To solve the quadratic equation \((3+i)w^2 - 2w + 3 - i = 0\), we use the quadratic formula:

\(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 3+i\), \(b = -2\), and \(c = 3-i\).

First, calculate the discriminant:

\(b^2 - 4ac = (-2)^2 - 4(3+i)(3-i)\)

\(= 4 - 4((3+i)(3-i))\)

\(= 4 - 4(9 - i^2)\)

\(= 4 - 4(9 + 1)\)

\(= 4 - 40\)

\(= -36\)

Now, substitute into the quadratic formula:

\(w = \frac{-(-2) \pm \sqrt{-36}}{2(3+i)}\)

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

Multiply numerator and denominator by the conjugate of the denominator:

\(w = \frac{(2 \pm 6i)(6-2i)}{(6+2i)(6-2i)}\)

\(= \frac{12 \mp 4i + 36i \mp 12i^2}{36 + 4}\)

\(= \frac{12 \mp 4i + 36i \pm 12}{40}\)

\(= \frac{24 \pm 32i}{40}\)

\(= \frac{3}{5} \pm \frac{4}{5}i\)

Thus, the solutions are \(w = \frac{3}{5} + \frac{4}{5}i\) and \(w = -i\).

(a) The equation \(|z + 3 - 2i| = 2\) represents a circle on the Argand diagram with center at \((-3, 2)\) and radius 2. To sketch this, plot the center at \((-3, 2)\) and draw a circle with radius 2.

(b) To find the least value of \(|z|\), calculate the distance from the origin \((0, 0)\) to the center \((-3, 2)\), which is \(\sqrt{(-3)^2 + 2^2} = \sqrt{13}\). The least value of \(|z|\) is the distance from the origin to the nearest point on the circle, which is \(\sqrt{13} - 2\).

(a) (i) To express \(z\) in the form \(x + iy\), multiply the numerator and denominator by the conjugate of the denominator:

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

\(= \frac{4 + 8i - 3i - 6i^2}{1 + 4}\)

\(= \frac{10 + 5i}{5}\)

\(= 2 + i\)

(ii) The modulus of \(z\) is \(\sqrt{2^2 + 1^2} = \sqrt{5}\) or 2.24.

The argument of \(z\) is \(\arctan\left(\frac{1}{2}\right) \approx 0.464\) radians or 26.6°.

(b) To find the square roots of \(5 - 12i\), let \(x + iy\) be a root. Then:

\((x + iy)^2 = 5 - 12i\)

Equate real and imaginary parts:

\(x^2 - y^2 = 5\)

\(2xy = -12\)

From \(2xy = -12\), \(xy = -6\).

Substitute \(y = -\frac{6}{x}\) into \(x^2 - y^2 = 5\):

\(x^2 - \left(-\frac{6}{x}\right)^2 = 5\)

\(x^4 - 36 = 5x^2\)

\(x^4 - 5x^2 - 36 = 0\)

Let \(u = x^2\), then \(u^2 - 5u - 36 = 0\).

Solving this quadratic gives \(u = 9\) or \(u = -4\).

Thus, \(x^2 = 9\) gives \(x = 3\) or \(x = -3\).

For \(x = 3\), \(y = -2\); for \(x = -3\), \(y = 2\).

So the roots are \(3 - 2i\) and \(-3 + 2i\).

To find \(u\), multiply the numerator and denominator by the conjugate of the denominator:

\(u = \frac{2}{-1+i} \times \frac{-1-i}{-1-i} = \frac{-2 - 2i}{1+1} = -1-i\)

The modulus of \(u\) is \(|u| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}\).

The argument of \(u\) is \(\text{arg}(u) = \arctan\left(\frac{-1}{-1}\right) = -\frac{3}{4}\pi\) or \(-135^\circ\).

For \(u^2\):

\(u^2 = (-1-i)^2 = 1 + 2i + i^2 = -1 - 2i\)

The modulus of \(u^2\) is \(|u^2| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{5}\).

The argument of \(u^2\) is \(\text{arg}(u^2) = \arctan\left(\frac{-2}{-1}\right) = \frac{1}{2}\pi\) or \(90^\circ\).

For the Argand diagram, plot \(u = -1-i\) and \(u^2 = -1-2i\). Draw a circle centered at the origin with radius 2. The line \(|z-u^2| < |z-u|\) is the perpendicular bisector of the line segment joining \(u\) and \(u^2\). Shade the region inside the circle and on the side of the bisector closer to \(u^2\).

(i) Multiply the numerator and denominator of \(\frac{3+i}{2-i}\) by the conjugate of the denominator, \(2+i\):

\(\frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6 + 3i + 2i + i^2}{4 + 1} = \frac{5 + 5i}{5} = 1 + i\)

Thus, \(u = 1 + i\).

(ii) The modulus of \(u = 1 + i\) is \(\sqrt{1^2 + 1^2} = \sqrt{2}\).

The argument of \(u = 1 + i\) is \(\arctan\left(\frac{1}{1}\right) = 45^{\circ}\) or \(\frac{\pi}{4}\).

(iii) On the Argand diagram, plot the point \(u = 1 + i\). Draw a circle centered at \(u\) with radius 1, representing the locus \(|z-u| = 1\).

(iv) The least value of \(|z|\) for points on this locus is the distance from the origin to the closest point on the circle. The center of the circle is at \(1 + i\) with radius 1, so the least distance is \(\sqrt{2} - 1\).