Key Concepts and Formulas

• Cube Roots of Unity ($\omega$): $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$. We have $\omega^3 = 1$.
• Polar Form of Complex Numbers: $z = x + iy = r(\cos\theta + i\sin\theta) = re^{i\theta}$, where $r = |z| = \sqrt{x^2+y^2}$ and $\theta = \arg(z)$.
• Powers of $i$: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

Step-by-Step Solution

Step 1: Simplify the Numerator

We aim to express the numerator, $-1 + i\sqrt{3}$, in terms of $\omega$. We can factor out a 2: $-1 + i\sqrt{3} = 2\left(\frac{-1 + i\sqrt{3}}{2}\right) = 2\omega$ This simplifies the numerator considerably.

Step 2: Simplify the Denominator

We have the denominator $1-i$. We want to find $(1-i)^{30}$. It's easier to first compute $(1-i)^2$: $(1-i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i$ Now, we can raise this to the 15th power: $(1-i)^{30} = ((1-i)^2)^{15} = (-2i)^{15}$

Step 3: Substitute and Simplify the Expression

Substitute the simplified numerator and denominator into the original expression: ${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}} = {\left( {{{2\omega } \over {1 - i}}} \right)^{30}} = \frac{(2\omega)^{30}}{(1-i)^{30}} = \frac{(2\omega)^{30}}{(-2i)^{15}}$ Now, we can rewrite the expression as: $\frac{2^{30} \omega^{30}}{(-2)^{15} i^{15}}$

Step 4: Evaluate the Powers of $\omega$ and $i$

We know that $\omega^3 = 1$, so $\omega^{30} = (\omega^3)^{10} = 1^{10} = 1$. For $i^{15}$, we divide 15 by 4: $15 = 4 \cdot 3 + 3$. Therefore, $i^{15} = i^3 = -i$. Substituting these values into the expression, we get: $\frac{2^{30} \cdot 1}{(-2)^{15} (-i)} = \frac{2^{30}}{-2^{15}(-i)} = \frac{2^{30}}{2^{15}i}$

Step 5: Simplify and Rationalize the Denominator

We can simplify the fraction: $\frac{2^{30}}{2^{15}i} = 2^{30-15} \cdot \frac{1}{i} = 2^{15} \cdot \frac{1}{i}$ To rationalize the denominator, multiply by $\frac{-i}{-i}$: $2^{15} \cdot \frac{1}{i} \cdot \frac{-i}{-i} = 2^{15} \cdot \frac{-i}{-i^2} = 2^{15} \cdot \frac{-i}{-(-1)} = 2^{15} \cdot \frac{-i}{1} = -2^{15}i$

Common Mistakes & Tips

• Remember the correct forms for $\omega$ and $\omega^2$. It's easy to mix up the signs.
• When simplifying powers of $i$, divide the exponent by 4 and use the remainder to determine the equivalent power of $i$.
• Be careful with signs, especially when raising negative numbers to odd powers.

Summary

By simplifying the numerator using cube roots of unity and strategically simplifying the denominator by squaring first, we were able to evaluate the complex expression. We then simplified the powers of $\omega$ and $i$, and rationalized the denominator. The final answer is $-2^{15}i$.

Final Answer The final answer is \boxed{-2^{15} i}, which corresponds to option (A).