1. Let $z=\frac{1+i\sqrt{3}}{1-i\sqrt{3}}.$ We need the least positive integer $n$ such that $z^n=1.$

2. Write numerator and denominator in polar form.

For $1+i\sqrt{3}$:

• modulus: $\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2$
• argument: $\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$ So, $1+i\sqrt{3}=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right).$

For $1-i\sqrt{3}$:

• modulus: $2$
• argument: $-\frac{\pi}{3}$ So, $1-i\sqrt{3}=2\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right).$

3. Divide the two complex numbers: $z=\frac{2\,\text{cis}(\pi/3)}{2\,\text{cis}(-\pi/3)}=\text{cis}\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right)=\text{cis}\left(\frac{2\pi}{3}\right).$ Thus, $z=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}.$

4. Now, $z^n=\text{cis}\left(\frac{2\pi n}{3}\right).$ For this to be equal to $1$, we need $\frac{2\pi n}{3}=2\pi k$ for some integer $k$.

Dividing by $2\pi$: $\frac{n}{3}=k$ So $n$ must be a multiple of $3$.

5. The least positive integer satisfying this is $n=3.$

6. Checking options:

• A: $2$ ❌
• B: $3$ ✅
• C: $5$ ❌
• D: $6$ is true but not least ❌

Therefore, the correct answer is B.