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

\(x + iy + \frac{i(x + iy)}{x - iy} - 2 = 0.\)

Multiply through by \(x - iy\) to clear the fraction:

\((x + iy)(x - iy) + i(x + iy) - 2(x - iy) = 0.\)

Expanding and simplifying using \(i^2 = -1\):

\(x^2 + y^2 + ix - iy - 2x + 2iy = 0.\)

Separate real and imaginary parts:

Real: \(x^2 + y^2 - 2x = 0\)

Imaginary: \(x + 2y = 0\)

From \(x + 2y = 0\), solve for \(x\):

\(x = -2y\)

Substitute into the real equation:

\((-2y)^2 + y^2 - 2(-2y) = 0\)

\(4y^2 + y^2 + 4y = 0\)

\(5y^2 + 4y = 0\)

\(y(5y + 4) = 0\)

\(y = 0\) or \(y = -\frac{4}{5}\)

If \(y = -\frac{4}{5}\), then \(x = \frac{8}{5}\).

Thus, \(z = \frac{6}{5} - \frac{3}{5}i\).

(b)(i) The locus \(|z - 2i| = 2\) is a circle with center \(2i\) and radius 2. The locus \(\text{Im} \, z = 3\) is a horizontal line at \(y = 3\).

(b)(ii) The intersection point \(P\) is \(\sqrt{3} + 3i\). The argument is:

\(\text{arg}(P) = \arctan\left(\frac{3}{\sqrt{3}}\right) = \frac{\pi}{3}\)