(i) To express \(\frac{i}{u}\) where \(u = 1 - i\), multiply numerator and denominator by the conjugate of the denominator:

\(\frac{i}{1-i} \times \frac{1+i}{1+i} = \frac{i(1+i)}{(1-i)(1+i)} = \frac{i + i^2}{1^2 - i^2} = \frac{i - 1}{1 + 1} = \frac{i - 1}{2}\)

Thus, \(\frac{i}{u} = -\frac{1}{2} + \frac{1}{2}i\).

(ii) The locus \(|z - u| = |z|\) is the perpendicular bisector of the line segment joining \(u\) to the origin. The locus \(|z - i| = 2\) is a circle centered at \(i\) with radius 2.

(iii) The points of intersection are \(2 + i\) and \(-2 + i\). The argument of \(2 + i\) is \(\arctan(\frac{1}{2}) \approx 0.464\) radians. The argument of \(-2 + i\) is \(-\frac{\pi}{2}\) radians.

(i) Let the square root of u be x + iy. Then, \((x + iy)^2 = -1 + 4\sqrt{3}i\).

Equate real and imaginary parts: \(x^2 - y^2 = -1\) and \(2xy = 4\sqrt{3}\).

From \(2xy = 4\sqrt{3}\), we have \(xy = 2\sqrt{3}\).

Substitute \(y = \frac{2\sqrt{3}}{x}\) into \(x^2 - y^2 = -1\):

\(x^2 - \left(\frac{2\sqrt{3}}{x}\right)^2 = -1\)

\(x^2 - \frac{12}{x^2} = -1\)

Multiply through by \(x^2\):

\(x^4 + x^2 - 12 = 0\)

Solving this quadratic in \(x^2\), we find \(x^2 = 3\) or \(x^2 = -4\).

Thus, \(x = \pm \sqrt{3}\) and \(y = \pm 2\).

The square roots are \(\pm (\sqrt{3} + 2i)\).

(ii) The locus \(|z - u| = 1\) is a circle with center \(-1 + 4\sqrt{3}i\) and radius 1.

The greatest value of \(arg z\) occurs when \(z\) is at the top of the circle.

Calculate the angle from the origin to the top of the circle: \(\arctan\left(\frac{4\sqrt{3} + 1}{-1}\right)\).

Using the mark scheme, the greatest value of \(arg z\) is approximately 1.86 or 106.4°.

(i) Expand \((2-i)^2\) to obtain \(3 - 4i\). Multiply by \(\frac{3 + 4i}{3 + 4i}\) and simplify to \(x + iy\) form to confirm \(w = 2 + 4i\).

(ii) Identify \(4 + 4i\) or \(-4 + 4i\) as points at either end. Use appropriate method to find both critical values of p. State \(-6 \leq p \leq 2\).

(iii) Identify equation as of form \(|z - a| = a\). Form correct equation for a not involving modulus, e.g., \((a-2)^2 + 4^2 = a^2\). State \(|z - 5| = 5\).

(a) Multiply numerator and denominator by \(3 - ai\):

\(z = \frac{(5a - 2i)(3 - ai)}{(3 + ai)(3 - ai)} = \frac{15a - 5ai - 6i + 2ai^2}{9 + a^2}\)

Since \(i^2 = -1\), this becomes:

\(z = \frac{15a + 2a - 6i - 5ai}{9 + a^2} = \frac{(15a + 2a) + (-6 - 5a)i}{9 + a^2}\)

\(z = \frac{13a + 6}{9 + a^2} - i\frac{5a + 6}{9 + a^2}\)

Given \(\arg z = -\frac{1}{4}\pi\), equate:

\(\tan\left(-\frac{1}{4}\pi\right) = \frac{-\frac{5a + 6}{9 + a^2}}{\frac{13a + 6}{9 + a^2}} = -1\)

\(5a + 6 = 13a + 6\)

\(8a = 0\)

\(a = 2\)

Substitute \(a = 2\) back into \(z\):

\(z = \frac{10 - 2i}{3 + 2i}\)

Multiply by conjugate:

\(z = \frac{(10 - 2i)(3 - 2i)}{9 + 4} = \frac{30 - 20i + 4}{13} = 2 - 2i\)

(b) \(z = 2 - 2i\)

\(z^3 = (2 - 2i)^3\)

\(r = |z|^3 = (\sqrt{2^2 + (-2)^2})^3 = (\sqrt{8})^3 = 16\sqrt{2}\)

\(\theta = 3 \times \arg(z) = 3 \times -\frac{\pi}{4} = -\frac{3}{4}\pi\)

(i) First, calculate \(iw = i(5 + 3i) = -3 + 5i\).

Now, divide by \(z = 4 + i\):

\(\frac{-3 + 5i}{4 + i}\).

Multiply numerator and denominator by the conjugate of the denominator:

\(\frac{(-3 + 5i)(4 - i)}{(4 + i)(4 - i)} = \frac{-12 + 3i + 20i - 5i^2}{16 + 1}\).

Simplify: \(\frac{-12 + 23i + 5}{17} = \frac{-7 + 23i}{17}\).

Thus, \(x = -\frac{7}{17}\) and \(y = \frac{23}{17}\).

(ii) Multiply \(w\) and \(z\):

\(w z = (5 + 3i)(4 + i) = 20 + 5i + 12i + 3i^2 = 20 + 17i - 3 = 17 + 17i\).

The argument of \(w\) is \(\arctan \left( \frac{3}{5} \right)\) and the argument of \(z\) is \(\arctan \left( \frac{1}{4} \right)\).

The argument of \(wz\) is the sum of the arguments: \(\arctan \left( \frac{3}{5} \right) + \arctan \left( \frac{1}{4} \right)\).

Since \(wz = 17 + 17i\), its argument is \(\arctan(1) = \frac{\pi}{4}\).

Thus, \(\arctan \left( \frac{3}{5} \right) + \arctan \left( \frac{1}{4} \right) = \frac{\pi}{4}\).