(a) To solve \(iz^2 + 2z - 3i = 0\), use the quadratic formula. Let \(z = x + iy\), then substitute and equate real and imaginary parts to zero.

Use \(i^2 = -1\).

Obtain \(-2xy + 2x = 0\) and \(x^2 - y^2 + 2y - 3 = 0\).

Solve for \(x\) and \(y\) to get \(\sqrt{2} + i\) and \(-\sqrt{2} + i\).

(b)(i) The locus \(|z| = |z - 4 - 3i|\) is the perpendicular bisector of the line segment joining the point \(4 + 3i\) to the origin.

(b)(ii) The point on the locus where \(|z|\) is least is \(2 + 1.5i\).

Calculate the modulus: \(\sqrt{2^2 + 1.5^2} = 2.5\).

Calculate the argument: \(\arctan(1.5/2) \approx 0.64\) radians or \(36.9^{\circ}\).

(a) Let the square root of \(7 - (6\sqrt{2})i\) be \(x + iy\). Then, \((x + iy)^2 = 7 - 6\sqrt{2}i\).

Expanding, we have \(x^2 - y^2 + 2xyi = 7 - 6\sqrt{2}i\).

Equating real and imaginary parts, we get:

\(x^2 - y^2 = 7\)

\(2xy = -6\sqrt{2}\)

From \(2xy = -6\sqrt{2}\), we have \(xy = -3\sqrt{2}\).

Substitute \(y = \frac{-3\sqrt{2}}{x}\) into \(x^2 - y^2 = 7\):

\(x^2 - \left(\frac{-3\sqrt{2}}{x}\right)^2 = 7\)

\(x^2 - \frac{18}{x^2} = 7\)

Multiply through by \(x^2\):

\(x^4 - 7x^2 - 18 = 0\)

Solving this quadratic in \(x^2\), we find \(x^2 = 9\) or \(x^2 = -2\).

Thus, \(x = \pm 3\) and \(y = \pm i\sqrt{2}\).

Therefore, the square roots are \(\pm (3 - i\sqrt{2})\).

(b)(i) The locus \(|w - 1 - 2i| = 1\) is a circle with center \((1, 2)\) and radius 1.

The locus \(\text{arg}(z - 1) = \frac{3}{4}\pi\) is a half-line starting from \((1, 0)\) making an angle of \(\frac{3}{4}\pi\) with the positive real axis.

(b)(ii) The least value of \(|w - z|\) is the perpendicular distance from the center of the circle \((1, 2)\) to the line \(\text{arg}(z - 1) = \frac{3}{4}\pi\).

This distance is \(\sqrt{2} - 1\) (or 0.414).

(a) Let \(z = x + iy\). Then \(z^* = x - iy\). Substitute into the equation:

\(x - iy + 1 = 2i(x + iy)\)

Equate real and imaginary parts:

Real: \(x + 1 = -2y\)

Imaginary: \(-y = 2x\)

From \(-y = 2x\), we have \(y = -2x\).

Substitute \(y = -2x\) into \(x + 1 = -2y\):

\(x + 1 = 4x\)

\(3x = 1\)

\(x = \frac{1}{3}\)

Then \(y = -2x = -\frac{2}{3}\).

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

(b)(i) The region is a circle with center \(-1 + 3i\) and radius 1, intersecting the line \(\text{Im } z = 3\). Shade the region inside the circle and above the line.

(b)(ii) The greatest and least values of \(\arg z\) are determined by the intersection points of the circle and the line. The difference in \(\arg z\) is 0.588 radians (accept 33.7°).

(a) To find \(w\), use the conjugate of \(1 + 3i\):

\(w = \frac{2 + 4i}{1 + 3i} \times \frac{1 - 3i}{1 - 3i} = \frac{7 - i}{5}\)

Square \(w\) to find \(w^2\):

\(w^2 = \left(\frac{7 - i}{5}\right)^2 = \frac{48 - 14i}{25}\)

Find the modulus of \(w^2\):

\(|w^2| = \sqrt{\left(\frac{48}{25}\right)^2 + \left(\frac{-14}{25}\right)^2} = 2\)

Find the argument of \(w^2\):

\(\arg(w^2) = \arctan\left(\frac{-14}{48}\right) = -0.284 \text{ radians or } -16.3^\circ\)

(b) The locus \(|z| = 5\) is a circle centered at the origin with radius 5. The locus \(|z - 5| = |z|\) is a perpendicular bisector of the line segment from the origin to the point (5,0), which is a vertical line through \(x = 2.5\).

The points common to both loci are where the circle intersects the line:

\(z = 5e^{\pm \frac{1}{3} \pi i}\)

(i) Plot the points on the Argand diagram: A at (3, -1), B at (3, 1), and C at (0, 2). The quadrilateral OABC is a parallelogram because opposite sides are parallel and equal in length.

(ii) Substitute \(u = 3 - i\) and \(u^* = 3 + i\). Then \(\frac{u^*}{u} = \frac{3+i}{3-i}\). Multiply numerator and denominator by the conjugate of the denominator: \(\frac{(3+i)(3+i)}{(3-i)(3+i)} = \frac{9 + 6i + i^2}{9 + 1} = \frac{8 + 6i}{10} = \frac{4}{5} + \frac{3}{5}i\).

(iii) The argument of \(\frac{u^*}{u}\) is \(\arg(u^*) - \arg(u) = \arctan\left(\frac{3}{4}\right)\). Using the identity \(\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)\) when \(ab < 1\), set \(a = \frac{1}{3}\) and \(b = \frac{1}{3}\). Then \(2 \arctan\left(\frac{1}{3}\right) = \arctan\left(\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}\right) = \arctan\left(\frac{3}{4}\right)\).