Putting $z = x + iy$,
$$|x - 3 + iy| = 2|x + i(y + 1)|$$ | M1, A1 | AO3

$$(x - 3)^2 + y^2 = 4\left(x^2 + (y+1)^2\right)$$ | m1 | AO3

$$x^2 - 6x + 9 + y^2 = 4x^2 + 4y^2 + 8y + 4$$

$$3x^2 + 3y^2 + 6x + 8y - 5 = 0$$ | A1 | AO3

This is the equation of a circle.

$$x^2 + y^2 + 2x + \frac{8}{3}y = \frac{5}{3}$$ | M1 | AO1

$$(x + 1)^2 + \left(y + \frac{4}{3}\right)^2 = \frac{5}{3} + 1 + \frac{16}{9}$$ | m1 | AO1

Centre $= \left(-1, -\frac{4}{3}\right)$ | A1 | AO1

Radius $= \sqrt{\frac{5}{3} + 1 + \frac{16}{9}} = 2.11 \left(\frac{2\sqrt{10}}{3}\right)$ | M1, A1 | AO1

| **Total: [9]** |

# Question 6(a)

Reflection matrix $= \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ | B1 | AO1

Translation matrix $= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}$ | B1 | AO1

Rotation matrix $= \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ | B1 | AO1

$$\mathbf{T} = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ | M1 | AO1

$$= \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ | A1 | AO1

$$= \begin{bmatrix} -1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$$ | A1 | AO1

# Question 6(b)

Fixed points satisfy
$$\begin{bmatrix} -1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$$ | M1 | AO1

$-x + 2 = x; y + 1 = y$ | A1 | AO1

The second equation is inconsistent so there are no fixed points. | A1 | AO2

| **Total: [9]** |

# Question 7(a)

Putting $z = x + iy$ and $w = u + iv$,
$$u + iv = (x + iy)(x + 1 + iy)$$ | M1, A1 | AO1

$$v = \text{imaginary part} = y(x + 1) + xy = y(1 + 2x)$$ | A1 | AO1

$$u = \text{real part} = x(x + 1) - y^2$$ | A1 | AO1

# Question 7(b)

$$u = x(x + 1) - (x + 1)^2 = -x - 1$$ | M1, A1 | AO3

$$v = (x + 1)(2x + 1) = -u(-2u - 1) = 2u^2 + u$$ | M1, A1, A1 | AO3

| **Total: [9]** |

# Question 8(a)(i)

$$\mathbf{AB} = (2\mathbf{i} + 3\mathbf{j} + \mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = \mathbf{i} + \mathbf{j} - 2\mathbf{k}$$ | M1, A1 | AO1

# Question 8(a)(ii)

Equation of line is
$$\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + \lambda(\mathbf{i} + \mathbf{j} - 2\mathbf{k})$$ | B1 | AO1

# Question 8(b)(i)

The line cuts the plane where
$$1 + \lambda + 3(2 + \lambda) - 2(3 - 2\lambda) = 5$$ | M1 | AO1

$$\lambda = \frac{1}{2}$$ | A1 | AO1

Point of intersection $= \left(\frac{3}{2}, \frac{5}{2}, 2\right)$ | A1 | AO1

# Question 8(b)(ii)

Direction of normal $= (1, 3, -2)$ | B1 | AO1

If $\theta$ denotes the angle between $L$ and the normal,
$$\cos\theta = \frac{(1,3,-2) \cdot (1,1,-2)}{|(1,3,-2)| |(1,1,-2)|}$$ | M1 | AO1

$$= \frac{8}{\sqrt{14}\sqrt{6}}$$ | A1, A1, A1 | AO1

$$\theta = 29.2(0593...)°$$

Angle between $L$ and plane $= 60.8°$. | A1 | AO1

| **Total: [12]** |