A2 June 2023 Q9

EdexcelCurrent spec5 marksFurther Complex Numbers

Figure 1: an arc of a circle from the point z1 on the right, rising up through z3 and continuing to z2 on the left — Figure 1

Figure 1 shows a locus in the complex plane.

The locus is an arc of a circle from the point represented by \(z_1 = 3 + 2\mathrm{i}\) to the point represented by \(z_2 = a + 4\mathrm{i}\), where \(a\) is a constant, \(a \neq 1\)

Given that

the point \(z_3 = 1 + 4\mathrm{i}\) also lies on the locus
the centre of the circle has real part equal to \(-1\)

(a) determine the value of \(a\). (2)

(b) Hence determine a complex equation for the locus, giving any angles in the equation as positive values. (3)

Mark scheme (a)

Scheme	Marks	AO
Centre lies on perpendicular bisector of \(z_2\) and \(z_3\), which is vertical as same imaginary components. Hence \(\tfrac{a + 1}{2} = -1 \Rightarrow a = \ldots\) Or \((a - (-1))^2 + (4 - b)^2 = (1 - (-1))^2 + (4 - b)^2 \Rightarrow a = \ldots\) \((a - (-1))^2 = (1 - (-1))^2 \Rightarrow a = \ldots\) Or \(-1 - a = 1 - (-1) \Rightarrow a = \ldots\)	M1	1.1b
\(a = -3\)	A1	2.2a
	(2)

Notes

M: Forms a correct strategy to find the value of \(a\), see scheme for various approaches

A1: Correct value

Alternative

Scheme	Marks	AO
\((x - (-1))^2 + (y - b)^2 = r^2\) \(\left.\begin{aligned} &(1 + 1)^2 + (4 - b)^2 = r^2 \\ &(3 + 1)^2 + (2 - b)^2 = r^2 \end{aligned}\right\} \Rightarrow 4 + (4 - b)^2 = 16 + (2 - b)^2 \Rightarrow b = \ldots\{0\}\) \(r^2 = 4 + (4 - \text{“}0\text{”})^2 = \ldots\{20\}\) \((a + 1)^2 + (4 - \text{“}0\text{”})^2 = \text{“}20\text{”} \Rightarrow a = \ldots\)	M1	1.1b
\(a = -3\) only the other root must be rejected	A1	2.2a
	(2)

Handwritten sketch: horizontal line from (−3, 4) to (1, 4), then a line down to (3, 2); the angle at (1, 4) is marked, with π/2 and π/4 labelled — Mark scheme (b)

Scheme	Marks	AO
Equation has form \(\arg\left(\dfrac{z - z_1}{z - z_2}\right) = \theta\)	B1	1.2
\[\arg\left(\frac{1 + 4\mathrm{i} - (3 + 2\mathrm{i})}{1 + 4\mathrm{i} - (\text{“}{-}3\text{”} + 4\mathrm{i})}\right) = \arg(-2 + 2\mathrm{i}) - \arg(4) = \ldots \left(= \frac{3\pi}{4}\right)\]Or\[\arg\left(\frac{1 + 4\mathrm{i} - 3 - 2\mathrm{i}}{1 + 4\mathrm{i} - a - 4\mathrm{i}}\right) = \arg\left(\frac{-2 + 2\mathrm{i}}{4}\right) = \arg\left(-\frac{1}{2} + \frac{1}{2}\mathrm{i}\right) = \frac{3\pi}{4}\]or\[\overrightarrow{z_3z_2} = \begin{pmatrix} -4 \\ 0 \end{pmatrix} \text{ and } \overrightarrow{z_3z_1} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}\]\[\cos\theta = \frac{\begin{pmatrix} -4 \\ 0 \end{pmatrix} \bullet \begin{pmatrix} 2 \\ -2 \end{pmatrix}}{4\sqrt{2^2 + (-2)^2}} \Rightarrow \theta = \ldots\]Or\[\cos\theta = \frac{4^2 + 8 - 40}{2 \times 4 \times \sqrt{8}} \Rightarrow \theta = \ldots\]or Using trigonometry	M1	3.1a
Equation is \(\arg\left(\tfrac{z - 3 - 2\mathrm{i}}{z + 3 - 4\mathrm{i}}\right) = \tfrac{3\pi}{4}\) o.e.e. Equation is \(\arg(z - 3 - 2\mathrm{i}) - \arg(z + 3 - 4\mathrm{i}) = \tfrac{3\pi}{4}\) o.e.e.	A1	2.1
	(3)
	(5 marks)