\((i) The complex number w is given by w = -1 + i. First, find w²:\)

\(w² = (-1 + i)² = 1 - 2i - 1 = -2i.\)

\(The modulus of w² is |w²| = |-2i| = 2.\)

\(The argument of w² is arg(w²) = -\frac{\pi}{2}.\)

Next, find w³:

\(w³ = w \cdot w² = (-1 + i)(-2i) = 2 - 2i.\)

\(The modulus of w³ is |w³| = \sqrt{2^2 + (-2)^2} = 2\sqrt{2}.\)

\(The argument of w³ is arg(w³) = \frac{\pi}{4}.\)

(ii) The center of the circle is the midpoint of w and w²:

\(Center = \left(\frac{-1 + 0}{2}, \frac{1 - 2}{2}\right) = \left(-\frac{1}{2}, -\frac{1}{2}\right).\)

The diameter is |w - w²| = |-1 + i + 2i| = |-1 + 3i| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}.

\(The radius is \frac{\sqrt{10}}{2}.\)

\(The equation of the circle is |z + \frac{1}{2} + \frac{1}{2}i| = \frac{1}{2}\sqrt{10}.\)

(a) To find the square roots of the complex number \(1 - 2\sqrt{6}i\), denote the square root as \(x + iy\). Then, \((x + iy)^2 = 1 - 2\sqrt{6}i\).

Equate real and imaginary parts: \(x^2 - y^2 = 1\) and \(2xy = -2\sqrt{6}\).

From \(2xy = -2\sqrt{6}\), we have \(xy = -\sqrt{6}\).

Substitute \(y = -\frac{\sqrt{6}}{x}\) into \(x^2 - y^2 = 1\):

\(x^2 - \left(-\frac{\sqrt{6}}{x}\right)^2 = 1\)

\(x^2 - \frac{6}{x^2} = 1\)

Multiply through by \(x^2\):

\(x^4 - 6 = x^2\)

\(x^4 - x^2 - 6 = 0\)

Let \(u = x^2\), then \(u^2 - u - 6 = 0\).

Factorize: \((u - 3)(u + 2) = 0\)

\(u = 3\) or \(u = -2\) (discard \(u = -2\) as \(x^2\) cannot be negative).

\(x^2 = 3\), so \(x = \pm \sqrt{3}\).

Substitute back to find \(y\):

\(y = -\frac{\sqrt{6}}{x} = -\frac{\sqrt{6}}{\pm \sqrt{3}} = \mp \sqrt{2}\).

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

(b) The inequality \(|z - 3i| \leq 2\) represents a circle centered at \(3i\) with radius 2 on the Argand diagram.

Shade the interior of this circle.

The greatest value of \(\arg z\) occurs at the topmost point of the circle, which is at \(3 + 2i\).

Calculate \(\arg(3 + 2i)\):

\(\arctan\left(\frac{2}{3}\right)\)

\(\arg z = 131.8^\circ\) or \(2.30\) radians.

(i) To find the roots of the equation \(z^2 + (2\sqrt{3})z + 4 = 0\), we use the quadratic formula:

\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1\), \(b = 2\sqrt{3}\), and \(c = 4\).

\(z = \frac{-(2\sqrt{3}) \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}\)

\(z = \frac{-2\sqrt{3} \pm \sqrt{12 - 16}}{2}\)

\(z = \frac{-2\sqrt{3} \pm \sqrt{-4}}{2}\)

\(z = \frac{-2\sqrt{3} \pm 2i}{2}\)

\(z = -\sqrt{3} \pm i\)

(ii) The modulus of each root \(z = x + iy\) is given by \(|z| = \sqrt{x^2 + y^2}\).

For \(z = -\sqrt{3} + i\), \(|z| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2\).

The argument of \(-\sqrt{3} + i\) is \(\arctan\left(\frac{1}{-\sqrt{3}}\right) = 150^\circ\) or \(\frac{5}{6}\pi\) radians.

For \(z = -\sqrt{3} - i\), \(|z| = \sqrt{(-\sqrt{3})^2 + (-1)^2} = 2\).

The argument of \(-\sqrt{3} - i\) is \(\arctan\left(\frac{-1}{-\sqrt{3}}\right) = -150^\circ\) or \(-\frac{5}{6}\pi\) radians.

(iii) To verify that each root satisfies \(z^6 = -64\), we calculate \((-\sqrt{3} + i)^6\) and \((-\sqrt{3} - i)^6\).

Using De Moivre's Theorem, \((re^{i\theta})^n = r^n e^{in\theta}\).

For \(z = -\sqrt{3} + i\), \(r = 2\) and \(\theta = 150^\circ\).

\(z^6 = 2^6 e^{i(6 \times 150^\circ)} = 64 e^{i900^\circ} = 64 e^{i180^\circ} = 64(-1) = -64\).

For \(z = -\sqrt{3} - i\), \(r = 2\) and \(\theta = -150^\circ\).

\(z^6 = 2^6 e^{i(6 \times -150^\circ)} = 64 e^{-i900^\circ} = 64 e^{-i180^\circ} = 64(-1) = -64\).

Both roots satisfy \(z^6 = -64\).

(a)(i) To express \(u\) in the form \(x + iy\), multiply the numerator and denominator by the conjugate of the denominator:

\(u = \frac{5}{a + 2i} \times \frac{a - 2i}{a - 2i} = \frac{5(a - 2i)}{a^2 + 4} = \frac{5a - 10i}{a^2 + 4}\)

Thus, \(x = \frac{5a}{a^2 + 4}\) and \(y = \frac{10}{a^2 + 4}\).

(a)(ii) The argument of \(u^*\) is \(-\arg(u)\). Given \(\arg(u^*) = \frac{3}{4}\pi\), we have \(\arg(u) = -\frac{3}{4}\pi\).

From \(u = \frac{5a}{a^2 + 4} + \frac{10i}{a^2 + 4}\), the argument is \(\arctan\left(\frac{10}{5a}\right) = -\frac{3}{4}\pi\).

Solving \(\tan\left(-\frac{3}{4}\pi\right) = \frac{10}{5a}\), we find \(a = -2\).

(b) The inequality \(|z| < 2\) represents a circle centered at the origin with radius 2. The inequality \(|z| < |z - 2 - 2i|\) represents the region closer to the origin than to the point \(2 + 2i\). This is the region below the perpendicular bisector of the line segment from the origin to \(2 + 2i\). Shade the region inside the circle and below this line.

(i) To find the modulus of \(u\), multiply numerator and denominator by the conjugate of the denominator: \((1 - 2i)\).

\(u = \frac{(6 - 3i)(1 - 2i)}{(1 + 2i)(1 - 2i)} = \frac{6 - 12i - 3i + 6i^2}{1 - 4i^2} = \frac{6 - 12i - 3i - 6}{1 + 4} = \frac{-3i}{5}\)

The modulus of \(u\) is \(\left| \frac{-3i}{5} \right| = \frac{3}{5} \times 5 = 3\).

The argument of \(u\) is \(-\frac{\pi}{2}\) because it lies on the negative imaginary axis.

(ii) For \(\text{arg}(z - u) = \frac{1}{4}\pi\), the line from \(u\) makes an angle of \(\frac{1}{4}\pi\) with the real axis. The least possible value of \(|z|\) is the distance from the origin to this line, which is \(\frac{3}{2}\sqrt{2}\).

(iii) The locus \(|z - (1 + i)u| = 1\) is a circle with center \((1 + i)u\) and radius 1. The greatest possible value of \(|z|\) is the distance from the origin to the furthest point on this circle, which is \(3\sqrt{2} + 1\).