What is \( \left( \frac{\sqrt{3}+i}{\sqrt{3}-i} \right)^3 \) equal to?

Concept:

The key concept involves simplifying complex numbers in the polar form. We use the argument (angle) of the complex number and De Moivre's theorem, which states:

  (r(cosθ + i sinθ))ⁿ = rⁿ(cos(nθ) + i sin(nθ))

Calculation:

⇒ \(\left( \frac{\sqrt{3} + i}{\sqrt{3} - i} \right)^3 \)

Multiply the numerator and the denominator by the conjugate of the denominator

⇒ \(\frac{\sqrt{3} + i}{\sqrt{3} - i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i} \)

⇒ \(\frac{(\sqrt{3} + i)^2}{(\sqrt{3} - i)(\sqrt{3} + i)} \)

 \(\frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2} \)

Convert to polar form. The modulus r is 

\(r = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \)

The argument θ is 

\(\theta = \tan^{-1}\left( \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} \right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \)

So the polar form of the complex number is

\(1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) \)

To cube the complex number, we use De Moivre’s Theorem

In our case, r = 0 so,

\(\left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)^3 = \cos \pi + i \sin \pi = -1 + 0i = -1 \)

Thus, the result of cubing the complex number is \(\boxed{-1} \)

Hence, the correct answer is Option 1.