FP1 June 2011 Q2

2. \[z_1 = -2 + \mathrm{i}\]

(a) Find the modulus of \(z_1\). (1)

(b) Find, in radians, the argument of \(z_1\), giving your answer to 2 decimal places. (2)

The solutions to the quadratic equation \[z^2 - 10z + 28 = 0\] are \(z_2\) and \(z_3\).

(c) Find \(z_2\) and \(z_3\), giving your answers in the form \(p \pm \mathrm{i}\sqrt{q}\), where \(p\) and \(q\) are integers. (3)

(d) Show, on an Argand diagram, the points representing your complex numbers \(z_1\), \(z_2\) and \(z_3\). (2)

Scheme	Marks
\(\|z_1\| = \sqrt{(-2)^2 + 1^2} = \sqrt{5} = 2.236\ldots\) \(\sqrt{5}\) or awrt 2.24	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(\arg z = \pi - \tan^{-1}\left(\tfrac{1}{2}\right)\) \(\tan^{-1}\left(\tfrac{1}{2}\right)\) or \(\tan^{-1}\left(\tfrac{2}{1}\right)\) or \(\cos^{-1}\left(\tfrac{2}{\sqrt{5}}\right)\) or \(\sin^{-1}\left(\tfrac{2}{\sqrt{5}}\right)\) or \(\sin^{-1}\left(\tfrac{1}{\sqrt{5}}\right)\) or \(\cos^{-1}\left(\tfrac{1}{\sqrt{5}}\right)\)	M1
\(= 2.677945045\ldots = 2.68\ (2\text{ dp})\) awrt 2.68	A1 oe
	(2)

Notes

Can work in degrees for the method mark (\(\arg z = 153.4349488^\circ\))

\(\arg z = \tan^{-1}\left(\tfrac{1}{-2}\right) = -0.46\) on its own is M0
but \(\pi + \tan^{-1}\left(\tfrac{1}{-2}\right) = 2.68\) scores M1A1
\(\pi - \tan^{-1}\left(\tfrac{1}{-2}\right) = \) is M0 as is \(\pi - \tan\left(\tfrac{1}{2}\right)\) (2.60)

Mark scheme (c)

Scheme	Marks
\(z^2 - 10z + 28 = 0\)
\(z = \dfrac{10 \pm \sqrt{100 - 4(1)(28)}}{2(1)}\) An attempt to use the quadratic formula (usual rules)	M1
\(= \dfrac{10 \pm \sqrt{100 - 112}}{2}\)
\(= \dfrac{10 \pm \sqrt{-12}}{2}\)
\(= \dfrac{10 \pm 2\sqrt{3}\,\mathrm{i}}{2}\) Attempt to simplify their \(\sqrt{-12}\) in terms of i. E.g. \(\mathrm{i}\sqrt{12}\) or \(\mathrm{i}\sqrt{3 \times 4}\)	M1
So, \(z = 5 \pm \sqrt{3}\,\mathrm{i}\).\(\quad \{p = 5,\ q = 3\}\) \(5 \pm \sqrt{3}\,\mathrm{i}\)	A1 oe
	(3)

Notes

If their \(b^2 - 4ac > 0\) then only the first M1 is available.

Correct answers with no working scores full marks.

See appendix for alternative solution by completing the square

Alternative (Way 2)

Scheme	Marks
\(z^2 - 10z + 28 = 0\)
\((z - 5)^2 - 25 + 28 = 0\) \((z \pm 5)^2 \pm 25 + 28 = 0\)	M1
\((z - 5)^2 = -3\)
\(z - 5 = \sqrt{-3}\)
\(z - 5 = \sqrt{3}\,\mathrm{i}\) Attempt to express their \(\sqrt{-3}\) in terms of i.	M1
So, \(z = 5 \pm \sqrt{3}\,\mathrm{i}\).\(\quad \{p = 5,\ q = 3\}\) \(5 \pm \sqrt{3}\,\mathrm{i}\)	A1 oe
	(3)

Alternative (Way 3)

Scheme	Marks
\(z^2 - 10z + 28 = 0\)
\(\left(z - \left(p + \mathrm{i}\sqrt{q}\right)\right)\left(z - \left(p - \mathrm{i}\sqrt{q}\right)\right) = z^2 - 2pz + p^2 + q\)
\(2p = \pm 10\) and \(p^2 \pm q = 28\) Uses sum and product of roots	M1
\(2p = \pm 10 \Rightarrow p = 5\) Attempt to solve for \(p\) (or \(q\))	M1
\(p = 5\) and \(q = 3\)	A1
	(3)

Argand diagram with one point in the second quadrant and two points symmetric about the real axis to the right — Mark scheme (d)

Scheme	Marks
Note that the points are \((-2,\ 1)\), \((5,\ \sqrt{3})\) and \((5,\ -\sqrt{3})\).
The point \((-2,\ 1)\) plotted correctly on the Argand diagram with/without label.	B1
The distinct points \(z_2\) and \(z_3\) plotted correctly and symmetrically about the \(x\)-axis on the Argand diagram with/without label.	B1ft
	(2)
	(8 marks)