(a) The inequality \(|z - 4 - 3i| \leq 2\) represents a circle centered at \((4, 3)\) with radius 2 on the Argand diagram. The inequality \(\text{Re} \, z \leq 3\) represents a vertical line at \(x = 3\). The region to be shaded is the intersection of the circle and the half-plane to the left of this line.

(b) To find the greatest value of \(\arg z\), consider the tangent to the circle from the origin. The equation of the circle is \((x - 4)^2 + (y - 3)^2 = 4\). The tangent line from the origin has the equation \(y = mx\). Solving for \(m\) using the condition that the line is tangent to the circle gives:

\((1 + m^2)x^2 - (8 + 6m)x + 21 = 0\)

The discriminant of this quadratic must be zero for tangency:

\(48m^2 - 96m + 20 = 0\)

Solving for \(m\), we find:

\(m = \frac{6 \pm \sqrt{21}}{6}\)

The larger value of \(m\) gives the greatest \(\arg z\). The required angle is \(\arctan m\).

Calculating this gives \(\arg z = 1.06\) radians or \(60.45^{\circ}\).

(i) The complex number \(w\) has modulus 1 and argument \(\frac{1}{3} \pi\). Therefore, \(w = \cos \frac{1}{3} \pi + i \sin \frac{1}{3} \pi\).

Calculating, \(\cos \frac{1}{3} \pi = \frac{1}{2}\) and \(\sin \frac{1}{3} \pi = \frac{\sqrt{3}}{2}\).

Thus, \(w = \frac{1}{2} + \frac{\sqrt{3}}{2}i\).

(ii) On the Argand diagram, plot \(u = 1 + 2i\) and \(v\) such that \(|v| = 2|u|\) and \(\arg v = \arg u + \frac{1}{3} \pi\).

(iii) Since \(|v| = 2|u|\), \(v\) can be expressed as \(2uw\) because \(|w| = 1\) and \(\arg v = \arg u + \arg w\).

Calculate \(2uw = 2(1 + 2i)\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\).

First, find \((1 + 2i)\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = \frac{1}{2} + \frac{\sqrt{3}}{2}i + i + \sqrt{3}i^2\).

Since \(i^2 = -1\), this becomes \(\frac{1}{2} + \frac{\sqrt{3}}{2}i + i - \sqrt{3}\).

Simplify to \(\left(\frac{1}{2} - \sqrt{3}\right) + \left(\frac{\sqrt{3}}{2} + 1\right)i\).

Multiply by 2: \(2uw = 1 - 2\sqrt{3} + (2 + \sqrt{3})i\).

(a) Let \(z = x + iy\), then \(z^* = x - iy\). Substitute into the equation:

\(x + iy + \frac{i(x + iy)}{x - iy} - 2 = 0.\)

Multiply through by \(x - iy\) to clear the fraction:

\((x + iy)(x - iy) + i(x + iy) - 2(x - iy) = 0.\)

Expanding and simplifying using \(i^2 = -1\):

\(x^2 + y^2 + ix - iy - 2x + 2iy = 0.\)

Separate real and imaginary parts:

Real: \(x^2 + y^2 - 2x = 0\)

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

From \(x + 2y = 0\), solve for \(x\):

\(x = -2y\)

Substitute into the real equation:

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

\(4y^2 + y^2 + 4y = 0\)

\(5y^2 + 4y = 0\)

\(y(5y + 4) = 0\)

\(y = 0\) or \(y = -\frac{4}{5}\)

If \(y = -\frac{4}{5}\), then \(x = \frac{8}{5}\).

Thus, \(z = \frac{6}{5} - \frac{3}{5}i\).

(b)(i) The locus \(|z - 2i| = 2\) is a circle with center \(2i\) and radius 2. The locus \(\text{Im} \, z = 3\) is a horizontal line at \(y = 3\).

(b)(ii) The intersection point \(P\) is \(\sqrt{3} + 3i\). The argument is:

\(\text{arg}(P) = \arctan\left(\frac{3}{\sqrt{3}}\right) = \frac{\pi}{3}\)

(a) Let the square root of \(u\) be \(a + ib\). Then \((a + ib)^2 = -3 - 2\sqrt{10}i\).

Expanding, we have \(a^2 - b^2 + 2abi = -3 - 2\sqrt{10}i\).

Equating real and imaginary parts, we get:

1. \(a^2 - b^2 = -3\)

2. \(2ab = -2\sqrt{10}\)

From equation 2, \(ab = -\sqrt{10}\).

Substitute \(b = -\frac{\sqrt{10}}{a}\) into equation 1:

\(a^2 - \left(-\frac{\sqrt{10}}{a}\right)^2 = -3\)

\(a^2 - \frac{10}{a^2} = -3\)

Multiply through by \(a^2\):

\(a^4 + 3a^2 - 10 = 0\)

Let \(x = a^2\), then \(x^2 + 3x - 10 = 0\).

Solving this quadratic equation using the quadratic formula:

\(x = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-10)}}{2 \times 1}\)

\(x = \frac{-3 \pm \sqrt{49}}{2}\)

\(x = \frac{-3 \pm 7}{2}\)

\(x = 2\) or \(x = -5\) (discard as \(x\) must be positive)

Thus, \(a^2 = 2\) and \(a = \pm \sqrt{2}\).

Substitute back to find \(b\):

\(b = -\frac{\sqrt{10}}{a} = -\frac{\sqrt{10}}{\sqrt{2}} = -\sqrt{5}\)

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

(b) The region is defined by:

- A circle centered at \((3, 1)\) with radius 3.

- A half-line from the origin making an angle of \(\frac{1}{4}\pi\) with the positive real axis.

- A horizontal line at \(y = 2\).

The shaded region is the intersection of these conditions on the Argand diagram.

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

\(u = \frac{4i}{1 - \sqrt{3}i} \times \frac{1 + \sqrt{3}i}{1 + \sqrt{3}i} = \frac{4i(1 + \sqrt{3}i)}{(1)^2 - (\sqrt{3}i)^2}\)

\(= \frac{4i + 4\sqrt{3}i^2}{1 + 3} = \frac{4i - 4\sqrt{3}}{4} = -\sqrt{3} + i\)

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

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

(iii) On the Argand diagram, draw a circle centered at the origin with radius 2. Plot \(u = -\sqrt{3} + i\) in the correct position. Draw the perpendicular bisector of the line joining \(u\) and the origin. Shade the region satisfying \(|z| < 2\) and \(|z - u| < |z|\).