AS June 2019 Paper 1 Q5

EdexcelCurrent spec9 marksComplex Numbers

Figure 1: an Argand diagram with origin O; z1 is on the negative real axis, z2 is in the second quadrant and z3 is in the first quadrant — Figure 1

The complex numbers \(z_1 = -2\), \(z_2 = -1 + 2\mathrm{i}\) and \(z_3 = 1 + \mathrm{i}\) are plotted in Figure 1, on an Argand diagram for the complex plane with \(z = x + \mathrm{i}y\)

(a) Explain why \(z_1\), \(z_2\) and \(z_3\) cannot all be roots of a quartic polynomial equation with real coefficients. (2)

(b) Show that \(\arg\left(\dfrac{z_2 - z_1}{z_3 - z_1}\right) = \dfrac{\pi}{4}\) (3)

A copy of Figure 1, labelled Diagram 1, is given below.

(d) Shade, on Diagram 1, the set of points of the complex plane that satisfy the inequality\[|z + 2| \leqslant |z - 1 - \mathrm{i}|\] (2)

Diagram 1: a copy of Figure 1, an Argand diagram showing the points z1 on the negative real axis, z2 in the second quadrant and z3 in the first quadrant — Diagram 1

Mark scheme (a)

Scheme	Marks	AO
Complex roots of a real polynomial occur in conjugate pairs	M1	1.2
so a polynomial with \(z_1\), \(z_2\) and \(z_3\) as roots also needs \(z_2^\) and \(z_3^\) as roots, so 5 roots in total, but a quartic has at most 4 roots, so no quartic can have \(z_1\), \(z_2\) and \(z_3\) as roots.	A1	2.4
	(2)

Notes

M1: Some evidence that complex roots occur as conjugate pairs shown, e.g. stated as in scheme, or e.g. identifying if \(-1 + 2\mathrm{i}\) is a root then so is \(-1 - 2\mathrm{i}\). Mere mention of complex conjugates is sufficient for this mark.

A1: A complete argument, referencing that a quartic has at most 4 roots, but would need at least 5 for all of \(z_1\), \(z_2\) and \(z_3\) as roots. There should be a clear statement about the number of roots of a quartic (e.g. a quartic has four roots), and that this is not enough for the two conjugate pairs and real root.

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{z_2 - z_1}{z_3 - z_1} = \dfrac{-1 + 2\mathrm{i} - (-2)}{1 + \mathrm{i} - (-2)} = \dfrac{1 + 2\mathrm{i}}{3 + \mathrm{i}} \times \dfrac{3 - \mathrm{i}}{3 - \mathrm{i}} = \ldots\)	M1	1.1b
\(= \dfrac{3 - \mathrm{i} + 6\mathrm{i} + 2}{9 + 1} = \dfrac{5 + 5\mathrm{i}}{10} = \dfrac{1}{2} + \dfrac{1}{2}\mathrm{i}\) oe	A1	1.1b
As \(\dfrac{1}{2} + \dfrac{1}{2}\mathrm{i}\) is in the first quadrant (may be shown by diagram), hence \(\arg\left(\dfrac{z_2 - z_1}{z_3 - z_1}\right) = \arctan\left(\dfrac{1/2}{1/2}\right) \bigl(= \arctan(1)\bigr) = \dfrac{\pi}{4}\) *	A1*	2.1
	(3)

Notes

M1: Substitutes the numbers in expression and attempts multiplication of numerator and denominator by the conjugate of their denominator or uses calculator to find the quotient. (May be implied.)
NB Applying the difference of arguments and using decimals is M0 here.

A1: Obtains \(\dfrac{1}{2} + \dfrac{1}{2}\mathrm{i}\). (May be from calculator.) Accepted equivalent Cartesian forms.

A1*: Uses arctan on their quotient and makes reference to first quadrant or draws diagram to show they are in the first quadrant, to justify the argument.

Mark scheme (c)

Scheme	Marks	AO
\(\arg\left(\dfrac{z_2 - z_1}{z_3 - z_1}\right) = \arg(z_2 - z_1) - \arg(z_3 - z_1) = \arg(1 + 2\mathrm{i}) - \arg(3 + \mathrm{i})\)	M1	1.1b
Hence \(\arctan(2) - \arctan\left(\dfrac{1}{3}\right) = \dfrac{\pi}{4}\) *	A1*	2.1
	(2)

Notes

M1: Applies the formula for the argument of a difference of complex numbers and substitutes values (may go directly to arctans if the arguments have already been established). If used in (b) it must be seen or referred to in (c) for this mark to be awarded. Allow for \(\arg(z_2 - z_1) - \arg(z_3 - z_1)\) if \(z_2 - z_1\) and \(z_3 - z_1\) have been clearly identified in earlier work.

A1*: Completes the proof clearly by identifying the required arguments and using the result of (b). Use of decimal approximations is A0.

Argand diagram: a straight line with negative gradient passes through z2 and crosses the axes just below and to the left of the origin, meeting the negative imaginary axis; the whole region to the left of and below this line, containing z1, is shaded; z3 lies outside the shaded region — Mark scheme (d)

Scheme	Marks	AO
Line passing through \(z_2\) and the negative imaginary axis drawn.	B1	1.1b
Area below and left of their line shaded, where the line must have negative gradient passing through negative imaginary axis but need not pass through \(z_2\)	B1	1.1b
Unless otherwise indicated by the student mark Diagram 1 (if used) if there are multiple attempts.
	(2)
	(9 marks)