The inequality \(|z + 1 - i| \leq 1\) represents a circle centered at \(-1 + i\) with radius 1. This is because the modulus \(|z + 1 - i|\) is the distance from \(z\) to the point \(-1 + i\).

The inequality \(\arg(z - 1) \leq \frac{3}{4}\pi\) represents the region below the line that makes an angle of \(\frac{3}{4}\pi\) with the positive real axis, starting from the point \(1\).

To solve the problem, draw the circle centered at \(-1 + i\) with radius 1 on the Argand diagram. Then, draw the line starting from \(1\) with an angle of \(\frac{3}{4}\pi\) to the positive real axis. The solution is the region inside the circle and below this line.

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

Substitute these values to get:

\(z = \frac{2pi \pm \sqrt{(-2pi)^2 - 4 \cdot 1 \cdot (-q)}}{2}\)

\(z = \frac{2pi \pm \sqrt{4p^2 + 4q}}{2}\)

\(z = pi \pm \sqrt{q - p^2}\)

(b) If \(A\) and \(B\) lie on the imaginary axis, their real parts are zero, implying \(q - p^2 < 0\), so \(q < p^2\).

(c) If triangle \(OAB\) is equilateral, the argument of one of the roots is \(\pm 60^\circ\). This implies:

\(\frac{p}{\sqrt{q - p^2}} = \pm \sqrt{3}\)

Squaring both sides gives:

\(\frac{p^2}{q - p^2} = 3\)

\(p^2 = 3(q - p^2)\)

\(p^2 = 3q - 3p^2\)

\(4p^2 = 3q\)

\(q = \frac{4}{3}p^2\)

(a) To find \(\frac{u}{v}\), multiply numerator and denominator by the conjugate of the denominator: \(\frac{-4 + 2i}{3 + i} \times \frac{3 - i}{3 - i} = \frac{(-4 + 2i)(3 - i)}{(3 + i)(3 - i)}\).

The denominator becomes \(3^2 - i^2 = 10\).

The numerator becomes \(-12 + 4i + 6i - 2i^2 = -12 + 10i + 2 = -10 + 10i\).

Thus, \(\frac{u}{v} = \frac{-10 + 10i}{10} = -1 + i\).

(b) The modulus \(r\) is \(\sqrt{(-1)^2 + 1^2} = \sqrt{2}\).

The argument \(\theta\) is \(\arctan\left(\frac{1}{-1}\right) = \frac{3}{4}\pi\).

Thus, \(\frac{u}{v} = \sqrt{2}e^{\frac{3}{4}\pi i}\).

\((c) Since C represents 2u + v, we have BC = 2OA. Both OA and BC are parallel.\)

(d) The angle AOB is given by \(\arg u - \arg v = \arg \left(\frac{u}{v}\right) = \frac{3}{4}\pi\).

(a) Multiply numerator and denominator by \(1 + i\):

\(u = \frac{(7+i)(1+i)}{(1-i)(1+i)}\)

\(= \frac{7 + 7i + i + i^2}{1 + i - i - i^2}\)

\(= \frac{7 + 8i - 1}{2}\)

\(= \frac{6 + 8i}{2}\)

\(= 3 + 4i\)

(b) Plot the points on an Argand diagram:

Point A at \((3, 4)\), point B at \((7, 1)\), and point C at \((1, -1)\).

(c) Calculate arguments:

\(\arg(1-i) = -\frac{1}{4}\pi\)

\(\arg(7+i) = \arctan\left(\frac{1}{7}\right)\)

\(\arg(u) = \arg(7+i) - \arg(1-i)\)

\(\arctan\left(\frac{4}{3}\right) = \arctan\left(\frac{1}{7}\right) + \frac{1}{4}\pi\)

(a) Substitute \(x = -1 + \sqrt{5}i\) into the equation:

\(2((-1 + \sqrt{5}i)^3) + ((-1 + \sqrt{5}i)^2) + 6(-1 + \sqrt{5}i) - 18 = 0\).

Calculate \((-1 + \sqrt{5}i)^2 = 1 - 2\sqrt{5}i - 5 = -4 - 2\sqrt{5}i\).

Calculate \((-1 + \sqrt{5}i)^3 = (-1 + \sqrt{5}i)(-4 - 2\sqrt{5}i) = 4 + 2\sqrt{5}i + 4\sqrt{5}i + 10 = 14 + 6\sqrt{5}i\).

Substitute back:

\(2(14 + 6\sqrt{5}i) + (-4 - 2\sqrt{5}i) + 6(-1 + \sqrt{5}i) - 18 = 0\).

Simplify to verify the equation holds.

(b) The conjugate root is \(-1 - \sqrt{5}i\).

Find the quadratic factor with roots \(-1 + \sqrt{5}i\) and \(-1 - \sqrt{5}i\):

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

Divide the original polynomial by \(x^2 + 2x + 6\) to find the remaining root:

\(2x^3 + x^2 + 6x - 18 = (x^2 + 2x + 6)(2x - 3)\).

The remaining root is \(x = \frac{3}{2}\).