AS June 2021 Paper 1 Q8

AQACurrent spec6 marksComplex Numbers Finding Roots of Polynomials

8 Stephen is correctly told that \((1 + \mathrm{i})\) and \(-1\) are two roots of the polynomial equation

\[z^3 - 2\mathrm{i}z^2 + pz + q = 0\]

where \(p\) and \(q\) are complex numbers.

(a) Stephen states that \((1 - \mathrm{i})\) must also be a root of the equation because roots of polynomial equations occur in conjugate pairs.

Explain why Stephen’s reasoning is wrong. [1 mark]

(b) Find \(p\) and \(q\) [5 marks]

Mark scheme (a)

Scheme	Marks	AO
Gives a correct explanation. Condone an explanation which suggests that conjugate pairs only occur if the coefficients are real.	B1	2.3
	(1)

Typical solution

Can only assume that complex roots of a polynomial equation are in conjugate pairs if the coefficients are all real.

Mark scheme (b)

Scheme	Marks	AO
Writes an equation for the 3^rd root using the sum of roots. Condone \(-2\mathrm{i}\) or \(2\) or \(-2\) for the sum of roots. Or writes an equation in \(p\) and \(q\) (may be unsimplified) from substitution of one known root into the equation.	M1	3.1a
Obtains correct 3^rd root. PI by a factor of \((z - \mathrm{i})\) Or writes two correct simultaneous equations in \(p\) and \(q\) from substitution of both known roots into the equation (any \(\mathrm{i}^2\) must be replaced with \(-1\)).	A1	1.1b
Writes an expression for \(p\) using the sum of pairwise products of roots with their 3^rd root and the two given roots. Or eliminates \(q\) from their simultaneous equations. Or multiplies \(\big(z - (1 + \mathrm{i})\big)(z + 1)(z - \alpha)\) to obtain a cubic expression where \(\alpha\) is their 3^rd root. May be unsimplified. Allow sign errors.	M1	3.1a
Writes an expression for \(q\) using the product of roots with their 3^rd root and the two given roots. Or eliminates \(p\) from their simultaneous equations. Or compares the coefficients of the given polynomial and their cubic expansion. Allow sign errors.	M1	3.1a
Obtains \(p = -2 - \mathrm{i}\) and \(q = \mathrm{i} - 1\)	A1	1.1b
	(5)
	(6 marks)

Typical solution

\[(1 + \mathrm{i}) + (-1) + \alpha = 2\mathrm{i}\]\[\alpha = \mathrm{i}\]\[p = (1 + \mathrm{i})(-1) + (-1)\mathrm{i} + \mathrm{i}(1 + \mathrm{i})\]\[p = -1 - \mathrm{i} - \mathrm{i} + \mathrm{i} + \mathrm{i}^2\]\[p = -2 - \mathrm{i}\]\[q = -(1 + \mathrm{i})(-1)\mathrm{i}\]\[q = \mathrm{i} + \mathrm{i}^2\]\[q = \mathrm{i} - 1\]