(a) To express \(u\) in the form \(x + iy\), multiply the numerator and denominator by the conjugate of the denominator \(1 - 2i\):

\(u = \frac{(\sqrt{2} - a\sqrt{2}i)(1 - 2i)}{(1 + 2i)(1 - 2i)}\)

The denominator becomes \(1^2 - (2i)^2 = 1 + 4 = 5\).

The numerator becomes \((1 - 2a)\sqrt{2} - (2 + a)\sqrt{2}i\).

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

(b) With \(a = 3\), \(u = \frac{1 - 6}{5}\sqrt{2} + \frac{5}{5}\sqrt{2}i = -\frac{1}{5}\sqrt{2} + \sqrt{2}i\).

Calculate \(r = \sqrt{\left(-\frac{1}{5}\sqrt{2}\right)^2 + (\sqrt{2})^2} = 2\).

Calculate \(\theta = \arctan\left(\frac{\sqrt{2}}{-\frac{1}{5}\sqrt{2}}\right) = -\frac{3}{4}\pi\).

(c) The square roots of \(u\) are \(\sqrt{r}e^{i\theta/2}\) and \(\sqrt{r}e^{i(\theta/2 + \pi)}\).

\(\sqrt{2}e^{\frac{3}{8}\pi i}\) and \(\sqrt{2}e^{\frac{5}{8}\pi i}\).

Let \(w = x + iy\), where \(x\) and \(y\) are real numbers. The conjugate \(w^* = x - iy\).

Substitute into the equation:

\((x + iy)^2 + 2i(x - iy) = 1\)

Expand \((x + iy)^2 = x^2 - y^2 + 2ixy\).

Thus, the equation becomes:

\(x^2 - y^2 + 2ixy + 2ix + 2y = 1\)

Equate real and imaginary parts:

Real: \(x^2 - y^2 + 2y = 1\)

Imaginary: \(2xy + 2x = 0\)

From the imaginary part, factor out \(2x\):

\(2x(y + 1) = 0\)

This gives \(x = 0\) or \(y = -1\).

Case 1: \(x = 0\)

Substitute into the real part:

\(-y^2 + 2y = 1\)

\(y^2 - 2y + 1 = 0\)

\((y - 1)^2 = 0\)

\(y = 1\)

Thus, \(w = i\).

Case 2: \(y = -1\)

Substitute into the real part:

\(x^2 - (-1)^2 + 2(-1) = 1\)

\(x^2 - 1 - 2 = 1\)

\(x^2 = 4\)

\(x = \\pm 2\)

Since \(\text{Re} \, w \leq 0\), \(x = -2\).

Thus, \(w = -2 - i\).

The solutions are \(w = -2 - i\) and \(w = i\).

The inequality \(|z + 2 - 3i| \leq 2\) represents a circle on the Argand diagram with center at \((-2, 3)\) and radius 2.

The inequality \(\text{arg} \, z \leq \frac{3}{4}\pi\) represents the region below the line with argument \(\frac{3}{4}\pi\), which is a half-line starting from the origin and making an angle of \(\frac{3}{4}\pi\) with the positive real axis.

To find the required region, shade the area inside the circle centered at \((-2, 3)\) with radius 2, and below the line \(\text{arg} \, z = \frac{3}{4}\pi\).

\((a) The modulus r of u = -\sqrt{3} + i is calculated as:\)

\(r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2.\)

\(The argument \theta is given by:\)

\(\theta = \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) = \frac{5}{6}\pi.\)

\(Thus, u = 2e^{i\frac{5}{6}\pi}.\)

(b) To find u^6, use De Moivre's Theorem:

\(u^6 = (2e^{i\frac{5}{6}\pi})^6 = 2^6 e^{i(6 \times \frac{5}{6}\pi)} = 64 e^{i5\pi} = 64(-1) = -64.\)

Thus, u^6 is real and equals -64.

(c)(i) On the Argand diagram, draw half-lines from -\sqrt{3} + i at angles 0 and \frac{1}{4}\pi. Also, draw the vertical line x = 2. Shade the region bounded by these lines.

(c)(ii) The greatest value of |z| occurs at the intersection of the line x = 2 and the boundary of the shaded region. Calculate |z| at this point to find the maximum value:

\(The maximum distance from the origin to the line x = 2 is \sqrt{(2 - (-\sqrt{3}))^2 + 1^2} = 5.14.\)

(a) The inequality \(|z - 3 - 2i| \leq 1\) represents a circle on the Argand diagram with center at \((3, 2)\) and radius 1. The inequality \(\text{Im} \, z \geq 2\) represents the region above the line \(y = 2\). The shaded region is the intersection of the circle and the half-plane above \(y = 2\).

(b) To find the greatest value of \(\arg z\), consider the point on the circle where the argument is maximized. This occurs at the point where the tangent to the circle is vertical. The point on the circle with the greatest argument is \((3, 3)\). The argument is given by \(\arg z = \arctan \left( \frac{2}{3} \right) + \sin^{-1} \left( \frac{1}{\sqrt{13}} \right)\). Calculating this gives \(\arg z \approx 49.8^{\circ}\).