Question:

Step 1: Convert the recurrence relation to complex form. Let $z_n = a_n + ib_n$ represent the complex number corresponding to the point $(a_n, b_n)$. The recurrence relation $(a_{n+1}, b_{n+1}) = (\sqrt{3}a_n - b_n, \sqrt{3}b_n + a_n)$ can be rewritten as: $$z_{n+1} = (\sqrt{3}a_n - b_n) + i(\sqrt{3}b_n + a_n)$$ Factoring out the common terms: $$z_{n+1} = \sqrt{3}(a_n + ib_n) + i(a_n + ib_n) = (\sqrt{3} + i)z_n$$ Step 2: Express the multiplier in polar form. We need to find the modulus and argument of $\sqrt{3} + i$. The modulus is: $$|\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$ The argument is: $$\arg(\sqrt{3} + i) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$ Therefore: $$z_{n+1} = 2e^{i\pi/6} \cdot z_n$$ Step 3: Find the general formula for $z_n$. Applying the recurrence relation repeatedly from $z_1$ to $z_{100}$: $$z_{100} = (2e^{i\pi/6})^{99} \cdot z_1 = 2^{99} e^{i \cdot 99\pi/6} \cdot z_1$$ Step 4: Solve for $z_1$ using the given condition. Given that $z_{100} = 2 + 4i$, we can write: $$z_1 = \frac{z_{100}}{2^{99} e^{i \cdot 99\pi/6}} = \frac{2 + 4i}{2^{99} e^{i \cdot 99\pi/6}}$$ Step 5: Simplify the exponential term. We need to reduce the angle $\frac{99\pi}{6}$: $$\frac{99\pi}{6} = \frac{96\pi + 3\pi}{6} = 16\pi + \frac{\pi}{2}$$ Since $e^{i \cdot 2\pi k} = 1$ for any integer $k$: $$e^{i \cdot 99\pi/6} = e^{i(16\pi + \pi/2)} = e^{i\pi/2} = i$$ Step 6: Calculate $z_1$ by dividing by the complex number $2^{99} \cdot i$. $$z_1 = \frac{2 + 4i}{2^{99} \cdot i}$$ To simplify, multiply numerator and denominator by $-i$: $$z_1 = \frac{(2 + 4i)(-i)}{2^{99} \cdot i \cdot (-i)} = \frac{-2i - 4i^2}{2^{99} \cdot (-i^2)} = \frac{-2i + 4}{2^{99} \cdot 1}$$ $$z_1 = \frac{4 - 2i}{2^{99}} = \frac{2(2 - i)}{2^{99}} = \frac{2 - i}{2^{98}}$$ Step 7: Extract the real and imaginary parts and find $a_1 + b_1$. From $z_1 = \frac{2 - i}{2^{98}}$, we have: $$a_1 = \frac{2}{2^{98}}, \quad b_1 = \frac{-1}{2^{98}}$$ Therefore: $$a_1 + b_1 = \frac{2}{2^{98}} + \frac{-1}{2^{98}} = \frac{2 - 1}{2^{98}} = \frac{1}{2^{98}}$$ The value of $(a_1 + b_1) = \frac{1}{2^{98}}$, which corresponds to **Option 3**.