Challenging +1.2 This is a standard Möbius transformation question requiring systematic application of the circle-mapping property. While it involves Further Maths content (making it inherently harder than Core), the method is algorithmic: find images of three points on |z|=2, then determine the circle through them. No novel insight required, just careful algebraic manipulation.
2. The transformation \(T\) from the \(z\)-plane, where \(z = x + \mathrm { i } y\), to the \(w\)-plane, where \(w = u + \mathrm { i } v\), is given by
$$w = \frac { z + 2 } { z - \mathrm { i } } \quad z \neq \mathrm { i }$$
The transformation \(T\) maps the circle \(| z | = 2\) in the \(z\)-plane onto a circle \(C\) in the \(w\)-plane.
Find (i) the centre of \(C\),
(ii) the radius of \(C\).
Attempt circle form. Coefficients for \(u^2\) and \(v^2\) must be 1. Depends on all 3 previous M marks
(i) centre is \(\left(\frac{4}{3}, -\frac{2}{3}\right)\)
A1
Correct centre given (no decimals)
(ii) radius is \(\frac{2\sqrt{5}}{3}\)
A1 [8]
Correct radius, any equivalent form (no decimals). NB: These 2 A marks can only be awarded if results deduced from a correct circle equation
ALT 1: Change \(w\) to \(u+iv\), rearrange to \(z=\ldots\) and use \(\
z\
=2\)
ALT 2: \(i\) maps to \(\infty \Rightarrow \pm 2i\) map to a diameter of \(C\)
M1A1; So \(\frac{2i+2}{i}\) and \(\frac{-2i+2}{-3i}\) are ends of a diameter
M2A1; Calculate centre and radius
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $w = \frac{z+2}{z-i}$, $z \neq i$; rearrange to $z = \frac{2+iw}{w-1}$ | M1 | Rearrange equation to $z = \ldots$ |
| $\|z\| = 2 \Rightarrow \left\|\frac{2+iw}{w-1}\right\| = 2 \Rightarrow \|2+iw\| = 2\|w-1\|$ | M1 A1 | Change $w$ to $u+iv$ and use $\|z\|=2$. Allow if a different pair of letters used. A1: Correct equation |
| $\|2+iu-v\| = 2\|u+iv-1\|$ | | |
| $(2-v)^2 + u^2 = 4\left((u-1)^2 + v^2\right)$ | M1 A1 | Correct use of Pythagoras on either side. Allow with 2 or 4 (RHS). A1: Correct unsimplified equation |
| $3u^2 + 3v^2 - 8u + 4v = 0$ | | |
| $\left(u - \frac{4}{3}\right)^2 + \left(v + \frac{2}{3}\right)^2 = \frac{20}{9}$ or $u^2+v^2-\frac{8}{3}u+\frac{4}{3}v=0$ | dM1 | Attempt circle form. Coefficients for $u^2$ and $v^2$ must be 1. Depends on all 3 previous M marks |
| (i) centre is $\left(\frac{4}{3}, -\frac{2}{3}\right)$ | A1 | Correct centre given (no decimals) |
| (ii) radius is $\frac{2\sqrt{5}}{3}$ | A1 [8] | Correct radius, any equivalent form (no decimals). NB: These 2 A marks can only be awarded if results deduced from a correct circle equation |
**ALT 1:** Change $w$ to $u+iv$, rearrange to $z=\ldots$ and use $\|z\|=2$ | M1, M1, A1 then as above.
**ALT 2:** $i$ maps to $\infty \Rightarrow \pm 2i$ map to a diameter of $C$ | M1A1; So $\frac{2i+2}{i}$ and $\frac{-2i+2}{-3i}$ are ends of a diameter | M2A1; Calculate centre and radius | M1A1A1
2. The transformation $T$ from the $z$-plane, where $z = x + \mathrm { i } y$, to the $w$-plane, where $w = u + \mathrm { i } v$, is given by
$$w = \frac { z + 2 } { z - \mathrm { i } } \quad z \neq \mathrm { i }$$
The transformation $T$ maps the circle $| z | = 2$ in the $z$-plane onto a circle $C$ in the $w$-plane.\\
Find (i) the centre of $C$,\\
(ii) the radius of $C$.\\
\hfill \mbox{\textit{Edexcel F2 2021 Q2 [8]}}