| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Region shading with multiple inequalities |
| Difficulty | Standard +0.3 This is a standard Core Pure AS question on Argand diagrams requiring students to shade a region defined by intersection of a circle and angular sector, then find a specific point satisfying both boundary conditions. Part (a) tests routine interpretation of modulus (circle) and argument (rays) inequalities. Part (b) requires solving simultaneous geometric conditions—straightforward using the circle equation and line from the argument—followed by calculating |w|². While multi-step, these are textbook techniques with no novel insight required, making it slightly easier than average. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| V349 SIHI NI IMIMM ION OC | VJYV SIHIL NI LIIIM ION OO | VJYV SIHIL NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Circle or arc with centre in first quadrant, circle spanning all 4 quadrants | M1 | Centre must be in first quadrant; arc in quadrants 1 and 2 acceptable |
| "V" shape with both branches above \(x\)-axis, vertex on positive real axis | M1 | Ignore branches below \(x\)-axis |
| Two half-lines meeting on positive real axis where right branch intersects circle in Q1 and left branch intersects circle in Q2 (not on \(y\)-axis) | A1 | Follows from both M1s |
| Shades region between half-lines and within circle | M1 | |
| Fully correct diagram with 2 marked at vertex on real axis, correct shading | A1 | cso; all previous marks required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-1)^2 + (y-1)^2 = 9\), \(y = x - 2 \Rightarrow x = \ldots\) or \(y = \ldots\) | M1 | Identify strategy; attempt to solve \((x\pm1)^2+(y\pm1)^2=9\) or \(3\) and \(y=\pm x\pm 2\) |
| \(x = 2 + \dfrac{\sqrt{14}}{2},\quad y = \dfrac{\sqrt{14}}{2}\) | A1 | Correct coordinates; may be implied by subsequent work. Allow as complex number \(2+\frac{\sqrt{14}}{2}+\frac{\sqrt{14}}{2}i\) |
| \(\ | w\ | ^2 = \left(2+\dfrac{\sqrt{14}}{2}\right)^2 + \left(\dfrac{\sqrt{14}}{2}\right)^2\) |
| \(= 11 + 2\sqrt{14}\) | A1 | Correct exact value; cso |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Circle or arc with centre in first quadrant, circle spanning all 4 quadrants | M1 | Centre must be in first quadrant; arc in quadrants 1 and 2 acceptable |
| "V" shape with both branches above $x$-axis, vertex on positive real axis | M1 | Ignore branches below $x$-axis |
| Two half-lines meeting on positive real axis where right branch intersects circle in Q1 and left branch intersects circle in Q2 (not on $y$-axis) | A1 | Follows from both M1s |
| Shades region between half-lines and within circle | M1 | |
| Fully correct diagram with 2 marked at vertex on real axis, correct shading | A1 | cso; all previous marks required |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-1)^2 + (y-1)^2 = 9$, $y = x - 2 \Rightarrow x = \ldots$ or $y = \ldots$ | M1 | Identify strategy; attempt to solve $(x\pm1)^2+(y\pm1)^2=9$ or $3$ and $y=\pm x\pm 2$ |
| $x = 2 + \dfrac{\sqrt{14}}{2},\quad y = \dfrac{\sqrt{14}}{2}$ | A1 | Correct coordinates; may be implied by subsequent work. Allow as complex number $2+\frac{\sqrt{14}}{2}+\frac{\sqrt{14}}{2}i$ |
| $\|w\|^2 = \left(2+\dfrac{\sqrt{14}}{2}\right)^2 + \left(\dfrac{\sqrt{14}}{2}\right)^2$ | M1 | Correct use of Pythagoras on coordinates (no $i$'s) |
| $= 11 + 2\sqrt{14}$ | A1 | Correct exact value; cso |
---\begin{enumerate}
\item (a) Shade on an Argand diagram the set of points
\end{enumerate}
$$\{ z \in \mathbb { C } : | z - 1 - \mathrm { i } | \leqslant 3 \} \cap \quad z \in \mathbb { C } : \frac { \pi } { 4 } \leqslant \arg ( z - 2 ) \leqslant \frac { 3 \pi } { 4 }$$
The complex number $w$ satisfies
$$| w - 1 - \mathrm { i } | = 3 \text { and } \arg ( w - 2 ) = \frac { \pi } { 4 }$$
(b) Find, in simplest form, the exact value of $| w | ^ { 2 }$
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\hfill \mbox{\textit{Edexcel CP AS 2018 Q3 [9]}}