FP1 June 2014 (R) Q4

4. The complex number \(z\) is given by \[z = \frac{p + 2\mathrm{i}}{3 + p\mathrm{i}}\] where \(p\) is an integer.

(a) Express \(z\) in the form \(a + b\mathrm{i}\) where \(a\) and \(b\) are real. Give your answer in its simplest form in terms of \(p\). (4)

(b) Given that \(\arg(z) = \theta\), where \(\tan\theta = 1\) find the possible values of \(p\). (5)

Mark scheme (a)

Scheme	Marks
\(z = \dfrac{p + 2\mathrm{i}}{3 + p\mathrm{i}} \cdot \dfrac{3 - p\mathrm{i}}{3 - p\mathrm{i}}\) Multiplying top and bottom by Conjugate	M1
\(= \dfrac{3p - p^2\mathrm{i} + 6\mathrm{i} + 2p}{9 + p^2}\) At least 3 correct terms in the numerator, evidence that \(\mathrm{i}^2 = -1\) and denominator real.	M1
\(= \dfrac{5p}{p^2 + 9}, \qquad + \dfrac{6 - p^2}{p^2 + 9}\mathrm{i}\) Real + imaginary with i factored out. Accept single denominator with numerator in correct form. Accept ‘a=’ and ‘b=’.	A1, A1
	(4)

Scheme	Marks
\(a + b\mathrm{i} = \dfrac{p + 2\mathrm{i}}{3 + p\mathrm{i}}\) Equate to \(a + b\mathrm{i}\) then rearrange and equate real and imaginary parts.	M1
\(3a - pb = p,\ ap + 3b = 2\) Two equations for \(a\) and \(b\) in terms of \(p\) and attempt to solve for \(a\) and \(b\) in terms of \(p\)	dM1
\(= \dfrac{5p}{p^2 + 9}, \qquad + \dfrac{6 - p^2}{p^2 + 9}\mathrm{i}\) Real + imaginary with i factored out. Accept single denominator with numerator in correct form. Accept ‘a=’ and ‘b=’.	A1, A1
	(4)

(corrected from the printed mark scheme: the total for this alternative is printed as (5))

Mark scheme (b)

Scheme	Marks
\(\arg(z) = \arctan\left(\dfrac{\frac{6 - p^2}{p^2 + 9}}{\frac{5p}{p^2 + 9}}\right)\) Correct method for the argument. Can be implied by correct equation for \(p\)	M1
\(\dfrac{6 - p^2}{5p} = 1\) Their \(\arg(z)\) in terms of \(p = 1\)	M1
\(p^2 + 5p - 6 = 0\) Correct 3TQ	A1
\((p + 6)(p - 1) = 0 \Rightarrow x =\) M1: Attempt to solve their quadratic in \(p\)	M1
\(p = 1, p = -6\) A1: both	A1
	(5)
	(9 marks)