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Step 1: Identify the Complex Number $\alpha$
We are given $\alpha = \frac{-1 + i \sqrt{3}}{2}$. This is commonly denoted by $\omega$, a non-real cube root of unity, satisfying $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Step 2: Express the Given Expression in Terms of $\omega$
Rewrite the problem as:
$(2 + \omega)^4 = a + b \,\omega$.
Our goal is to find $a + b$.
Step 3: Expand $(2 + \omega)^4$ Using the Binomial Theorem
By the binomial theorem:
$(2 + \omega)^4
= \sum_{k=0}^{4} \binom{4}{k} \, 2^{4-k} \,\omega^k
= \binom{4}{0}2^4 \omega^0 + \binom{4}{1}2^3 \omega^1 + \binom{4}{2}2^2 \omega^2 + \binom{4}{3}2^1 \omega^3 + \binom{4}{4}2^0 \omega^4.$
Substitute the binomial coefficients:
$= 16 + 32\omega + 24\omega^2 + 8\omega^3 + \omega^4.$
Step 4: Simplify Powers of $\omega$
Since $\omega^3 = 1$, we have $\omega^4 = \omega \cdot \omega^3 = \omega \cdot 1 = \omega.$
Thus,
$(2 + \omega)^4 = 16 + 32\omega + 24\omega^2 + 8 \cdot 1 + \omega.$
Combine like terms:
$= 24 + 32\omega + \omega + 24\omega^2
= 24 + 33\omega + 24\omega^2.$
Step 5: Use the Relation $1 + \omega + \omega^2 = 0$
From $1 + \omega + \omega^2 = 0,$ we get
$\omega^2 = -1 - \omega.$
Hence,
$24\omega^2 = 24(-1 - \omega) = -24 - 24\omega.$
So,
$24 + 33\omega + 24\omega^2
= 24 + 33\omega + \bigl(-24 - 24\omega\bigr)
= (24 - 24) + (33\omega - 24\omega)
= 0 + 9\omega
= 9\omega.$
Step 6: Compare to Identify $a$ and $b$
We have $(2 + \omega)^4 = 9\omega.$ Thus it must match $a + b\omega.$ By comparison,
$a = 0, \quad b = 9.
Step 7: Find $a + b$
$a + b = 0 + 9 = 9.$
Therefore, the required value of $a + b$ is $\boxed{9}.$
