(a) To solve the equation \(z^2 - 2piz - q = 0\), use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -2pi\), and \(c = -q\).

Substitute these values to get:

\(z = \frac{2pi \pm \sqrt{(-2pi)^2 - 4 \cdot 1 \cdot (-q)}}{2}\)

\(z = \frac{2pi \pm \sqrt{4p^2 + 4q}}{2}\)

\(z = pi \pm \sqrt{q - p^2}\)

(b) If \(A\) and \(B\) lie on the imaginary axis, their real parts are zero, implying \(q - p^2 < 0\), so \(q < p^2\).

(c) If triangle \(OAB\) is equilateral, the argument of one of the roots is \(\pm 60^\circ\). This implies:

\(\frac{p}{\sqrt{q - p^2}} = \pm \sqrt{3}\)

Squaring both sides gives:

\(\frac{p^2}{q - p^2} = 3\)

\(p^2 = 3(q - p^2)\)

\(p^2 = 3q - 3p^2\)

\(4p^2 = 3q\)

\(q = \frac{4}{3}p^2\)