The least positive integer n for which $$\left(\frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}\right)^n = 1$$ is:

We start with the complex number

$$z=\dfrac{1+i\sqrt{3}}{1-i\sqrt{3}}.$$

To work comfortably, we first eliminate the complex number in the denominator. The standard trick is to multiply the fraction by the conjugate of the denominator. The conjugate of $$1-i\sqrt{3}$$ is $$1+i\sqrt{3}.$$ So we write

$$\dfrac{1+i\sqrt{3}}{1-i\sqrt{3}}\;=\;\dfrac{1+i\sqrt{3}}{1-i\sqrt{3}}\times \dfrac{1+i\sqrt{3}}{1+i\sqrt{3}}.$$

Now we evaluate the denominator first. A conjugate pair satisfies $$(a-bi)(a+bi)=a^2+b^2.$$ Here $$a=1$$ and $$b=\sqrt{3},$$ so

$$(1-i\sqrt{3})(1+i\sqrt{3})=1^2+(\sqrt{3})^2=1+3=4.$$

Next we expand the numerator:

$$(1+i\sqrt{3})^2 =1^2+2\cdot1\cdot i\sqrt{3}+(i\sqrt{3})^2 =1+2i\sqrt{3}+i^2\cdot3 =1+2i\sqrt{3}-3 =-2+2i\sqrt{3} =2\!\left(-1+i\sqrt{3}\right).$$

Putting numerator and denominator together gives

$$z=\dfrac{2\!\left(-1+i\sqrt{3}\right)}{4} =\dfrac{-1+i\sqrt{3}}{2}.$$

So

$$z=\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}\,i.$$

We now convert this cartesian form to polar (trigonometric) form. Recall the identities

$$\cos\theta=\dfrac{-1}{2},\qquad \sin\theta=\dfrac{\sqrt{3}}{2}.$$

These cosine and sine values correspond to the angle

$$\theta=\dfrac{2\pi}{3}\;(120^\circ)$$

because for that angle $$\cos\dfrac{2\pi}{3}=-\tfrac12$$ and $$\sin\dfrac{2\pi}{3}=+\tfrac{\sqrt3}{2}.$$

We also check the modulus (magnitude) of $$z$$:

$$|z|=\sqrt{\left(-\tfrac12\right)^2+\left(\tfrac{\sqrt3}{2}\right)^2} =\sqrt{\tfrac14+\tfrac34}=\sqrt1=1.$$

Since the modulus is $$1,$$ the polar form is simply

$$z=\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3} =e^{\,i\frac{2\pi}{3}}.$$

Now we have to find the smallest positive integer $$n$$ such that

$$z^{\,n}=1.$$

We use De Moivre’s Theorem, which states:

$$\bigl(r(\cos\theta+i\sin\theta)\bigr)^{\,n} =r^{\,n}\bigl(\cos n\theta+i\sin n\theta\bigr).$$

Applying it with $$r=1$$ and $$\theta=\dfrac{2\pi}{3},$$ we get

$$z^{\,n}=1^{\,n}\bigl(\cos(n\cdot\tfrac{2\pi}{3})+ i\sin(n\cdot\tfrac{2\pi}{3})\bigr) =\cos\Bigl(\dfrac{2\pi n}{3}\Bigr)+ i\sin\Bigl(\dfrac{2\pi n}{3}\Bigr).$$

For this to equal $$1,$$ the real part must be $$\cos0=1$$ and the imaginary part must be $$\sin0=0.$$ Hence the angle $$\dfrac{2\pi n}{3}$$ must differ from $$0$$ by a whole multiple of $$2\pi.$$ So we set

$$\dfrac{2\pi n}{3}=2\pi k,\qquad k\in\mathbb{Z}.$$

Dividing both sides by $$2\pi$$ gives

$$\dfrac{n}{3}=k.$$

So

$$n=3k.$$

The smallest positive integer value occurs when $$k=1,$$ giving

$$n=3.$$

Hence, the correct answer is Option D.