(a) Let the square root of \(u\) be \(a + ib\). Then \((a + ib)^2 = -3 - 2\sqrt{10}i\).

Expanding, we have \(a^2 - b^2 + 2abi = -3 - 2\sqrt{10}i\).

Equating real and imaginary parts, we get:

1. \(a^2 - b^2 = -3\)

2. \(2ab = -2\sqrt{10}\)

From equation 2, \(ab = -\sqrt{10}\).

Substitute \(b = -\frac{\sqrt{10}}{a}\) into equation 1:

\(a^2 - \left(-\frac{\sqrt{10}}{a}\right)^2 = -3\)

\(a^2 - \frac{10}{a^2} = -3\)

Multiply through by \(a^2\):

\(a^4 + 3a^2 - 10 = 0\)

Let \(x = a^2\), then \(x^2 + 3x - 10 = 0\).

Solving this quadratic equation using the quadratic formula:

\(x = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-10)}}{2 \times 1}\)

\(x = \frac{-3 \pm \sqrt{49}}{2}\)

\(x = \frac{-3 \pm 7}{2}\)

\(x = 2\) or \(x = -5\) (discard as \(x\) must be positive)

Thus, \(a^2 = 2\) and \(a = \pm \sqrt{2}\).

Substitute back to find \(b\):

\(b = -\frac{\sqrt{10}}{a} = -\frac{\sqrt{10}}{\sqrt{2}} = -\sqrt{5}\)

Thus, the square roots are \(\pm (\sqrt{2} - \sqrt{5}i)\).

(b) The region is defined by:

- A circle centered at \((3, 1)\) with radius 3.

- A half-line from the origin making an angle of \(\frac{1}{4}\pi\) with the positive real axis.

- A horizontal line at \(y = 2\).

The shaded region is the intersection of these conditions on the Argand diagram.