AS June 2025 Paper 1 Q8

OCR MEICurrent spec8 marksComplex Numbers Using Roots of Polynomials

8 The three distinct roots of the equation \(z^3 - 4z^2 + pz + q = 0\), where \(p\) and \(q\) are real, are drawn on an Argand diagram. The three points which represent these roots do not lie on a straight line but instead form a triangle T.

(a) Show that T is isosceles. [3]

(b) In this question you must show detailed reasoning.
You are given the following information.

The area of T is 10 square units.
One of the roots of the equation \(z^3 - 4z^2 + pz + q = 0\) is \(z = -2\).

Find the other roots of the equation. [5]

Mark scheme (a)

Scheme	Marks	AO
If all roots are real, they would lie on real axis and so would not form a triangle	B1	2.1
The two complex roots are complex conjugates	M1	2.1
These points are reflections in the real axis, so triangle is isosceles.	A1	2.2a
	[3]

Notes

B1: Allow: to form a triangle, only one of the roots is real

A1: Allow a correct diagram to illustrate symmetry about real axis

Mark scheme (b)

Scheme	Marks	AO
DR Suppose complex roots are \(\alpha = a + b\mathrm{i}\) and \(\beta = a - b\mathrm{i}\)	M1	3.1a
Sum of roots \(= \alpha + \beta - 2 = 4\)	M1	1.1
\(\Rightarrow 2a = 6\), \(a = 3\)	A1	1.1
Area of triangle \(= \frac{1}{2}(a + 2).2b = 10\) \(\Rightarrow b = 2\)	M1	3.1a
[so other roots are \(3 + 2\mathrm{i}\) and \(3 - 2\mathrm{i}\)]	A1	3.2a
	[5]

Notes

M1: Allow M1 for sum of roots \(= -4\)

M1: Forming equation in \(a\) and \(b\) using given area of the triangle

Alternative solution

Scheme	Marks
\(z + 2\) is a factor \(\Rightarrow \mathrm{f}(z) = (z + 2)(z^2 - 6z + p + 12)\)	M1
roots \(3 \pm \mathrm{i}\sqrt{p + 3}\)	M1 A1
Area of triangle \(= \frac{1}{2}(3 + 2).2\sqrt{(p + 3)} = 10\) \(\Rightarrow p = 1\)	M1
so other roots are \(3 + 2\mathrm{i}\) and \(3 - 2\mathrm{i}\)	A1

M1: Long division (oe) to get \(z^2 - 6z \ldots\)

M1: Solving quadratic

A1: roots \(3 \pm \mathrm{i}k\) or Re(roots) \(= 3\)

M1: or \(\frac{1}{2}(3 + 2).2k = 10 \Rightarrow k = 2\)