A2 June 2025 Paper 2 Q5

OCR ACurrent spec6 marksComplex Numbers

5 In this question you must show detailed reasoning.

(a) Use an algebraic method to determine the two square roots of \(-3 + \left(4\sqrt{7}\right)\mathrm{i}\).
Give your answers in the form \(a + b\mathrm{i}\) where \(a\) and \(b\) are exact. [5]

(b) State the relationship between the two arguments of the two square roots found in part (a). [1]

Mark scheme (a)

Scheme	Marks	AO
DR \((a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = -3 + \left(4\sqrt{7}\right)\mathrm{i}\)	M1	1.1
\(a^2 - b^2 = -3\) and \(2ab = 4\sqrt{7}\)	M1	1.1
\(a^2 - \left(\frac{2\sqrt{7}}{a}\right)^2 = -3\) \(\therefore (a^2)^2 + 3a^2 - 28 = 0\)	M1	1.1
\((a^2 + 7)(a^2 - 4) = 0 \Rightarrow a^2 = -7\) or \(a^2 = 4\) (\(a\) is real so) \(a^2 \neq -7\)	A1	2.3
\(a = 2 \Rightarrow b = \sqrt{7}\) and \(a = -2 \Rightarrow b = -\sqrt{7}\) (So square roots are \(\pm\left(2 + \sqrt{7}\mathrm{i}\right)\))	A1	1.1
	[5]

Notes

M1: Setting up the square root, squaring and equating. May be implied by correctly equated real and imaginary parts.

M1: Equating real and imaginary parts

M1: Using one equation to eliminate one unknown and rearranging to correct quadratic in the square of the other (condone \(a^4\) or \(b^4\)).
\(\left(\frac{2\sqrt{7}}{b}\right)^2 - b^2 = -3 \quad \therefore (b^2)^2 - 3b^2 - 28 = 0\)

A1: Solving and rejecting negative root (may be implicit e.g. by asserting that \(a^2 \gt 0\) or not using negative root in subsequent working).
\((b^2 - 7)(b^2 + 4) = 0 \Rightarrow b^2 = 7\) or \(b^2 = -4\) (\(b\) is real so) \(b^2 \neq -4\)

A1: Root need not be assembled but values must be clearly paired. i.e. \(a = \pm 2, b = \pm\sqrt{7}\) is insufficient for A1.
\(b = \sqrt{7} \Rightarrow a = 2\) and \(b = -\sqrt{7} \Rightarrow a = -2\)
Correct answers without working is 0/5. Use of polar form 0/5 unless via an equivalent algebraic method.

Mark scheme (b)

Scheme	Marks	AO
DR Their arguments differ by \(\pi\) (radians) oe	B1	2.4
	[1]

Notes

B1: Accept 180°. Answer must give a connection that could be used to find one argument from the other. Do not accept e.g.: ‘they each make the same angle with the real axis’ B0; ‘one is \(-1 \times\) the other’ B0
Accept e.g.: ‘Arguments have the same tangents’ B1; ‘They are \(\pi\) apart’ B1; ‘The argument of one is \(\pi\) away from the argument of the other’ B1; ‘Rotation by \(\pi\) or 180°’ B1