FP1 January 2011 Q4

4. Given that \(2 - 4\mathrm{i}\) is a root of the equation \[z^2 + pz + q = 0,\] where \(p\) and \(q\) are real constants,

(a) write down the other root of the equation, (1)

(b) find the value of \(p\) and the value of \(q\). (3)

Mark scheme (b)

Scheme	Marks
\((z - 2 + 4\mathrm{i})(z - 2 - 4\mathrm{i}) = 0\)
\(\Rightarrow z^2 - 2z - 4\mathrm{i}z - 2z + 4 + 8\mathrm{i} + 4\mathrm{i}z - 8\mathrm{i} + 16 = 0\) An attempt to multiply out brackets of two complex factors and no \(\mathrm{i}^2\).	M1
\(\Rightarrow z^2 - 4z + 20 = 0\) Any one of \(p = -4,\ q = 20\).	A1
Both \(p = -4,\ q = 20\). \(\Rightarrow z^2 - 4z + 20 = 0\) only 3/3	A1
	(3)
	[4]

(corrected from the printed mark scheme: the second line printed “\(+ 4 - 8\mathrm{i}\)” for \(+ 4 + 8\mathrm{i}\))

Scheme	Marks
\(z_2 = 2 + 4\mathrm{i}\) \(2 + 4\mathrm{i}\)	B1
Product of roots \(= (2 - 4\mathrm{i})(2 + 4\mathrm{i})\) \(= 4 + 16 = 20\) or \(b^2 - 4ac = (8\mathrm{i})^2\) Sum of roots \(= (2 - 4\mathrm{i}) + (2 + 4\mathrm{i}) = 4\) No \(\mathrm{i}^2\). Attempt Sum and Product of roots or Sum and discriminant	M1
\(= z^2 - 4z + 20 = 0\) Any one of \(p = -4,\ q = 20\).	A1
Both \(p = -4,\ q = 20\).	A1
	(4)

Scheme	Marks
\(z_2 = 2 + 4\mathrm{i}\) \(2 + 4\mathrm{i}\)	B1
\((2 - 4\mathrm{i})^2 + p(2 - 4\mathrm{i}) + q = 0\) An attempt to substitute either \(z_1\) or \(z_2\) into \(z^2 + pz + q = 0\) and no \(\mathrm{i}^2\).	M1
\(-12 - 16\mathrm{i} + p(2 - 4\mathrm{i}) + q = 0\) Imaginary part: \(-16 - 4p = 0\) Real part: \(-12 + 2p + q = 0\)
\(4p = -16 \Rightarrow p = -4\) Any one of \(p = -4,\ q = 20\).	A1
\(q = 12 - 2p \Rightarrow q = 12 - 2(-4) = 20\) Both \(p = -4,\ q = 20\).	A1
	(4)