A sequence $ (a_1,b_1)$, $ (a_2,b_2)$, $ (a_3,b_3)$, $ \ldots$ of points in the coordinate plane satisfies \[ (a_{n \plus{} 1}, b_{n \plus{} 1}) \equal{} (\sqrt {3}a_n \minus{} b_n, \sqrt {3}b_n \plus{} a_n)\hspace{3ex}\text{for}\hspace{3ex}  n \equal{} 1,2,3,\ldots.\] Suppose that $ (a_{100},b_{100}) \equal{} (2,4)$.  What is $ a_1 \plus{} b_1$?

$ \textbf{(A)}\\minus{} \frac {1}{2^{97}}  \qquad
\textbf{(B)}\\minus{} \frac {1}{2^{99}} \qquad
\textbf{(C)}\ 0 \qquad
\textbf{(D)}\ \frac {1}{2^{98}}  \qquad
\textbf{(E)}\ \frac {1}{2^{96}}$
1. Define the complex number \( z_n = a_n + i b_n \). The given recurrence relation can be rewritten in terms of \( z_n \):
   \[
   z_{n+1} = (\sqrt{3}a_n - b_n) + i(\sqrt{3}b_n + a_n)
   \]
   This can be expressed as:
   \[
   z_{n+1} = (\sqrt{3} + i)z_n
   \]

2. By induction, we can generalize this to:
   \[
   z_n = z_1 (\sqrt{3} + i)^{n-1}
   \]

3. Given \( z_{100} = 2 + 4i \), we need to find \( z_1 \). Using the general form:
   \[
   z_{100} = z_1 (\sqrt{3} + i)^{99}
   \]

4. To find \( (\sqrt{3} + i)^{99} \), we use De Moivre's Theorem. First, express \( \sqrt{3} + i \) in polar form:
   \[
   \sqrt{3} + i = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)
   \]
   Therefore:
   \[
   (\sqrt{3} + i)^{99} = \left( 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \right)^{99} = 2^{99} \left( \cos \frac{99\pi}{6} + i \sin \frac{99\pi}{6} \right)
   \]

5. Simplify the angle:
   \[
   \frac{99\pi}{6} = 16\pi + \frac{3\pi}{2} = 16\pi + \frac{3\pi}{2} = 8 \cdot 2\pi + \frac{3\pi}{2} = \frac{3\pi}{2}
   \]
   Thus:
   \[
   \cos \frac{3\pi}{2} = 0 \quad \text{and} \quad \sin \frac{3\pi}{2} = -1
   \]
   Therefore:
   \[
   (\sqrt{3} + i)^{99} = 2^{99} (0 - i) = -2^{99} i
   \]

6. Substitute back into the equation for \( z_{100} \):
   \[
   2 + 4i = z_1 (-2^{99} i)
   \]
   Solving for \( z_1 \):
   \[
   z_1 = \frac{2 + 4i}{-2^{99} i} = \frac{2 + 4i}{-2^{99} i} \cdot \frac{-i}{-i} = \frac{(2 + 4i)(-i)}{2^{99}} = \frac{-2i - 4(-1)}{2^{99}} = \frac{-2i + 4}{2^{99}} = 2^{-97} - 2^{-98}i
   \]

7. Finally, we need to find \( a_1 + b_1 \):
   \[
   a_1 + b_1 = 2^{-97} - 2^{-98}
   \]
   Simplify:
   \[
   2^{-97} - 2^{-98} = 2^{-98} (2 - 1) = 2^{-98}
   \]

The final answer is \( \boxed{\frac{1}{2^{98}}} \).