Challenging +1.2 This is a Further Maths F2 question requiring substitution of z = iy into a Möbius transformation, algebraic manipulation to separate real and imaginary parts, and elimination of the parameter. While it involves complex numbers and rational expressions requiring careful algebra, it follows a standard procedure taught in F2 with no novel insight needed. The multi-step nature and algebraic complexity place it slightly above average difficulty.
3. A transformation maps points from the \(z\)-plane, where \(z = x + \mathrm { i } y\), to the \(w\)-plane, where \(w = u + \mathrm { i } v\). The transformation is given by
$$w = \frac { ( 2 + \mathrm { i } ) z + 4 } { z - \mathrm { i } } \quad z \neq \mathrm { i }$$
The transformation maps the imaginary axis in the \(z\)-plane onto the line \(l\) in the \(w\)-plane.
Determine a Cartesian equation of \(l\), giving your answer in the form \(a u + b v + c = 0\) where \(a , b\) and \(c\) are integers to be found.
(6)
\(w = \frac{(2+i)z+4}{z-i} \Rightarrow wz - wi = (2+i)z+4 \Rightarrow z = \ldots\)
M1
Attempts to make \(z\) the subject
\(z = \frac{wi+4}{w-2-i}\)
A1
Correct equation in any form
\(z = \frac{(u+iv)i+4}{u+iv-2-i}\), then multiply numerator and denominator by conjugate of denominator
M1
Introduces \(w = u+iv\) and multiplies by conjugate of denominator
\(u(v-1)+(4-v)(u-2)=0\)
dM1
Sets real part \(= 0\) (with or without denominator). Depends on both M marks above
(correct equation)
A1
Any correct equation
\(3u + 2v - 8 = 0\)
A1
Correct equation in required form (allow any integer multiple)
Way 2:
Answer
Marks
Guidance
Working/Answer
Mark
Notes
\(w = \frac{(2+i)z+4}{z-i},\ z=yi \Rightarrow w = \frac{(2+i)yi+4}{yi-i} \times \frac{i}{i}\)
M1
Solves simultaneously and multiplies numerator and denominator by \(i\)
\(u = \frac{2y}{y-1},\ v = \frac{y-4}{y-1}\)
A1
Correct real and imaginary parts
\(u = \frac{2y}{y-1} \Rightarrow y = \frac{u}{u-2}\)
M1
Attempts \(y\) in terms of \(u\) or \(v\)
\(y = \frac{u}{u-2} \Rightarrow v = \frac{\frac{u}{u-2}-4}{\frac{u}{u-2}-1}\)
M1
Obtains an equation connecting \(u\) and \(v\)
(correct equation)
A1
Any correct equation
\(3u + 2v - 8 = 0\)
A1
Correct equation in required form (allow any integer multiple)
Way 3:
Answer
Marks
Guidance
Working/Answer
Mark
Notes
Apply transformation to any point on imaginary axis
M1
e.g. \((0,0)\to(0,4)\), \((0,1)\to(4,-2)\)
Apply transformation to a second point on imaginary axis
M1
Second M mark
Both transformations correct
A1
First A mark
Complete method to obtain equation for line through 2 points in \(w\)-plane
M1
Correct equation in any form
A1
\(3u + 2v - 8 = 0\)
A1
Correct equation in required form (allow any integer multiple)
## Question 3:
**Way 1:**
| Working/Answer | Mark | Notes |
|---|---|---|
| $w = \frac{(2+i)z+4}{z-i} \Rightarrow wz - wi = (2+i)z+4 \Rightarrow z = \ldots$ | M1 | Attempts to make $z$ the subject |
| $z = \frac{wi+4}{w-2-i}$ | A1 | Correct equation in any form |
| $z = \frac{(u+iv)i+4}{u+iv-2-i}$, then multiply numerator and denominator by conjugate of denominator | M1 | Introduces $w = u+iv$ and multiplies by conjugate of denominator |
| $u(v-1)+(4-v)(u-2)=0$ | dM1 | Sets real part $= 0$ (with or without denominator). Depends on both M marks above |
| (correct equation) | A1 | Any correct equation |
| $3u + 2v - 8 = 0$ | A1 | Correct equation in required form (allow any integer multiple) |
**Way 2:**
| Working/Answer | Mark | Notes |
|---|---|---|
| $w = \frac{(2+i)z+4}{z-i},\ z=yi \Rightarrow w = \frac{(2+i)yi+4}{yi-i} \times \frac{i}{i}$ | M1 | Solves simultaneously and multiplies numerator and denominator by $i$ |
| $u = \frac{2y}{y-1},\ v = \frac{y-4}{y-1}$ | A1 | Correct real and imaginary parts |
| $u = \frac{2y}{y-1} \Rightarrow y = \frac{u}{u-2}$ | M1 | Attempts $y$ in terms of $u$ or $v$ |
| $y = \frac{u}{u-2} \Rightarrow v = \frac{\frac{u}{u-2}-4}{\frac{u}{u-2}-1}$ | M1 | Obtains an equation connecting $u$ and $v$ |
| (correct equation) | A1 | Any correct equation |
| $3u + 2v - 8 = 0$ | A1 | Correct equation in required form (allow any integer multiple) |
**Way 3:**
| Working/Answer | Mark | Notes |
|---|---|---|
| Apply transformation to any point on imaginary axis | M1 | e.g. $(0,0)\to(0,4)$, $(0,1)\to(4,-2)$ |
| Apply transformation to a second point on imaginary axis | M1 | Second M mark |
| Both transformations correct | A1 | First A mark |
| Complete method to obtain equation for line through 2 points in $w$-plane | M1 | |
| Correct equation in any form | A1 | |
| $3u + 2v - 8 = 0$ | A1 | Correct equation in required form (allow any integer multiple) |
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3. A transformation maps points from the $z$-plane, where $z = x + \mathrm { i } y$, to the $w$-plane, where $w = u + \mathrm { i } v$. The transformation is given by
$$w = \frac { ( 2 + \mathrm { i } ) z + 4 } { z - \mathrm { i } } \quad z \neq \mathrm { i }$$
The transformation maps the imaginary axis in the $z$-plane onto the line $l$ in the $w$-plane.\\
Determine a Cartesian equation of $l$, giving your answer in the form $a u + b v + c = 0$ where $a , b$ and $c$ are integers to be found.\\
(6)
\hfill \mbox{\textit{Edexcel F2 2021 Q3 [6]}}