Question:

Step 1: Express the given complex number $z_{n+1}$ in terms of $a_n$ and $b_n$. The problem provides the complex numbers $z_n = a_n + i b_n$ and $z_{n+1} = (a_{n+1} + i b_{n+1})$. The transformation relating $z_{n+1}$ to $a_n$ and $b_n$ is given as: $$z_{n+1} = (\sqrt{3}\, a_n - b_n) + i(\sqrt{3}\, b_n + a_n)$$ Step 2: Factor the expression for $z_{n+1}$ to relate it to $z_n$. We can rearrange and factor the terms in the expression for $z_{n+1}$ to identify the complex number $z_n$: $$z_{n+1} = \sqrt{3}\, a_n - b_n + i\sqrt{3}\, b_n + i a_n$$ Group the terms such that $a_n + i b_n$ becomes apparent: $$z_{n+1} = \sqrt{3} a_n + i\sqrt{3} b_n + i a_n - b_n$$ Factor out $\sqrt{3}$ from the first two terms and $i$ from the last two terms (noticing $i^2 = -1$): $$z_{n+1} = \sqrt{3}(a_n + i b_n) + i(a_n + i b_n)$$ Since $z_n = a_n + i b_n$, we can substitute $z_n$ into the equation: $$z_{n+1} = (\sqrt{3} + i)z_n$$ Step 3: Convert the complex number $(\sqrt{3} + i)$ into polar form. To express the complex number $(\sqrt{3} + i)$ in polar form, $r(\cos\theta + i\sin\theta)$, we first calculate its modulus $r$ and argument $\theta$. The modulus $r$ is given by: $$r = |\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$ The argument $\theta$ is found using $\cos\theta = \frac{\text{real part}}{r}$ and $\sin\theta = \frac{\text{imaginary part}}{r}$: $$\cos\theta = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin\theta = \frac{1}{2}$$ The angle $\theta$ in the first quadrant that satisfies these conditions is $\frac{\pi}{6}$. Step 4: Express $(\sqrt{3} + i)$ in trigonometric (polar) form. Using the modulus $r=2$ and argument $\theta=\frac{\pi}{6}$, we can write the complex number $(\sqrt{3} + i)$ in its polar form: $$\sqrt{3} + i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$$ Step 5: Substitute the polar form into the recursive relation for $z_{n+1}$. Now, substitute the polar form of $(\sqrt{3} + i)$ back into the recursive relation $z_{n+1} = (\sqrt{3} + i)z_n$: $$z_{n+1} = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)z_n$$ This can be rewritten as: $$z_{n+1} = 2z_n\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$$ Step 6: State the final recursive relation and identify the correct option. The derived recursive relation is $z_{n+1} = 2z_n\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$. Comparing this expression with the given options, it matches Option 1. The final answer is $\boxed{\text{Option 1}}$.