Edexcel FP2 2013 June — Question 8 8 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeComplex transformations and mappings
DifficultyChallenging +1.2 This is a standard FP2 transformation question requiring manipulation of modulus conditions and algebraic rearrangement to identify a circle. While it involves multiple steps and complex number manipulation, the techniques (substituting z = e^(iθ), cross-multiplying, and completing the square or using |w-a|=r form) are well-practiced in FP2. The 'show that' structure provides guidance, making it moderately above average but not requiring novel insight.
Spec4.02k Argand diagrams: geometric interpretation4.02m Geometrical effects: multiplication and division4.02o Loci in Argand diagram: circles, half-lines

8. A complex number \(z\) is represented by the point \(P\) on an Argand diagram.
  1. Given that \(| z | = 1\), sketch the locus of \(P\). The transformation \(T\) from the \(z\)-plane to the \(w\)-plane is given by $$w = \frac { z + 7 \mathrm { i } } { z - 2 \mathrm { i } }$$
  2. Show that \(T\) maps \(| z | = 1\) onto a circle in the \(w\)-plane.
  3. Show that this circle has its centre at \(w = - 5\) and find its radius.

Question 8:
Part (a):
AnswerMarks Guidance
Circle centred at origin, radius 1 (sketch)B1 (1 mark)
Part (b):
AnswerMarks Guidance
\(w(z-2i) = z+7i\) so \(z(w-1) = 7i + 2iw\) and \(z = \frac{7i+2iw}{w-1}\)M1 A1
So \(7i + 2iw =
Using \(w = u+iv\): \((-2v)^2 + (2u+7)^2 = (u-1)^2 + v^2\)M1
So \(3u^2 + 3v^2 + 30u + 48 = 0\), which is a circle equationA1 (5 marks)
As \((u+5)^2 + v^2 = 3^2\), so centre is \(-5\) and radius is \(3\)M1 A1 (2 marks, 8 total)
## Question 8:

### Part (a):
| Circle centred at origin, radius 1 (sketch) | B1 | (1 mark) |

### Part (b):
| $w(z-2i) = z+7i$ so $z(w-1) = 7i + 2iw$ and $z = \frac{7i+2iw}{w-1}$ | M1 A1 | |
| So $|7i + 2iw| = |w-1|$ | M1 | |
| Using $w = u+iv$: $(-2v)^2 + (2u+7)^2 = (u-1)^2 + v^2$ | M1 | |
| So $3u^2 + 3v^2 + 30u + 48 = 0$, which is a circle equation | A1 | (5 marks) |
| As $(u+5)^2 + v^2 = 3^2$, so centre is $-5$ and radius is $3$ | M1 A1 | (2 marks, 8 total) |

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8. A complex number $z$ is represented by the point $P$ on an Argand diagram.
\begin{enumerate}[label=(\alph*)]
\item Given that $| z | = 1$, sketch the locus of $P$.

The transformation $T$ from the $z$-plane to the $w$-plane is given by

$$w = \frac { z + 7 \mathrm { i } } { z - 2 \mathrm { i } }$$
\item Show that $T$ maps $| z | = 1$ onto a circle in the $w$-plane.
\item Show that this circle has its centre at $w = - 5$ and find its radius.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2 2013 Q8 [8]}}