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

\(\frac{5(x + iy)}{1 + 2i} - (x + iy)(x - iy) + 30 + 10i = 0\)

Multiply numerator and denominator by the conjugate of the denominator:

\(\frac{5(x + iy)(1 - 2i)}{(1 + 2i)(1 - 2i)} - (x^2 + y^2) + 30 + 10i = 0\)

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

\(\frac{5(x + iy)(1 - 2i)}{5} - x^2 - y^2 + 30 + 10i = 0\)

\(5(x + iy)(1 - 2i) = 5x - 10y + i(5y + 10x)\)

\(x - 2ix + iy + 2y - x^2 - y^2 + 30 + 10i = 0\)

Equate real and imaginary parts:

Real: \(x + 2y - x^2 - y^2 + 30 = 0\)

Imaginary: \(-2x + y + 10 = 0\)

Solve the quadratic equations:

\(x^2 - 9x + 18 = 0\) gives \(x = 3\) or \(x = 6\)

\(y^2 + 2y - 8 = 0\) gives \(y = -4\) or \(y = 2\)

Thus, the solutions are \(3 - 4i\) and \(6 + 2i\).

(a) The region is defined by the argument inequalities \(-\frac{1}{3}\pi \leq \arg(z - 1 - 2i) \leq \frac{1}{3}\pi\), which represent half-lines from the point \((1, 2)\) on the Argand diagram. These lines are symmetrical about the line \(y = 2\) and are drawn between angles \(\frac{\pi}{4}\) and \(\frac{5\pi}{12}\). Additionally, the line \(x = 3\) is drawn, extending in both quadrants. The correct region is shaded between these lines.

(b) To find the least value of \(\arg z\), we use the formula \(-\arctan\left(\frac{2\sqrt{3} - 2}{3}\right)\). Calculating this gives \(-0.454\) radians, correct to 3 decimal places.

To solve the quadratic equation \((1 - 3i)z^2 - (2 + i)z + i = 0\), we use the quadratic formula:

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

where \(a = 1 - 3i\), \(b = -(2 + i)\), and \(c = i\).

First, calculate \(b^2 - 4ac\):

\(b^2 = (-(2 + i))^2 = (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i\)

\(4ac = 4(1 - 3i)(i) = 4(i - 3i^2) = 4(i + 3) = 12 + 4i\)

\(b^2 - 4ac = (3 + 4i) - (12 + 4i) = -9\)

Now, substitute into the quadratic formula:

\(z = \frac{2 + i \pm \sqrt{-9}}{2(1 - 3i)}\)

\(\sqrt{-9} = 3i\), so:

\(z = \frac{2 + i \pm 3i}{2(1 - 3i)}\)

This gives two solutions:

\(z = \frac{2 + 4i}{2(1 - 3i)} \quad \text{and} \quad z = \frac{2 - 2i}{2(1 - 3i)}\)

Multiply numerator and denominator by the conjugate \((1 + 3i)\):

For \(z = \frac{2 + 4i}{2(1 - 3i)}\):

\(z = \frac{(2 + 4i)(1 + 3i)}{2((1 - 3i)(1 + 3i))} = \frac{2 + 6i + 4i + 12i^2}{2(1 + 9)} = \frac{2 + 10i - 12}{20} = \frac{-10 + 10i}{20} = -\frac{1}{2} + \frac{1}{2}i\)

For \(z = \frac{2 - 2i}{2(1 - 3i)}\):

\(z = \frac{(2 - 2i)(1 + 3i)}{2((1 - 3i)(1 + 3i))} = \frac{2 + 6i - 2i - 6i^2}{20} = \frac{2 + 4i + 6}{20} = \frac{8 + 4i}{20} = \frac{2}{5} + \frac{1}{5}i\)

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

(b) To find the greatest value of \(\arg z\), consider the point on the circle \(|z + 2| = 2\) that is tangent to the line \(y = 1\). This point is where the argument is maximized. The argument \(\arg z\) is given by \(\arctan\left(\frac{y}{x}\right)\). The correct point is identified, and the calculation yields \(\frac{11}{12}\pi\), which is approximately 2.88 radians or 165°.

(a) To solve the equation \(z^2 - 6iz - 12 = 0\), use the quadratic formula or complete the square. Completing the square gives \((z - 3i)^2 - 3 = 0\). Solving for \(z\), we find the roots \(z = \sqrt{3} + 3i\) and \(z = -\sqrt{3} + 3i\).

(b) On the Argand diagram, plot the points \(A(\sqrt{3}, 3)\) and \(B(-\sqrt{3}, 3)\) to represent the roots.

(c) The modulus of each root is \(\sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}\). The argument of \(\sqrt{3} + 3i\) is \(\arctan\left(\frac{3}{\sqrt{3}}\right) = \frac{\pi}{3}\), and the argument of \(-\sqrt{3} + 3i\) is \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\).

(d) To show that triangle \(OAB\) is equilateral, note that the distances \(OA\), \(OB\), and \(AB\) are all equal to \(2\sqrt{3}\), and the angles between them are all \(\frac{\pi}{3}\), confirming that \(OAB\) is equilateral.