FP2 June 2013 (R) Q1

1. A transformation \(T\) from the \(z\)-plane to the \(w\)-plane is given by \[w = \frac{z + 2\mathrm{i}}{\mathrm{i}z} \qquad z \neq 0\]

The transformation maps points on the real axis in the \(z\)-plane onto a line in the \(w\)-plane.

Find an equation of this line. (4)

Mark scheme

Scheme	Marks
\(z = x\qquad w = \dfrac{x + 2\mathrm{i}}{\mathrm{i}x}\)	M1A1
\(w = \dfrac{1}{\mathrm{i}} + \dfrac{2\mathrm{i}}{\mathrm{i}x}\)
\(u + \mathrm{i}v = -\mathrm{i} + \dfrac{2}{x}\)
\(\left(u = \dfrac{2}{x}\right)\qquad v = -1\)	M1
\(\therefore w\) is on the line \(v + 1 = 0\)	A1
	(4 marks)

M1 for replacing at least one \(z\) with \(x\) to obtain (ie show an appreciation that \(y = 0\))

A1 \(w = \dfrac{x + 2\mathrm{i}}{\mathrm{i}x}\)

M1 for writing \(w\) as \(u + \mathrm{i}v\) and equating real or imaginary parts to obtain either \(u\) or \(v\) in terms of \(x\) or just a number

A1 for giving the equation of the line \(v + 1 = 0\) oe must be in terms of \(v\)

Scheme	Marks
\(w = \dfrac{x + \mathrm{i}y + 2\mathrm{i}}{\mathrm{i}(x + \mathrm{i}y)}\) Replacing \(z\) with \(x + \mathrm{i}y\)
\(w = \dfrac{x + \mathrm{i}y + 2\mathrm{i}}{-y + \mathrm{i}x} \times \dfrac{-y - \mathrm{i}x}{-y - \mathrm{i}x}\)
\(w = \dfrac{\left(x + \mathrm{i}(y + 2)\right)(-y - \mathrm{i}x)}{y^2 + x^2}\)
\(w = \dfrac{2x - \mathrm{i}\left(x^2 + y^2 + 2y\right)}{y^2 + x^2}\)
\(w = \dfrac{2x - \mathrm{i}x^2}{x^2} = \dfrac{2}{x} - \mathrm{i}\) Using \(y = 0\). This is where the first M1 may be awarded. A1 if correct even if expression is unsimplified but denominator must be real	M1A1
\(v = -1\) M1, A1 as in main scheme	M1A1

Scheme	Marks
\(z = \dfrac{2\mathrm{i}}{\mathrm{i}w - 1}\) Writing the transformation as a function of \(w\)
\(z = \dfrac{2\mathrm{i}}{\mathrm{i}(u + \mathrm{i}v) - 1}\)
\(z = \dfrac{2\mathrm{i}}{(-v - 1) + \mathrm{i}u} \times \dfrac{(-v - 1) - \mathrm{i}u}{(-v - 1) - \mathrm{i}u}\)
\(z = \dfrac{2u + 2\mathrm{i}(-v - 1)}{(-v - 1)^2 + u^2} = \dfrac{2u}{(-v - 1)^2 + u^2} + \mathrm{i}\left(\dfrac{2(-v - 1)}{(-v - 1)^2 + u^2}\right)\)
\(\left(\dfrac{2(-v - 1)}{(v - 1)^2 + u^2}\right) = 0\) or simply \(-2(v + 1) = 0\) Using \(y = 0\). This is where the first M1 may be awarded. A1 if correct even if expression is unsimplified but denominator must be real	M1A1
\(v = -1\) M1, A1 as in main mark scheme above	M1A1