a) Express \( z \) in the form \( r e^{i \theta} \), where \( -\pi < \theta \leq \pi \):

Given: \( z = -1 + i \).

Modulus: \( r = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \).

Argument: \( \tan \theta = \frac{\text{Im}}{\text{Re}} = \frac{1}{-1} = -1 \). Since \( z \) is in the second quadrant (\( x < 0 \), \( y > 0 \)), \( \theta = \pi – \tan^{-1}(1) = \pi – \frac{\pi}{4} = \frac{3\pi}{4} \).

Check: \( -\pi < \frac{3\pi}{4} \approx 2.3562 \leq \pi \).

Thus: \( z = \sqrt{2} e^{i \frac{3\pi}{4}} \). [2]

Final Answer: \( z = \sqrt{2} e^{i \frac{3\pi}{4}} \). [2]

b) (i) Describe the mapping of A onto B, stating the scale factor and angle of rotation:

Points: A represents \( z = -1 + i \), B represents \( z^2 \).

Compute: \( z^2 = (-1 + i)^2 = 1 – 2i + i^2 = 1 – 2i – 1 = -2i \).

Modulus: \( |z| = \sqrt{2} \), \( |z^2| = |-2i| = 2 \). Scale factor: \( \frac{|z^2|}{|z|} = \frac{2}{\sqrt{2}} = \sqrt{2} \).

Argument: \( \arg(z) = \frac{3\pi}{4} \), \( \arg(z^2) = \arg(-2i) = -\frac{\pi}{2} \) (fourth quadrant).

Rotation angle: \( \arg(z^2) – \arg(z) = -\frac{\pi}{2} – \frac{3\pi}{4} = -\frac{2\pi}{4} – \frac{3\pi}{4} = -\frac{5\pi}{4} \). Adjust to \( -\pi < \theta \leq \pi \): \( -\frac{5\pi}{4} + 2\pi = \frac{3\pi}{4} \).

Thus, mapping is an enlargement by scale factor \( \sqrt{2} \), followed by a rotation of \( \frac{3\pi}{4} \) radians counterclockwise about the origin. [1]

Final Answer: Enlargement by scale factor \( \sqrt{2} \), rotation by \( \frac{3\pi}{4} \) radians counterclockwise about the origin. [2]

b) (ii) Find and simplify the matrix that maps A onto B:

Enlargement matrix (scale factor \( \sqrt{2} \)): \( \begin{pmatrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{pmatrix} \).

Rotation matrix (\( \theta = \frac{3\pi}{4} \)): \( \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \), \( \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \).

Rotation matrix: \( \begin{pmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{pmatrix} \).

Combined (rotation then enlargement): \( \begin{pmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 1 & -1 \end{pmatrix} \).

Verify: \( \begin{pmatrix} -1 & -1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 – 1 \\ -1 – 1 \end{pmatrix} = \begin{pmatrix} 0 \\ -2 \end{pmatrix} \), matches \( z^2 = -2i \). [1]

Final Answer: \( \begin{pmatrix} -1 & -1 \\ 1 & -1 \end{pmatrix} \). [2]

c) Find the smallest positive integer \( n \) for which \( z^n \) is real and positive:

From part (a): \( z = \sqrt{2} e^{i \frac{3\pi}{4}} \).

\( z^n = (\sqrt{2})^n e^{i \frac{3n\pi}{4}} \).

For \( z^n \) to be real and positive: argument must be \( 2k\pi \), and modulus \( (\sqrt{2})^n > 0 \).

Solve: \( \frac{3n\pi}{4} = 2k\pi \).

\( \frac{3n}{4} = 2k \).

\( n = \frac{8k}{3} \).

For integer \( n \), \( k = 3m \), so \( n = 8m \).

Smallest positive \( n \): \( m = 1 \), \( n = 8 \).

Verify: \( z^8 = (\sqrt{2})^8 e^{i \frac{3 \cdot 8 \pi}{4}} = 16 e^{i 6\pi} = 16 \cdot 1 = 16 \), real and positive.

Smaller \( n \): \( n = 4 \), \( z^4 = (\sqrt{2})^4 e^{i \frac{3 \cdot 4 \pi}{4}} = 4 e^{i 3\pi} = 4(-1) = -4 \), not positive.

Final Answer: \( n = 8 \). [3]

Total: [9 marks]