FP2 June 2012 Q8

EdexcelOld spec14 marksComplex Numbers

8. The point \(P\) represents a complex number \(z\) on an Argand diagram such that \[|z - 6\mathrm{i}| = 2|z - 3|\]

(a) Show that, as \(z\) varies, the locus of \(P\) is a circle, stating the radius and the coordinates of the centre of this circle. (6)

The point \(Q\) represents a complex number \(z\) on an Argand diagram such that \[\arg(z - 6) = -\frac{3\pi}{4}\]

(b) Sketch, on the same Argand diagram, the locus of \(P\) and the locus of \(Q\) as \(z\) varies. (4)

Mark scheme (a)

Scheme	Marks
\(\|x + \mathrm{i}y - 6\mathrm{i}\| = 2\|x + \mathrm{i}y - 3\|\)	M1
\(x^2 + (y - 6)^2 = 4\left[(x - 3)^2 + y^2\right]\)	M1 A1
\(x^2 + y^2 - 12y + 36 = 4x^2 - 24x + 36 + 4y^2\) \(3x^2 + 3y^2 - 24x + 12y = 0\)
\((x - 4)^2 + (y + 2)^2 = 20\)	M1
Centre \((4, -2)\), Radius \(\sqrt{20} = 2\sqrt{5} =\) awrt 4.47	A1 A1
	(6)

Notes

1^st M Substituting \(z = x + \mathrm{i}y\) oe

2^nd M implementing modulus of both sides and squaring. Require \(\mathrm{Re}^2\) plus \(\mathrm{Im}^2\) on both sides & no terms in i. Condone 2 instead of 4 here.

3^rd M1 for gathering terms and attempting to find centre and / or radius

2^nd A1 for centre, 3^rd A1 for radius

Scheme	Marks
Centre in correct quad for their circle	M1
Passes through O centre in 4^th quad.	A1cao
Half line with positive gradient	B1
Correct position, clearly through \((6, 0)\)	B1
	(4)

Mark scheme (c)

Scheme	Marks
Equation of line \(y = x - 6\)	B1
Attempting simultaneous solution of \((x - 4)^2 + (y + 2)^2 = 20\) and \(y = x - 6\)	M1
\(x = 4 \pm \sqrt{10}\)	A1
\(\left(4 - \sqrt{10}\right) + \mathrm{i}\left(-2 - \sqrt{10}\right)\)	A1cao
	(4)
	(14 marks)

Alt 8(c)

For geometric approach in this part.

Scheme	Marks
Centre \((4, -2)\) on line, can be implied.	B1
Use of Pythagoras or trigonometry to find lengths of isosceles triangle	M1
\(x = 4 - \sqrt{10}\)	A1
\(\left(4 - \sqrt{10}\right) + \mathrm{i}\left(-2 - \sqrt{10}\right)\)	A1cao