AS June 2023 Paper 1 Q8

OCR ACurrent spec9 marksComplex Numbers

(a) Solve the equation \(\omega + 2 + 7\mathrm{i} = 3\omega^* - \mathrm{i}\). [4]

(b) Prove algebraically that, for non-zero \(z\), \(z = -z^*\) if and only if \(z\) is purely imaginary. [2]

(c) The complex numbers \(z\) and \(z^*\) are represented on an Argand diagram by the points \(A\) and \(B\) respectively.

(i) State, for any \(z\), the single transformation which transforms \(A\) to \(B\). [1]

(ii) Use a geometric argument to prove that \(z = z^*\) if and only if \(z\) is purely real. [2]

Mark scheme (a)

Scheme	Marks	AO
\(\omega = a + \mathrm{i}b\) oe	M1	3.1a
\(a + 2 = 3a\) \(b + 7 = -3b - 1\)	M1	1.1
\(a = 1\) or \(b = -2\)	A1	1.1
\((\omega =)\ 1 - 2\mathrm{i}\)	A1	1.1
	[4]

Notes

M1: (1st) Writing \(\omega\) in a form which allows 2 equations to be found oe (eg taking Re and Im of both sides)

M1: (2nd) Equating real and imaginary parts. Aef
This might happen after some simplification.

A1: (2nd) Need to see answer as a complex number.
Ignore presence of \(\omega^* = 1 + 2\mathrm{i}\) as long as \(1 - 2\mathrm{i}\) identified as \(\omega\).

Mark scheme (b)

Scheme	Marks	AO
If \(z\) is purely imaginary, so \(z = k\mathrm{i}\) for some real \(k\), then \(z^* = -k\mathrm{i} = -z\) as required	B1	2.1
If \(z = r + s\mathrm{i}\) (\(r\), \(s\) real) then \(z^* = r - s\mathrm{i}\) so \(z = -z^* \Rightarrow r + s\mathrm{i} = -(r - s\mathrm{i}) = -r + s\mathrm{i} \Rightarrow r = 0\) (so \(s \ne 0\) since \(z\) is non-zero) so \(z\) is purely imaginary	B1	2.1
	[2]

Notes

B1: (1st) \(\Leftarrow\). This could, with care, be included in the \(\Rightarrow\) proof.

B1: (2nd) \(\Rightarrow\). \(z\) being non-zero does not have to be rigorously dealt with.
Could instead consider if \(z\) is not purely imaginary and show this means \(z^* \ne -z\)

Mark scheme (c)

Scheme	Marks	AO
(i) Reflection in the real axis	B1	1.2
	[1]
(ii) \(z = z^*\) means that \(A\) and \(B\) are coincident so \(A\) is an invariant point so \(A\) must lie on the mirror line, which is the real axis, so \(A\) must represent a purely real number so \(z\) is purely real.	B1	2.4
If \(z\) is purely real then \(A\) lies on the real axis so it is invariant under a reflection in the real axis so the conjugate \(z^\) is also represented by the same point so \(z = z^\)	B1	2.4
	[2]

Notes

(c)(i)

B1: Must be real, rather than \(x\), axis.
If mention of real axis, can ignore \(x\)

(c)(ii)

B1: (1st) \(\Rightarrow\). Could, with care, be included in the \(\Leftarrow\) proof.
Needs to be a geometric explanation.

B1: (2nd) \(\Leftarrow\).