(i) The modulus of \(u = 1 - (\sqrt{3})i\) is given by \(\sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2\).

The argument is \(\arctan\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}\) or \(-60^\circ\).

(ii) To show \(u^3 + 8 = 0\), calculate \(u^3 = (1 - (\sqrt{3})i)^3\).

Using the binomial theorem or De Moivre's theorem, \(u^3 = 2^3 \left(\cos(-\pi) + i\sin(-\pi)\right) = -8\).

Thus, \(u^3 + 8 = -8 + 8 = 0\).

(iii) On the Argand diagram, draw a circle centered at \(1 - (\sqrt{3})i\) with radius 2.

Draw the vertical line \(\text{Re } z = 2\).

Shade the region to the right of the line and within the circle.

(a) Let the square root of \(u\) be \(x + iy\). Then, \((x + iy)^2 = 8 - 15i\).

Expanding, we have \(x^2 - y^2 + 2xyi = 8 - 15i\).

Equating real and imaginary parts gives:

\(x^2 - y^2 = 8\)

\(2xy = -15\)

From \(2xy = -15\), we have \(xy = -\frac{15}{2}\).

Substitute \(y = -\frac{15}{2x}\) into \(x^2 - y^2 = 8\):

\(x^2 - \left(-\frac{15}{2x}\right)^2 = 8\)

\(x^2 - \frac{225}{4x^2} = 8\)

Multiply through by \(4x^2\):

\(4x^4 - 225 = 32x^2\)

\(4x^4 - 32x^2 - 225 = 0\)

Let \(z = x^2\), then:

\(4z^2 - 32z - 225 = 0\)

Solving this quadratic equation gives \(z = \frac{25}{4}\) or \(z = -9\).

Since \(z = x^2\), \(x^2 = \frac{25}{4}\), so \(x = \pm\frac{5}{2}\).

Substitute back to find \(y\):

\(y = -\frac{15}{2x} = \mp\frac{3}{2}\).

Thus, the square roots are \(\pm \frac{1}{\sqrt{2}}(5 - 3i)\).

(b) The inequality \(|z - 2 - i| \leq 2\) represents a circle centered at \(2+i\) with radius 2.

The inequality \(0 \leq \arg(z - i) \leq \frac{1}{4}\pi\) represents the region within an angle \(\frac{1}{4}\pi\) from the positive real axis.

Shade the intersection of these regions on the Argand diagram.

(a) Solve for \(z\) or \(w\):

Use \(i^2 = -1\).

Obtain \(w = \frac{i}{2-i}\) or \(= \frac{2+i}{2-i}\).

Multiply numerator and denominator by the conjugate of the denominator.

Obtain \(w = -\frac{1}{5} + \frac{4}{5}i\).

Obtain \(z = \frac{3}{5} + \frac{4}{5}i\).

(b) EITHER:

Find \(\pm \left[ 2 + (2 - 2\sqrt{3})i \right]\).

Multiply by \(2i\) (or \(-2i\)).

Add result to \(v\).

Obtain answer \(4\sqrt{3} - 1 + 6i\).

OR:

State \(\frac{z-v}{v-u} = ki\), or equivalent.

State \(k = 2\).

Substitute and solve for \(z\) even if \(i\) omitted.

Obtain answer \(4\sqrt{3} - 1 + 6i\).

(i) To find \(a\) and \(b\), substitute \(x = 2 - i\) into the equation \(x^3 + ax^2 - 3x + b = 0\). Calculate \((2-i)^3\) and \((2-i)^2\), and equate real and imaginary parts to zero.

\((2-i)^2 = 4 - 4i + i^2 = 3 - 4i\)

\((2-i)^3 = (2-i)(3-4i) = 6 - 8i - 3i + 4i^2 = 2 - 11i\)

Substitute into the equation: \((2 - 11i) + a(3 - 4i) - 3(2 - i) + b = 0\)

Equate real parts: \(2 + 3a - 6 + b = 0\)

Equate imaginary parts: \(-11 - 4a + 3 = 0\)

Solve these equations to find \(a = -2\) and \(b = 10\).

(ii) On the Argand diagram, draw a circle with center \(2 - i\) and radius 1. Also, draw the perpendicular bisector of the line segment joining \(0\) to \(-i\). Shade the region inside the circle and on the side of the bisector closer to \(0\).

(i) To find u - 2w, calculate:

\(u - 2w = (-1 + 7i) - 2(3 + 4i) = -1 + 7i - 6 - 8i = -7 - i\)

To find \(\frac{u}{w}\), multiply numerator and denominator by the conjugate of the denominator:

\(\frac{u}{w} = \frac{-1 + 7i}{3 + 4i} \times \frac{3 - 4i}{3 - 4i} = \frac{(-1 + 7i)(3 - 4i)}{(3 + 4i)(3 - 4i)}\)

Calculate the denominator:

\((3 + 4i)(3 - 4i) = 9 + 16 = 25\)

Calculate the numerator:

\((-1 + 7i)(3 - 4i) = -3 + 4i + 21i - 28i^2 = -3 + 25i + 28 = 25 + 25i\)

Thus, \(\frac{u}{w} = \frac{25 + 25i}{25} = 1 + i\)

(ii) The argument of \(\frac{u}{w} = 1 + i\) is \(\arctan(1) = \frac{\pi}{4}\).

\((iii) Since u - 2w = -7 - i, OB and CA are parallel and CA = 2OB.\)