(a) Expand \(iw^2 = (2 - 2i)^2\) to get \(iw^2 = 4 - 8i + 4i^2 = 4 - 8i - 4 = -8i\). Thus, \(w^2 = -8\). Solving for \(w\), we have \(w = \pm i\sqrt{8}\).

(b) (i) The region \(R\) is a circle centered at \((4, 4)\) with radius 2. Sketch this circle on an Argand diagram.

(ii) The modulus \(|z|\) ranges from the distance from the origin to the nearest point on the circle to the distance to the farthest point. The center is at \((4, 4)\), so the distance from the origin to the center is \(\sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}\). The nearest point is \(4\sqrt{2} - 2\) and the farthest is \(4\sqrt{2} + 2\). Thus, \(p = 3.66\) and \(q = 7.66\).

The angles \(\alpha\) and \(\beta\) are found by considering the tangents from the origin to the circle. The angle \(\theta\) is given by \(\sin^{-1}\left(\frac{1}{4\sqrt{2}}\right)\). Thus, \(\alpha = \frac{1}{4}\pi - \sin^{-1}\left(\frac{1}{4\sqrt{2}}\right) \approx 0.424\) and \(\beta = \frac{1}{4}\pi + \sin^{-1}\left(\frac{1}{4\sqrt{2}}\right) \approx 1.15\).

(i) To verify that \(u = 1 + \sqrt{2}i\) is a root of \(p(x) = 0\), substitute \(x = 1 + \sqrt{2}i\) into \(p(x)\):

\(p(1 + \sqrt{2}i) = (1 + \sqrt{2}i)^4 + (1 + \sqrt{2}i)^2 + 2(1 + \sqrt{2}i) + 6\).

Calculate \((1 + \sqrt{2}i)^2 = 1 + 2\sqrt{2}i - 2 = -1 + 2\sqrt{2}i\).

Calculate \((1 + \sqrt{2}i)^4 = ((1 + \sqrt{2}i)^2)^2 = (-1 + 2\sqrt{2}i)^2 = -1 - 4 + 8i^2 = -5 - 8 = -13\).

Substitute back: \(-13 + (-1 + 2\sqrt{2}i) + 2 + 2\sqrt{2}i + 6 = 0\).

Thus, \(u = 1 + \sqrt{2}i\) is a root. The second complex root is \(1 - \sqrt{2}i\).

(ii) To find the other two roots, factor \(p(x)\) using the quadratic factor \(x^2 - 2x + 3\) obtained from the roots \(1 \pm \sqrt{2}i\).

Divide \(p(x)\) by \(x^2 - 2x + 3\) to get a quotient \(x^2 + kx + c\).

Perform polynomial division to find \(x^2 - 2x + 2\) as the other factor.

Find the roots of \(x^2 - 2x + 2 = 0\) using the quadratic formula: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i\).

Thus, the other two roots are \(-1 + i\) and \(-1 - i\).

(a) To solve for u and w, we start with the equations:

\(u - w = 4i\)

\(uw = 5\)

Eliminate w by expressing it in terms of u: \(w = u - 4i\).

Substitute into the second equation: \(u(u - 4i) = 5\).

This gives the quadratic equation: \(u^2 - 4iu - 5 = 0\).

Solving this quadratic equation, we find:

\(u = 1 + 2i\) and \(u = -1 + 2i\).

Substituting back to find w:

If \(u = 1 + 2i\), then \(w = 1 - 2i\).

If \(u = -1 + 2i\), then \(w = -1 - 2i\).

Thus, the solutions are \(u = 1 + 2i, w = 1 - 2i\) or \(u = -1 + 2i, w = -1 - 2i\).

(b)(i) On the Argand diagram, plot the point \(2 - 2i\) and draw a circle centered at \(2 - 2i\) with radius 2.

Draw the line for \(\text{arg } z = -\frac{1}{4}\pi\) and the line \(\text{Re } z = 1\).

Shade the region satisfying all inequalities.

(b)(ii) The greatest possible value of \(\text{Re } z\) for points in the shaded region is \(2 + \sqrt{2}\).

(i) Multiply the numerator and denominator of \(u = \frac{1 + 2i}{1 - 3i}\) by the conjugate of the denominator, \(1 + 3i\):

\(u = \frac{(1 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)}\).

Calculate the denominator: \((1 - 3i)(1 + 3i) = 1^2 - (3i)^2 = 1 + 9 = 10\).

Calculate the numerator: \((1 + 2i)(1 + 3i) = 1 + 3i + 2i + 6i^2 = 1 + 5i - 6 = -5 + 5i\).

Thus, \(u = \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i\).

(ii) On an Argand diagram, plot the points: A at \(-\frac{1}{2} + \frac{1}{2}i\), B at 1 + 2i, and C at 1 - 3i.

(iii) The argument of 1 + 2i is \(\arctan 2\) and the argument of 1 - 3i is \(\arctan 3\).

The argument of \(u\) is \(\arg(1 + 2i) - \arg(1 - 3i) = \arctan 2 - \arctan 3\).

Given \(\arg u = \frac{3}{4} \pi\), we have \(\arctan 2 + \arctan 3 = \frac{3}{4} \pi\).

(i) First, expand \((1 + 2i)^2\):

\((1 + 2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i\).

Now, divide by \(2 + i\) by multiplying the numerator and denominator by the conjugate \(2 - i\):

\(\frac{-3 + 4i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{(-3 + 4i)(2 - i)}{(2 + i)(2 - i)}\).

The denominator is:

\((2 + i)(2 - i) = 4 + 1 = 5\).

The numerator is:

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

Thus, \(u = \frac{-2 + 11i}{5} = -\frac{2}{5} + \frac{11}{5}i\).

(ii) The locus \(|z-u| = |u|\) represents a circle centered at \(u\) with radius \(|u|\). Since \(u = -\frac{2}{5} + \frac{11}{5}i\), the circle is centered at this point and passes through the origin.