1. The inequality \(|z| \geq 2\) represents the region outside a circle centered at the origin with radius 2.

2. The inequality \(|z - 1 + i| \leq 1\) represents the region inside a circle centered at \(1 - i\) with radius 1.

3. On the Argand diagram, plot the circle centered at the origin with radius 2.

4. Plot the circle centered at \(1 - i\) with radius 1.

5. Shade the region that is outside the first circle and inside the second circle.

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

\(u - w = 2i\)

\(uw = 6\)

Substitute \(u = w + 2i\) into the second equation:

\((w + 2i)w = 6\)

\(w^2 + 2iw - 6 = 0\)

This is a quadratic equation in w. Solving using the quadratic formula:

\(w = \frac{-2i \pm \sqrt{(2i)^2 + 4 \times 6}}{2}\)

\(w = \frac{-2i \pm \sqrt{-4 + 24}}{2}\)

\(w = \frac{-2i \pm \sqrt{20}}{2}\)

\(w = \frac{-2i \pm 2\sqrt{5}}{2}\)

\(w = -i \pm \sqrt{5}\)

Thus, \(w = \sqrt{5} - i\) or \(w = -\sqrt{5} - i\).

Substituting back to find u:

If \(w = \sqrt{5} - i\), then \(u = w + 2i = \sqrt{5} + i\).

If \(w = -\sqrt{5} - i\), then \(u = w + 2i = -\sqrt{5} + i\).

(b) On the Argand diagram:

1. Plot the point \(2 + 2i\).

2. Draw a circle centered at \(2 + 2i\) with radius 2.

3. Draw a half-line from the origin at an angle of \(45^\circ\) to the positive x-axis.

4. Draw a vertical line at \(\text{Re } z = 3\).

5. Shade the region that satisfies all the inequalities.

(a) Substitute \(w = x + iy\) and \(w^* = x - iy\) into the equation:

\((1 + 2i)(x + iy) + i(x - iy) = 3 + 5i\)

Expand and simplify:

\((1 + 2i)(x + iy) = x + 2ix + iy - 2y = (x - 2y) + i(2x + y)\)

\(i(x - iy) = ix + y\)

Combine terms:

\((x - 2y) + i(2x + y) + ix + y = 3 + 5i\)

Equate real and imaginary parts:

Real: \(x - 2y + y = 3\)

Imaginary: \(2x + y + x = 5\)

Simplify to get two equations:

\(x - y = 3\)

\(3x + y = 5\)

Solve these equations to find \(x = 2\) and \(y = -1\).

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

(b)(ii) The region is a circle centered at \(2 + 2i\) with radius 1, intersecting the line \(\arg(z - 4i) = -\frac{1}{4}\pi\).

The least value of \(\text{Im } z\) is at the lowest point of the circle, which is \(2 - \frac{1}{2}\sqrt{2}\).

(a)(i) To express \(u = \frac{3i}{a + 2i}\) in Cartesian form, multiply the numerator and denominator by the conjugate of the denominator: \(a - 2i\).

\(u = \frac{3i(a - 2i)}{(a + 2i)(a - 2i)} = \frac{3ai - 6i^2}{a^2 + 4} = \frac{3ai + 6}{a^2 + 4}\)

Thus, \(u = \frac{6}{a^2 + 4} + \frac{3ai}{a^2 + 4}\).

(a)(ii) Given \(\arg u^* = \frac{1}{3} \pi\), we need to find \(a\).

\(\arg u = \arg \left( \frac{6}{a^2 + 4} + \frac{3ai}{a^2 + 4} \right) = \arctan \left( \frac{3a}{6} \right) = \frac{1}{3} \pi\).

\(\frac{3a}{6} = \sqrt{3} \Rightarrow a = -2\sqrt{3}\).

(b)(i) The region is defined by the perpendicular bisector of points representing \(2i\) and \(1 + i\), and a circle with radius 2 centered at \(2 + i\).

(b)(ii) The least value of \(\arg z\) is at the critical point \(2 + 3i\), which is 56.3° or 0.983 radians.

(a) To solve for \(v\) and \(w\), use the equations:

1. \(v + iw = 5\)

2. \((1 + 2i)v - w = 3i\)

Substitute \(w = \frac{5 - v}{i}\) into the second equation:

\((1 + 2i)v - \frac{5 - v}{i} = 3i\)

Simplify and solve for \(v\):

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

Multiply numerator and denominator by the conjugate of the denominator:

\(v = -1 - i\)

Substitute \(v\) back to find \(w\):

\(w = 1 - 6i\)

(b)(i) The equation \(|z - 2 - 3i| = 1\) represents a circle with center \(2 + 3i\) and radius 1.

(b)(ii) The least value of \(\arg z\) is calculated using the geometry of the circle. The answer is \(40.2^\circ\) or \(0.702\) radians.