The value of $$\left(\frac{1 + \sin\frac{2\pi}{9} + i\cos\frac{2\pi}{9}}{1 + \sin\frac{2\pi}{9} - i\cos\frac{2\pi}{9}}\right)^3$$ is

Let $$\theta = \frac{2\pi}{9}$$. Note that $$\sin\theta = \cos(\pi/2 - \theta)$$ and $$\cos\theta = \sin(\pi/2 - \theta)$$.

$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}$$

Let $$a = 1 + \sin\theta$$, $$b = \cos\theta$$. Then:

$$\frac{a + ib}{a - ib} = \frac{(a+ib)^2}{a^2+b^2}$$

Note: $$\pi/2 - \theta = \pi/2 - 2\pi/9 = 5\pi/18$$

$$1 + \sin\theta = 1 + \cos(5\pi/18)$$, $$\cos\theta = \sin(5\pi/18)$$

Using $$1 + \cos\phi = 2\cos^2(\phi/2)$$:

$$a = 2\cos^2(5\pi/36)$$, $$b = 2\sin(5\pi/36)\cos(5\pi/36)$$

$$\frac{a + ib}{a - ib}$$ = write in polar form. $$a + ib = 2\cos(5\pi/36)[\cos(5\pi/36) + i\sin(5\pi/36)]$$

$$= 2\cos(5\pi/36) \cdot e^{i \cdot 5\pi/36}$$

Similarly $$a - ib = 2\cos(5\pi/36) \cdot e^{-i \cdot 5\pi/36}$$

$$\frac{a+ib}{a-ib} = e^{i \cdot 10\pi/36} = e^{i \cdot 5\pi/18}$$

Cubing: $$\left(\frac{a+ib}{a-ib}\right)^3 = e^{i \cdot 15\pi/18} = e^{i \cdot 5\pi/6}$$

$$= \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{i}{2} = \frac{1}{2}(-\sqrt{3} + i) = \frac{-1}{2}(\sqrt{3} - i)$$

The correct answer is Option 3: $$\frac{-1}{2}(\sqrt{3} - i)$$.