FP2 June 2018 Q2

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

The transformation maps points on the real axis in the \(z\)-plane onto the line \(l\) in the \(w\)-plane.
Find an equation of the line \(l\). (4)

Mark scheme

Scheme	Marks
\(z = x + \mathrm{i}y\) and \(w = u + \mathrm{i}v\) used. Candidates may use any suitable letters.
\(z = x \Rightarrow w = \dfrac{1 - \mathrm{i}x}{x}\) Replaces at least one \(z\) with \(x\) ie indicate that \(y = 0\) (may be done later )	M1
\(w = \dfrac{1}{x} - \mathrm{i}\) or \(w = \dfrac{1 - \mathrm{i}x}{x}\) oe Reach this statement somewhere	A1
\(u + \mathrm{i}v = \dfrac{1}{x} - \mathrm{i}\) \(w = u + \mathrm{i}v\) and equating real or imaginary parts to obtain either \(u\) or \(v\) in terms of \(x\) or just a (real) number	M1
\(v = -1\) oe \(\left(u = \dfrac{1}{x}\right.\) need not be shown\(\left.\right)\) \(v = -1\) or \(v + 1 = 0\) oe ie equation of the line	A1
NB If \(x + \mathrm{i}y\) has been used for \(z\) and then also for \(w\) allow M1A1M1A0 max.
	(4 marks)

Scheme	Marks
\(z = \dfrac{1}{w + \mathrm{i}} = \dfrac{1}{u + \mathrm{i}v + \mathrm{i}} = \dfrac{u - \mathrm{i}(v + 1)}{u^2 + (v + 1)^2}\) Multiplies numerator and denominator by complex conjugate.	M1
\(\dfrac{u - \mathrm{i}(v + 1)}{u^2 + (v + 1)^2}\)	A1
\((y = 0 \Rightarrow)\ \dfrac{(v + 1)}{u^2 + (v + 1)^2} = 0 \Rightarrow v + 1 = 0\) Uses \(y = 0\) and equates real or imaginary parts to obtain either \(u\) or \(v\) in terms of \(x\) or just a number	M1
\(v = -1\) or \(v + 1 = 0\) oe	A1
NB 1 If \(x + \mathrm{i}y\) has been used for \(z\) and then also for \(w\) allow M1A1M1A0 max.
2. M1A0M1A1 is possible
	(4)

Scheme	Marks
\(\|z + \mathrm{i}\| = \|z - \mathrm{i}\|\)
\(\left\|\dfrac{1}{w + \mathrm{i}} + \mathrm{i}\right\| = \left\|\dfrac{1}{w + \mathrm{i}} - \mathrm{i}\right\|\) M1: Use of real line and attempt to substitute A1: Correct substitution	M1 A1
\(\left\|\dfrac{1 + w\mathrm{i} - 1}{w + \mathrm{i}}\right\| = \left\|\dfrac{1 - w\mathrm{i} + 1}{w + \mathrm{i}}\right\|\)
\(\|w\mathrm{i}\| = \|2 - w\mathrm{i}\|\) Common denominator and equate numerators	M1
\(\|w\| = \|w + 2\mathrm{i}\|\) Equation of the line – any form accepted	A1
	(4)

Scheme	Marks
\(z = \dfrac{1}{w + \mathrm{i}}\)
\(z\) lies on real axis \(\Rightarrow \dfrac{1}{w + \mathrm{i}}\) is real Re-arrange equation and state that \(\dfrac{1}{w + \mathrm{i}}\) is real	M1
\(\Rightarrow w + \mathrm{i}\) is real Deduce that \(w + \mathrm{i}\) is real	A1
\(w = u + \mathrm{i}v,\quad u + \mathrm{i}(v + 1)\) is real Replace \(w\) with \(u + \mathrm{i}v\) (any letters inc \(x + \mathrm{i}y\) allowed here)	M1
\(v + 1 = 0\) Deduce equation of the line	A1
	(4)

Scheme	Marks
Choose any 2 points on the real axis in the \(z\)-plane:
\(z = a:\ w_a = \dfrac{1 - \mathrm{i}a}{a}\) Any one point	M1
\(z = b:\ w_b = \dfrac{1 - \mathrm{i}b}{b}\) Any two points	A1
\(w_a = \dfrac{1}{a} - \mathrm{i} \quad w_b = \dfrac{1}{b} - \mathrm{i}\) Simplify both	M1
\(v = -1\) oe Any letter (inc \(y\)) allowed here	A1

NB The work can be done using arguments to find the equation. If seen, send to review.

(Corrected from the printed mark scheme: in ALT 4 the second point is printed as \(w_a = \dfrac{1 - \mathrm{i}b}{b}\); it is \(w_b\).)