(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 \(36.9^{\circ}\).