(a) The given equation is \((1 + i)z^2 - (4 + 3i)z + 5 + i = 0\).

Use the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1+i\), \(b = -(4+3i)\), \(c = 5+i\).

Calculate \(b^2 = (4+3i)^2 = 16 + 24i + 9i^2 = 16 + 24i - 9 = 7 + 24i\).

Calculate \(4ac = 4(1+i)(5+i) = 4(5 + 6i - 1) = 16 + 24i\).

Then \(b^2 - 4ac = (7 + 24i) - (16 + 24i) = -9\).

\(\sqrt{-9} = 3i\).

Substitute into the quadratic formula:

\(z = \frac{4+3i \pm 3i}{2(1+i)}\).

\(z = \frac{4+6i}{2+2i}\) or \(z = \frac{4}{2+2i}\).

Multiply numerator and denominator by \(1-i\):

\(z = \frac{(4+6i)(1-i)}{2} = 1-i\).

\(z = \frac{4(1-i)}{2} = \frac{5}{2} + \frac{1}{2}i\).

(b) Plot the point \(u = -1 - i\) on the Argand diagram.

Draw the horizontal line through \(z = i\).

Draw half-lines from \(u\) with gradient 1 and vertical.

Shade the region satisfying \(|z| < |z - 2i|\) and \(\frac{1}{4}\pi < \text{arg}(z - u) < \frac{1}{2}\pi\).