Key Concepts and Formulas

• Polar Form of Complex Numbers: A complex number $z = x + iy$ can be represented as $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, where $r = |z| = \sqrt{x^2+y^2}$ is the modulus and $\theta = \arg(z)$ is the argument.
• De Moivre's Theorem: For any real number $\theta$ and integer $n$, $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$. In exponential form, $(e^{i\theta})^n = e^{in\theta}$.
• Euler's Formula: $e^{i\theta} = \cos\theta + i\sin\theta$. A complex number $e^{i\phi} = 1$ if and only if $\phi = 2k\pi$ for some integer $k$.

Step-by-Step Solution

We are looking for the least positive integer $n$ such that: ${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1$

Step 1: Convert the numerator and denominator to polar form.

• Reasoning: Converting to polar form simplifies the division and exponentiation operations.

• Numerator $z_1 = 1 + i\sqrt{3}$:

  • Modulus: $r_1 = |1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$.
  • Argument: $\theta_1 = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.
  • Polar form: $1 + i\sqrt{3} = 2e^{i\frac{\pi}{3}}$.

• Denominator $z_2 = 1 - i\sqrt{3}$:

  • Modulus: $r_2 = |1 - i\sqrt{3}| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{4} = 2$.
  • Argument: $\theta_2 = \arctan\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$.
  • Polar form: $1 - i\sqrt{3} = 2e^{-i\frac{\pi}{3}}$.

Step 2: Simplify the fraction.

• Reasoning: Dividing complex numbers in polar form is straightforward.
• $\frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} = \frac{2e^{i\frac{\pi}{3}}}{2e^{-i\frac{\pi}{3}}} = e^{i\frac{\pi}{3} - \left(-i\frac{\pi}{3}\right)} = e^{i\frac{2\pi}{3}}$

Step 3: Substitute the simplified fraction into the original equation and apply De Moivre's Theorem.

• Reasoning: De Moivre's Theorem makes it easy to raise a complex number in polar form to a power.
• The original equation becomes: ${\left(e^{i\frac{2\pi}{3}}\right)^n} = 1$
• Applying De Moivre's Theorem: $e^{i\frac{2n\pi}{3}} = 1$

Step 4: Solve for the least positive integer $n$.

• Reasoning: $e^{i\theta} = 1$ if and only if $\theta$ is a multiple of $2\pi$.
• We must have $\frac{2n\pi}{3} = 2k\pi$ for some integer $k$.
• Dividing by $2\pi$, we get $\frac{n}{3} = k$, so $n = 3k$.
• The least positive integer $n$ occurs when $k = 1$, so $n = 3$.

Tips and Common Mistakes

• Argument Calculation: Be careful when finding the argument of a complex number; consider the quadrant in which the complex number lies.
• Using Euler's Formula: Remember Euler's formula ($e^{i\theta} = \cos\theta + i\sin\theta$) and its implications, especially when dealing with unity.
• Least Positive Integer: Always check if the question asks for the smallest positive integer, which often requires setting the parameter (like $k$ in this case) to its smallest possible positive value.

Summary We simplified the expression inside the parentheses by converting the numerator and denominator to polar form and then performing the division. Then, using De Moivre's Theorem, we raised the result to the power of $n$. Finally, we solved for the smallest positive integer $n$ such that the expression equals 1, which occurs when $n = 3$.

The final answer is \boxed{3}, which corresponds to option (B).