appropriate factorisation; For fully correct proof to AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1+e^{i\theta}}{1-e^{i\theta}} \times \frac{1+e^{i\theta}}{1+e^{i\theta}} = \frac{1+2e^{i\theta}+e^{2i\theta}}{1-e^{2i\theta}}\) | For multiplying top and bottom by \(1+e^{i\theta}\) OR multiplying top and bottom by \(1+e^{-i\theta}\) | |
| \(= \frac{2+e^{i\theta}+e^{-i\theta}}{e^{-i\theta}-e^{i\theta}}\) | M1 | |
| \(= \frac{2(1+\cos\theta)}{-2i\sin\theta} = \frac{2\cos^2\frac{\theta}{2}}{-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \frac{\cos\frac{\theta}{2}}{-i\sin\frac{\theta}{2}} = i\cot\frac{\theta}{2}\) | M1 | For using both double angle formulae correctly |
| \(= i\cot\frac{\theta}{2}\) | A1 3 | For fully correct proof to AG |
| Answer | Marks | Guidance |
|---|---|---|
| [Diagram showing circle in z-plane with center O and radius marked as 1 on re-axis] | M1 A1 B1 3 | For a circle centre O; For indication of radius = 1 and anticlockwise arrow shown; For locus of \(w\) shown as imaginary axis described downwards |
## (i) METHOD 1
EITHER $\frac{1+e^{i\theta}}{1-e^{i\theta}} = \frac{e^{-\frac{i\theta}{2}}+e^{\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}}$ | M1 | EITHER For changing LHS terms to $e^{\pm i\theta}$
$= \frac{2\cos\frac{\theta}{2}}{-2i\sin\frac{\theta}{2}} = i\cot\frac{\theta}{2}$ | M1 | OR in reverse For using $\cot\frac{\theta}{2} = \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}$
OR in reverse with similar working | |
$\frac{e^{\frac{i\theta}{2}} \pm e^{-i\theta}}{(2)(i)}$ soi | M1 A1 3 | For either of $\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = \frac{e^{\frac{i\theta}{2}} \pm e^{-i\theta}}{(2)(i)}$; For fully correct proof to AG; SR If factors of 2 or $i$ are not clearly seen, award M1 M1 A0
## METHOD 2
EITHER $\frac{1+e^{i\theta}}{1-e^{i\theta}} = \frac{1-e^{-i\theta}}{1-e^{i\theta}} \cdot \frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}$ | M1 | For multiplying top and bottom by complex conjugate in exp or trig form
$\text{OR} \frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta} \cdot \frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}$ | |
$= \frac{2\sin\theta}{2-2\cos\theta} \cdot \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = i\cot\frac{\theta}{2}$ | M1 A1 | For using both double angle formulae correctly; For fully correct proof to AG
## METHOD 3
$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta} = \frac{2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}$ | M1 | For using both double angle formulae correctly
$= \frac{2\cos\frac{\theta}{2}(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})}{2\sin\frac{\theta}{2}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})}$ | M1 | For appropriate factorisation
$= i\cot\frac{\theta}{2} \cdot \frac{(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})}{(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})} = i\cot\frac{\theta}{2}$ | A1 | For fully correct proof to AG
## METHOD 4
$\frac{1+e^{i\theta}}{1-e^{i\theta}} = \frac{1+\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}}{1-\frac{1-t^2}{1+t^2}-i\frac{2t}{1+t^2}}$ | M1 | For substituting both $t$ formulae correctly
$= \frac{2+2it}{2t^2-2it} = \frac{1+1it}{t(t-i)} = i\cot\frac{\theta}{2}$ | M1 A1 | For appropriate factorisation; For fully correct proof to AG
## METHOD 5
$\frac{1+e^{i\theta}}{1-e^{i\theta}} \times \frac{1+e^{i\theta}}{1+e^{i\theta}} = \frac{1+2e^{i\theta}+e^{2i\theta}}{1-e^{2i\theta}}$ | | For multiplying top and bottom by $1+e^{i\theta}$ OR multiplying top and bottom by $1+e^{-i\theta}$
$= \frac{2+e^{i\theta}+e^{-i\theta}}{e^{-i\theta}-e^{i\theta}}$ | M1 |
$= \frac{2(1+\cos\theta)}{-2i\sin\theta} = \frac{2\cos^2\frac{\theta}{2}}{-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \frac{\cos\frac{\theta}{2}}{-i\sin\frac{\theta}{2}} = i\cot\frac{\theta}{2}$ | M1 | For using both double angle formulae correctly
$= i\cot\frac{\theta}{2}$ | A1 3 | For fully correct proof to AG
## (ii)
[Diagram showing circle in z-plane with center O and radius marked as 1 on re-axis] | M1 A1 B1 3 | For a circle centre O; For indication of radius = 1 and anticlockwise arrow shown; For locus of $w$ shown as imaginary axis described downwards
---It is given that $z = e^{i\theta}$, where $0 < \theta < 2\pi$, and $w = \frac{1+z}{1-z}$.
\begin{enumerate}[label=(\roman*)]
\item Prove that $w = i \cot \frac{1}{2}\theta$. [3]
\item Sketch separate Argand diagrams to show the locus of $z$ and the locus of $w$. You should show the direction in which each locus is described when $\theta$ increases in the interval $0 < \theta < 2\pi$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q2 [6]}}