| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Complex transformations (Möbius) |
| Difficulty | Challenging +1.2 This is a standard Möbius transformation question requiring students to find how a circle maps under the transformation. While it involves Further Maths content (making it inherently harder than single maths), the technique is routine: substitute points on |z|=1, use the property that circles map to circles, and find centre/radius by testing specific points (e.g., z=1, i, -1). It's a textbook exercise with a well-established method, but requires more sophistication than typical single maths questions. |
| Spec | 4.02k Argand diagrams: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w(z+2i) = z-3i\), \(z = \frac{i(2w+3)}{1-w}\) | M1 | Re-arrange to \(z = \ldots\) |
| \( | z | =1 \Rightarrow \left |
| \( | i(2w+3) | = |
| \(w=u+iv\): \((2u+3)^2 + 4v^2 = (1-u)^2 + v^2\) | ddM1 | Dep on both previous M marks; use \(w=u+iv\) and find moduli (or square) |
| \(4u^2+12u+9+4v^2 = 1-2u+u^2+v^2\) | ||
| \(3u^2+3v^2+14u+8=0\) | dddM1 | Dep on all previous M marks; re-arrange to circle form (same coeffs for squared terms) |
| \(u^2+v^2+\frac{14}{3}u+\frac{8}{3}=0\) | A1 | Correct equation in \(u\) and \(v\) with coefficients of \(u^2\) and \(v^2\) both 1 |
| \(\left(u+\frac{7}{3}\right)^2+v^2 = -\frac{8}{3}+\frac{49}{9}=\frac{25}{9}\) | ||
| (i) Centre \(\left(-\frac{7}{3}, 0\right)\) | A1 | Correct centre in coordinate brackets; completion of square need not be shown |
| (ii) Radius \(\frac{5}{3}\) | A1 (7) | Correct radius; centre and radius must come from a correct circle equation |
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w(z+2i) = z-3i$, $z = \frac{i(2w+3)}{1-w}$ | M1 | Re-arrange to $z = \ldots$ |
| $|z|=1 \Rightarrow \left|\frac{i(2w+3)}{1-w}\right|=1$ | dM1 | Dep on first M1, using $|z|=1$ with previous result |
| $|i(2w+3)| = |1-w|$ | | |
| $w=u+iv$: $(2u+3)^2 + 4v^2 = (1-u)^2 + v^2$ | ddM1 | Dep on both previous M marks; use $w=u+iv$ and find moduli (or square) |
| $4u^2+12u+9+4v^2 = 1-2u+u^2+v^2$ | | |
| $3u^2+3v^2+14u+8=0$ | dddM1 | Dep on all previous M marks; re-arrange to circle form (same coeffs for squared terms) |
| $u^2+v^2+\frac{14}{3}u+\frac{8}{3}=0$ | A1 | Correct equation in $u$ and $v$ with coefficients of $u^2$ and $v^2$ both 1 |
| $\left(u+\frac{7}{3}\right)^2+v^2 = -\frac{8}{3}+\frac{49}{9}=\frac{25}{9}$ | | |
| **(i)** Centre $\left(-\frac{7}{3}, 0\right)$ | A1 | Correct centre in coordinate brackets; completion of square need not be shown |
| **(ii)** Radius $\frac{5}{3}$ | A1 (7) | Correct radius; centre and radius must come from a correct circle equation |
---5. The transformation $T$ from the $z$-plane to the $w$-plane is given by
$$w = \frac { z - 3 \mathrm { i } } { z + 2 \mathrm { i } } \quad z \neq - 2 \mathrm { i }$$
The circle with equation $| z | = 1$ in the $z$-plane is mapped by $T$ onto the circle $C$ in the $w$-plane.
Determine\\
(i) the centre of $C$,\\
(ii) the radius of $C$.\\
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\hfill \mbox{\textit{Edexcel F2 2020 Q5 [7]}}