FP2 January 2006 Q8

8. In the Argand diagram the point \(P\) represents the complex number \(z\).

Given that \(\arg\left(\dfrac{z - 2\mathrm{i}}{z + 2}\right) = \dfrac{\pi}{2}\),

(a) sketch the locus of \(P\), (4)

(b) deduce the value of \(|z + 1 - \mathrm{i}|\). (2)

The transformation \(T\) from the \(z\)-plane to the \(w\)-plane is defined by \[w = \frac{2(1 + \mathrm{i})}{z + 2}, \qquad z \neq -2\]

(c) Show that the locus of \(P\) in the \(z\)-plane is mapped to part of a straight line in the \(w\)-plane, and show this in an Argand diagram. (6)

Mark scheme (a)

Scheme	Marks
Relating lines and angle (generous) [angle between \(\pm 2\mathrm{i}\) to \(P\) and \(\pm 2\) to \(P\)]	M1 A1
Angle between correct lines is \(\dfrac{\pi}{2}\)	M1 A1
Circle Selecting correct (“top half”) semi-circle.
	(4)

Scheme	Marks
Method for finding Cartesian equation	M1
Correct equation, any form, \(\Rightarrow x(x + 2) + y(y - 2) = 0\)	A1
Sketch: showing circle	M1
Correct circle {centre \((-1, 1)\)}, choosing only “top half”	A1]

Scheme	Marks
\(z = \dfrac{2(1 + \mathrm{i}) - 2\omega}{\omega} \qquad \left(= \dfrac{2(1 + \mathrm{i})}{\omega} - 2\right)\)	M1
\(\dfrac{z - 2\mathrm{i}}{z + 2} = \dfrac{2(1 + \mathrm{i}) - 2(1 + \mathrm{i})\omega}{2(1 + \mathrm{i})} \qquad (= 1 - \omega)\)	M1 A1
\(\arg(1 - \omega) = \dfrac{\pi}{2}\) is line segment, passing through \((1, 0)\)	A1, A1
	A1
	(6)
	(12 marks)

Scheme	Marks
\(u + \mathrm{i}v = \dfrac{2 + 2\mathrm{i}}{(x + 2) + \mathrm{i}y} = \dfrac{(2x + 2y + 4) + \mathrm{i}(x + 2 - y)}{(x + 2)^2 + y^2}\)	M1
\(x = -1 + \sqrt{2}\cos\theta,\ y = 1 + \sqrt{2}\sin\theta\)	M1
\(\Rightarrow w = \dfrac{(2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta + 4) + \mathrm{i}\ldots\ldots}{(2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta + 4)}\ \{= 1 + \mathrm{i}\,\mathrm{f}(\theta)\}\)	A1,
\(\Rightarrow\) part of line \(u = 1\), show lower “half” of line	A1, A1

Scheme	Marks
\(\|z + 1 - \mathrm{i}\|\) is radius; \(\ = \sqrt{2}\)	M1 A1
	(2)