AS June 2018 Paper 1 Q3

EdexcelCurrent spec9 marksComplex Numbers

(a) Shade on an Argand diagram the set of points\[\left\{z \in \mathbb{C} : |z - 1 - \mathrm{i}| \leqslant 3\right\} \cap \left\{z \in \mathbb{C} : \frac{\pi}{4} \leqslant \arg(z - 2) \leqslant \frac{3\pi}{4}\right\}\] (5)

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 \(|w|^2\) (4)

Argand diagram: a circle with centre in the first quadrant, enclosing the origin, and two half lines from the point 2 on the real axis at 45 degrees either side of the vertical, forming a V shape above the real axis; the region inside the circle and between the half lines is shaded — Mark scheme (a)

Candidate sketch: a circle centred near the origin with a V shape whose vertex is near 2 on the real axis but whose lower branch goes below the real axis; a small region near the vertex is shaded — Mark scheme (a)

Scheme	Marks	AO
	M1 M1 A1 M1 A1	1.1b 1.1b 2.2a 3.1a 1.1b
	(5)

Mark scheme (b)

Scheme	Marks	AO
\((x - 1)^2 + (y - 1)^2 = 9,\ \ y = x - 2 \Rightarrow x = \ldots,\) or \(y = \ldots\)	M1	3.1a
\(x = 2 + \dfrac{\sqrt{14}}{2},\ y = \dfrac{\sqrt{14}}{2}\)	A1	1.1b
\(\|w\|^2 = \left(2 + \dfrac{\sqrt{14}}{2}\right)^2 + \left(\dfrac{\sqrt{14}}{2}\right)^2\)	M1	1.1b
\(= 11 + 2\sqrt{14}\)	A1	1.1b
	(4)
	(9 marks)

Notes

M1: Identifies a suitable strategy for finding the \(x\) or \(y\) coordinate of the point of intersection.
Look for an attempt to solve equations of the form \((x \pm 1)^2 + (y \pm 1)^2 = 9 \text{ or } 3\) and \(y = \pm x \pm 2\)

A1: Correct coordinates for the intersection (there may be other points but allow this mark if the correct coordinates are seen). (The correct coordinates may be implied by subsequent work.)
Allow equivalent exact forms and allow as a complex number e.g. \(2 + \dfrac{\sqrt{14}}{2} + \dfrac{\sqrt{14}}{2}\mathrm{i}\)

M1: Correct use of Pythagoras on their coordinates (There must be no i’s)

A1: Correct exact value by cso

Note that solving \((x - 1)^2 + (y - 1)^2 = 9,\ \ y = x + 2\) gives \(x = \dfrac{\sqrt{14}}{2},\ y = 2 + \dfrac{\sqrt{14}}{2}\) and hence the correct answer fortuitously so scores M1A0M1A0

Notes

Example marking for 3(a)

Notes