Edexcel FP2 2004 June — Question 12 14 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2004
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeComplex transformations (Möbius)
DifficultyChallenging +1.2 This is a standard Further Maths FP2 Möbius transformation question requiring systematic application of known techniques: substituting parametric forms, algebraic manipulation to show circle mappings, and sketching. While it involves multiple parts and requires careful algebra with complex numbers, the methods are direct applications of syllabus content without requiring novel insight or particularly sophisticated reasoning.
Spec4.02k Argand diagrams: geometric interpretation4.02l Geometrical effects: conjugate, addition, subtraction4.02m Geometrical effects: multiplication and division

12. The transformation \(T\) from the complex \(z\)-plane to the complex \(w\)-plane is given by $$w = \frac { z + 1 } { z + \mathrm { i } } , \quad z \neq - \mathrm { i }$$
  1. Show that \(T\) maps points on the half-line \(\arg ( z ) = \frac { \pi } { 4 }\) in the \(z\)-plane into points on the circle \(| w | = 1\) in the \(w\)-plane.
  2. Find the image under \(T\) in the \(w\)-plane of the circle \(| Z | = 1\) in the \(z\)-plane.
  3. Sketch on separate diagrams the circle \(| \mathbf { Z } | = 1\) in the \(z\)-plane and its image under \(T\) in the \(w\)-plane.
  4. Mark on your sketches the point \(P\), where \(z = \mathrm { i }\), and its image \(Q\) under \(T\) in the \(w\)-plane.

(a) \(\arg z = \frac{\pi}{4} \Rightarrow z = \lambda + \lambda i\) (or putting \(x\) and \(y\) equal at some stage) B1
\(w = \frac{(\lambda+1)+\lambda i}{\lambda+(\lambda+1)i}\), and attempt modulus of numerator or denominator. M1
(Could still be in terms of \(x\) and \(y\))
AnswerMarks Guidance
\((\lambda+1)+\lambda i =
(b) \(w = \frac{z+1}{z+i} \Rightarrow zw + wi = z + 1 \Rightarrow z = \frac{1-wi}{w-1}\) M1
AnswerMarks Guidance
\(z = 1 \Rightarrow
For \(w = a + ib\), \((1+b) - ai = (a-1) + ib\) M1
$(1+b)^2 + a^2 =
(a) $\arg z = \frac{\pi}{4} \Rightarrow z = \lambda + \lambda i$ (or putting $x$ and $y$ equal at some stage) B1

$w = \frac{(\lambda+1)+\lambda i}{\lambda+(\lambda+1)i}$, and attempt modulus of numerator or denominator. M1

(Could still be in terms of $x$ and $y$)

$|(\lambda+1)+\lambda i| = |\lambda+(\lambda+1)i| = \sqrt{(\lambda+1)^2 + \lambda^2}$, $\therefore w = 1$ (*) A1 A1 (cso)

(b) $w = \frac{z+1}{z+i} \Rightarrow zw + wi = z + 1 \Rightarrow z = \frac{1-wi}{w-1}$ M1

$|z| = 1 \Rightarrow |1-wi| = |w-1|$ M1 A1

For $w = a + ib$, $(1+b) - ai = (a-1) + ib$ M1

$(1+b)^2 + a^2 =
12. The transformation $T$ from the complex $z$-plane to the complex $w$-plane is given by

$$w = \frac { z + 1 } { z + \mathrm { i } } , \quad z \neq - \mathrm { i }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $T$ maps points on the half-line $\arg ( z ) = \frac { \pi } { 4 }$ in the $z$-plane into points on the circle $| w | = 1$ in the $w$-plane.
\item Find the image under $T$ in the $w$-plane of the circle $| Z | = 1$ in the $z$-plane.
\item Sketch on separate diagrams the circle $| \mathbf { Z } | = 1$ in the $z$-plane and its image under $T$ in the $w$-plane.
\item Mark on your sketches the point $P$, where $z = \mathrm { i }$, and its image $Q$ under $T$ in the $w$-plane.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2 2004 Q12 [14]}}