A2 June 2019 Q7

EdexcelCurrent spec6 marksFurther Complex Numbers

7. A transformation from the \(z\)-plane to the \(w\)-plane is given by

\[w = \frac{3\mathrm{i}z - 2}{z + \mathrm{i}} \qquad z \ne -\mathrm{i}\]

(a) Show that the circle \(C\) with equation \(|z + \mathrm{i}| = 1\) in the \(z\)-plane is mapped to a circle \(D\) in the \(w\)-plane, giving a Cartesian equation for \(D\). (4)

(b) Sketch \(C\) and \(D\) on Argand diagrams. (2)

Mark scheme (a)

Scheme	Marks	AO
\(w = \dfrac{3\mathrm{i}z - 2}{z + \mathrm{i}} \Rightarrow z = \dfrac{2 + w\mathrm{i}}{3\mathrm{i} - w}\)	M1	2.1
\(\|z + \mathrm{i}\| = 1 \Rightarrow \left\|\dfrac{2 + w\mathrm{i}}{3\mathrm{i} - w} + \mathrm{i}\right\| = 1\)	M1	2.1
\(\left\|\dfrac{2 + w\mathrm{i} - 3 - \mathrm{i}w}{3\mathrm{i} - w}\right\| = 1 \Rightarrow \left\|\dfrac{1}{w - 3\mathrm{i}}\right\| = 1\)	A1	1.1b
\(\|w - 3\mathrm{i}\| = 1 \Rightarrow u^2 + (v - 3)^2 = 1\)	A1	2.2a
	(4)

Notes

M1: Attempts to solve the problem by attempting to make \(z\) the subject of the formula

M1: Uses the given locus to obtain an equation in \(w\)

A1: Obtains a correct simplified equation in terms of \(w\)

A1: Deduces the correct form of the equation (allow \(x\), \(y\) or \(u\), \(v\) etc.)

Alt 1 (a)

Scheme	Marks	AO
\(w = \dfrac{3\mathrm{i}z - 2}{z + \mathrm{i}} \Rightarrow w(z + \mathrm{i}) = 3\mathrm{i}(z + \mathrm{i}) + 3 - 2\) Attempts to isolate \(z + \mathrm{i}\) terms	M1	2.1
\((z + \mathrm{i})(w - 3\mathrm{i}) = 1 \Rightarrow \|(z + \mathrm{i})(w - 3\mathrm{i})\| = 1 \Rightarrow \|(w - 3\mathrm{i})\| = 1\) Gathers \(z + \mathrm{i}\) terms and applies \(\|(z + \mathrm{i})\| = 1\)	M1	2.1
As main scheme	A1	1.1b
As main scheme	A1	2.2a

Alt 2 (a)

Scheme	Marks	AO
\(w = \dfrac{3\mathrm{i}z - 2}{z + \mathrm{i}} \Rightarrow z = \dfrac{2 + w\mathrm{i}}{3\mathrm{i} - w}\) as main scheme	M1	2.1
\(\displaystyle x + y\mathrm{i} = \frac{2 - v + u\mathrm{i}}{-u - (v - 3)\mathrm{i}} \times \frac{-u + (v - 3)\mathrm{i}}{-u + (v - 3)\mathrm{i}} = \ldots = \frac{u - (u^2 + v^2 - 5v + 6)\mathrm{i}}{u^2 + (v - 3)^2}\) \(\displaystyle \Rightarrow \left(\frac{u}{u^2 + (v - 3)^2}\right)^2 + \left(\frac{-(u^2 + v^2 - 5v + 6)}{u^2 + (v - 3)^2} + 1\right)^2 = 1\) Applies Cartesian coordinates to both sides, extracts \(x\) and \(y\) terms and attempts to apply \(x^2 + (y + 1)^2 = 1\)	M1	2.1
\(\displaystyle \left(\frac{u}{u^2 + (v - 3)^2}\right)^2 + \left(\frac{3 - v}{u^2 + (v - 3)^2}\right)^2 = 1\) Correct expression with \(y + 1\) term combined and simplified.	A1	1.1b
\(\Rightarrow u^2 + (v - 3)^2 = 1\)	A1	2.2a

Alt 3 (a)

Scheme	Marks	AO
\(\displaystyle u + \mathrm{i}v = \frac{3\mathrm{i}(x + \mathrm{i}y) - 2}{x + \mathrm{i}y + \mathrm{i}} \times \frac{x - (y + 1)\mathrm{i}}{x - (y + 1)\mathrm{i}} = \frac{\mathrm{f}(x, y) + \mathrm{g}(x, y)\mathrm{i}}{x^2 + (y + 1)^2}\) Applies Cartesian coordinates to expression and use complex conjugate of denominator to reach Cartesian form.	M1	2.1
\(x^2 + (y + 1)^2 = 1 \Rightarrow u + \mathrm{i}v = x + \left(3x^2 + 3y^2 + 5y + 2\right)\mathrm{i}\) \(\Rightarrow u = x\) and \(v = 3x^2 + 3(y + 1)^2 - y - 1 = a + by\) Uses \(x^2 + (y + 1)^2 = 1\) in their equation and extract \(u\) and \(v\) as linear terms in \(x\) and \(y\)	M1	2.1
\(u = x\) and \(v = 2 - y\) Correct \(u\) and \(v\)	A1	1.1b
\(\Rightarrow u^2 + (2 - v + 1)^2 = 1 \Rightarrow u^2 + (v - 3)^2 = 1\) Uses \(x^2 + (y + 1)^2 = 1\) again to find correct equation.	A1	2.2a

Note that there may be attempts via identifying images of points on a diameter. If seen, send to review.

Sketch: circle C centred at -1 on the imaginary axis, passing through the origin — Mark scheme (b)

Sketch: circle D centred at 3 on the imaginary axis, above the real axis — Mark scheme (b)

Scheme	Marks	AO
	B1	1.1b
	B1ft	1.1b
	(2)
	(6 marks)