| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2021 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Geometric properties in Argand diagram |
| Difficulty | Challenging +1.2 This is a geometric proof in the Argand diagram requiring understanding that the diagonal of a rhombus bisects the angle, combined with the parallelogram law for complex addition. While it requires geometric insight about rhombi and connecting this to complex number arguments, it's a fairly standard Further Maths question with a clear geometric setup and well-known properties. The proof is relatively short and follows naturally from the geometry. |
| Spec | 4.02k Argand diagrams: geometric interpretation |
| Answer | Marks |
|---|---|
| 12 | Let D O A , C O A = = |
| Answer | Marks |
|---|---|
| so a r g ( z + w ) = 12 ( a r g z + a r g w ) | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | finds a r g z + a r g w in terms of , |
Question 12:
12 | Let D O A , C O A = =
z + w is represented by C
a r g ( z w ) + = +
BOC = (diagonal of rhombus bisects BOA)
a r g z a r g w ( 2 ) + = + +
2 ( ) = +
so a r g ( z + w ) = 12 ( a r g z + a r g w ) | B1
M1
M1
A1
[4] | 3.1a
2.1
2.1
2.2a | finds a r g z + a r g w in terms of ,
AG12 Fig. 12 shows a rhombus OACB in an Argand diagram. The points A and B represent the complex numbers $z$ and $w$ respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{82808722-0abc-411a-9aa3-c0f368a4c95e-4_641_659_1201_242}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
Prove that $\arg ( z + w ) = \frac { 1 } { 2 } ( \arg z + \arg w )$.\\[0pt]
[A copy of Fig. 12 is provided in the Printed Answer Booklet.]
\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q12 [4]}}