(i) On the Argand diagram, plot u = 2 + i and u* = 2 - i. The point A represents u, and B represents u*. The point C represents u + u* = 4. The points O, A, B, and C form a parallelogram. Since OA = OB and AC = BC, OACB is a rhombus.

(ii) To express \(\frac{u}{u^*}\), multiply numerator and denominator by the conjugate of the denominator: \(\frac{2+i}{2-i} \times \frac{2+i}{2+i} = \frac{(2+i)^2}{(2-i)(2+i)} = \frac{4 + 4i + i^2}{4 + 1} = \frac{3 + 4i}{5} = \frac{3}{5} + \frac{4}{5}i\).

(iii) The argument of \(\frac{u}{u^*}\) is \(2 \arg(u)\). Since \(\arg(u) = \arctan\left(\frac{1}{2}\right)\), we have \(2 \arg(u) = 2 \arctan\left(\frac{1}{2}\right)\). Therefore, \(\arctan\left(\frac{4}{3}\right) = 2 \arctan\left(\frac{1}{2}\right)\).

(i) To verify that \(1 + 2i\) is a root, substitute \(x = 1 + 2i\) into the equation \(2x^3 + x^2 + 25 = 0\).

Calculate \((1 + 2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i\).

Calculate \((1 + 2i)^3 = (1 + 2i)(-3 + 4i) = -3 + 4i - 6i + 8i^2 = -3 - 2i - 8 = -11 - 2i\).

Substitute into the equation: \(2(-11 - 2i) + (-3 + 4i) + 25 = -22 - 4i - 3 + 4i + 25 = 0\).

Thus, \(1 + 2i\) is verified as a root.

(ii) Since the coefficients of the polynomial are real, the complex roots occur in conjugate pairs. Therefore, the other complex root is \(1 - 2i\).

(iii) On the Argand diagram, plot the point \(1 + 2i\). The equation \(|z| = |z - 1 - 2i|\) represents the perpendicular bisector of the line segment joining the origin \(O\) and the point \(A = 1 + 2i\). This is a straight line passing through the midpoint of \(OA\) and intersecting \(OA\) at right angles.

(i) To solve \(z^2 - 2iz - 5 = 0\), use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -2i\), and \(c = -5\).

Calculate the discriminant: \(b^2 - 4ac = (-2i)^2 - 4(1)(-5) = -4 - (-20) = 16\).

Thus, \(z = \frac{2i \pm \sqrt{16}}{2} = \frac{2i \pm 4}{2}\).

The roots are \(z = 2 + i\) and \(z = -2 + i\).

(ii) The modulus of each root \(z = x + iy\) is \(\sqrt{x^2 + y^2}\).

For \(z = 2 + i\), modulus is \(\sqrt{2^2 + 1^2} = \sqrt{5}\).

For \(z = -2 + i\), modulus is also \(\sqrt{5}\).

The argument of \(z = 2 + i\) is \(\arctan(\frac{1}{2}) \approx 26.6^\circ\).

The argument of \(z = -2 + i\) is \(180^\circ - \arctan(\frac{1}{2}) \approx 153.4^\circ\).

(iii) Plot the points \((2, 1)\) and \((-2, 1)\) on the Argand diagram, representing the roots \(z = 2 + i\) and \(z = -2 + i\).

(i) To find \(u - v\), subtract the complex numbers: \((1 + 3i) - (4 + 2i) = -3 + i\).

To find \(\frac{u}{v}\), multiply numerator and denominator by the conjugate of the denominator: \(\frac{1 + 3i}{4 + 2i} \times \frac{4 - 2i}{4 - 2i} = \frac{(1 + 3i)(4 - 2i)}{(4 + 2i)(4 - 2i)} = \frac{4 - 2i + 12i - 6i^2}{16 + 4} = \frac{10 + 10i}{20} = \frac{1}{2} + \frac{1}{2}i\).

(ii) The argument of \(\frac{u}{v}\) is \(\frac{1}{4}\pi\) radians.

(iii) OC and BA are equal in length and parallel, as they represent the same complex number \(u - v\).

(iv) To prove angle AOB = \(\frac{1}{4}\pi\), note that \(\text{arg}(u) - \text{arg}(v) = \text{arg}\left(\frac{u}{v}\right) = \frac{1}{4}\pi\).

(i) To find the roots of \(z^2 - z + 1 = 0\), use the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = 1\).

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 1 \times 1 = 1 - 4 = -3\).

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

The roots are \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\) and \(\frac{1}{2} - i\frac{\sqrt{3}}{2}\).

(ii) The modulus of each root \(z = x + iy\) is \(\sqrt{x^2 + y^2}\).

For \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\), the modulus is \(\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\).

The argument is \(\arctan\left(\frac{\sqrt{3}/2}{1/2}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3}\).

For \(\frac{1}{2} - i\frac{\sqrt{3}}{2}\), the modulus is also 1, and the argument is \(-\frac{\pi}{3}\).

(iii) To show that each root satisfies \(z^3 = -1\), calculate \(z^3\) for each root.

For \(z = \frac{1}{2} + i\frac{\sqrt{3}}{2}\), \(z^3 = \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^3 = -1\).

Similarly, for \(z = \frac{1}{2} - i\frac{\sqrt{3}}{2}\), \(z^3 = \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)^3 = -1\).