The points \((a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n), (a_{n+1}, b_{n+1})\) are given with \((a_{n+1} + i b_{n+1}) = z_{n+1}\), where \(z_{n+1} = (\sqrt{3}\, a_n - b_n) + i(\sqrt{3}\, b_n + a_n)\). If \(z_n = a_n + i b_n\), then \(z_{n+1} = z_n(\sqrt{3} + i) = 2z_n\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right)\). Which option correctly represents the recursive relation?
Let \(a, b \in \mathbb{R}\) and \(a^2 + b^2 \neq 0\). Suppose \(S = \left\{z \in C : z = \dfrac{1}{a + ibt},\, t \in \mathbb{R},\, t \neq 0\right\}\), where \(i = \sqrt{-1}\). If \(z = x + iy\) and \(z \in S\), then \((x, y)\) lies on