(i) Substitute \(z = 1 + i\) into the equation:

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

Multiply numerator and denominator by the conjugate of the denominator:

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

(ii) Substitute \(w = z\) into the equation:

\(z = \frac{z + i}{iz + 2}\).

Cross-multiply to obtain:

\(z(iz + 2) = z + i\).

\(iz^2 + 2z = z + i\).

\(iz^2 + z - i = 0\).

Substitute \(z = x + iy\) and equate real and imaginary parts:

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

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

Solve these equations to find \(x = -\frac{\sqrt{3}}{2}\) and \(y = \frac{1}{2}\).

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

(a) To express \(u = \frac{3 - 5i}{1 + 4i}\) in the form \(x + iy\), multiply the numerator and denominator by the conjugate of the denominator:

\(u = \frac{(3 - 5i)(1 - 4i)}{(1 + 4i)(1 - 4i)}\)

Calculate the denominator: \((1 + 4i)(1 - 4i) = 1 - 16i^2 = 1 + 16 = 17\).

Calculate the numerator: \((3 - 5i)(1 - 4i) = 3 - 12i - 5i + 20i^2 = 3 - 17i - 20 = -17 - 17i\).

Thus, \(u = \frac{-17 - 17i}{17} = -1 - i\).

(b)(i) On the Argand diagram, plot the point \(2 + i\) and draw a circle with center \(2 + i\) and radius 1. The inequality \(|z - 2 - i| \leq 1\) represents this circle.

The inequality \(|z - i| \leq |z - 2|\) represents the region on or closer to the line perpendicular bisector of the segment joining \(i\) and \(2\). Shade the region inside the circle and on the correct side of the bisector.

(b)(ii) The maximum value of \(\arg z\) is the angle between the tangents from the origin to the circle. This angle is \(0.927\) radians (or \(53.1^\circ\)).

(a) Since \(-1 + \sqrt{5}i\) is a root, its complex conjugate \(-1 - \sqrt{5}i\) is also a root because the coefficients of the polynomial are real. Let the third root be \(r\). By Vieta's formulas, the sum of the roots is zero:

\((-1 + \sqrt{5}i) + (-1 - \sqrt{5}i) + r = 0\)

\(-2 + r = 0\)

\(r = 2\)

The polynomial can be expressed as \((z + 1 - \sqrt{5}i)(z + 1 + \sqrt{5}i)(z - 2)\).

Expanding \((z + 1 - \sqrt{5}i)(z + 1 + \sqrt{5}i)\) gives:

\((z + 1)^2 - (\sqrt{5}i)^2 = z^2 + 2z + 1 + 5 = z^2 + 2z + 6\)

Thus, the polynomial is \((z^2 + 2z + 6)(z - 2)\).

Expanding gives:

\(z^3 - 2z^2 + 2z^2 - 4z + 6z - 12 = z^3 + 2z - 12\)

Comparing with \(z^3 + 2z + a = 0\), we find \(a = -12\).

(b) Let \(w = e^{i2\theta}\). Then:

\(\frac{w-1}{w+1} = \frac{e^{i2\theta} - 1}{e^{i2\theta} + 1}\)

Multiply numerator and denominator by \(e^{-i\theta}\):

\(\frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = \frac{2i \sin \theta}{2 \cos \theta} = i \tan \theta\)

(i) Multiply the numerator and denominator by \(\sqrt{3} + i\) to rationalize the denominator:

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

\(= \frac{18 + 18\sqrt{3}i}{4}\)

\(= \frac{9}{2} + \frac{9}{2}\sqrt{3}i\)

Find the modulus \(r\) and argument \(\theta\):

\(r = \sqrt{\left(\frac{9}{2}\right)^2 + \left(\frac{9}{2}\sqrt{3}\right)^2} = 9\)

\(\theta = \arctan\left(\frac{\frac{9}{2}\sqrt{3}}{\frac{9}{2}}\right) = \frac{\pi}{3}\)

Thus, \(z = 9e^{i\frac{\pi}{3}}\).

(ii) The square roots of \(z\) are given by \(\sqrt{9}e^{i\frac{\pi}{6}}\) and \(\sqrt{9}e^{i\left(\frac{\pi}{3} - \pi\right)}\).

\(= 3e^{i\frac{\pi}{6}}\) and \(3e^{-i\frac{5\pi}{6}}\).

(a) The quadratic equation is \((2 - i)z^2 + 2z + 2 + i = 0\). Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2 - i\), \(b = 2\), and \(c = 2 + i\), we find:

\(z = \frac{-2 \pm \sqrt{2^2 - 4(2-i)(2+i)}}{2(2-i)}\)

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

\(z = \frac{-2 \pm \sqrt{4 - 4(4 + 1)}}{2(2-i)}\)

\(z = \frac{-2 \pm \sqrt{4 - 20}}{2(2-i)}\)

\(z = \frac{-2 \pm \sqrt{-16}}{2(2-i)}\)

\(z = \frac{-2 \pm 4i}{2(2-i)}\)

\(z = \frac{-2 \pm 4i}{4 - 2i}\)

Multiply numerator and denominator by the conjugate of the denominator:

\(z = \frac{(-2 \pm 4i)(2+i)}{(4-2i)(2+i)}\)

\(z = \frac{-4 - 2i \pm 8i + 4i^2}{8 + 2i - 2i - 2i^2}\)

\(z = \frac{-4 - 2i \pm 8i - 4}{8 + 2}\)

\(z = \frac{-8 + 6i}{10}\)

\(z = -\frac{4}{5} + \frac{3}{5}i\)

For the other root:

\(z = \frac{-2 - 4i}{4 - 2i}\)

Multiply numerator and denominator by the conjugate of the denominator:

\(z = \frac{(-2 - 4i)(2+i)}{(4-2i)(2+i)}\)

\(z = \frac{-4 - 8i - 2i - 4i^2}{8 + 2i - 2i - 2i^2}\)

\(z = \frac{-4 - 8i - 2i + 4}{8 + 2}\)

\(z = \frac{-10i}{10}\)

\(z = -i\)

(b) The complex number \(w = 2e^{\frac{1}{4}\pi i}\) has modulus 2 and argument \(\frac{\pi}{4}\). Thus, \(w = 2(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}) = \sqrt{2} + i\sqrt{2}\).

\(w^3 = (2e^{\frac{1}{4}\pi i})^3 = 8e^{\frac{3}{4}\pi i}\) has modulus 8 and argument \(\frac{3\pi}{4}\).

\(w^* = \sqrt{2} - i\sqrt{2}\).

To find the area of triangle \(ABC\), use the formula for the area of a triangle in the complex plane: \(\text{Area} = \frac{1}{2} |\text{Im}(w \cdot \overline{w^3} + w^3 \cdot \overline{w^*} + w^* \cdot \overline{w})|\).

Calculate the area: \(\text{Area} = 10\).