Key Concepts and Formulas

• De Moivre's Theorem: For any complex number $z = r(\cos \theta + i \sin \theta)$ and any real number $n$, $z^n = r^n(\cos(n\theta) + i \sin(n\theta))$.
• Polar Form of a Complex Number: A complex number $z = x + iy$ can be written as $z = r(\cos \theta + i \sin \theta)$, where $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}\left(\frac{y}{x}\right)$.
• Quadratic Equation from Roots: A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1 + r_2)x + r_1r_2 = 0$.

Step-by-Step Solution

Step 1: Convert the complex number $\sqrt{3} + i$ to its polar form.

Why? Converting to polar form allows us to easily apply De Moivre's Theorem.

• Calculate the modulus $r$: The modulus is $r = \sqrt{x^2 + y^2}$, where $x = \sqrt{3}$ and $y = 1$. $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$
• Calculate the argument $\theta$: The argument is $\theta = \tan^{-1}\left(\frac{y}{x}\right)$. Since both $x$ and $y$ are positive, $\theta$ is in the first quadrant. $\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$
• Write in polar form: $\sqrt{3} + i = r(\cos \theta + i \sin \theta) = 2\left(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}\right)$.

Step 2: Apply De Moivre's Theorem to calculate $(\sqrt{3} + i)^{100}$.

Why? De Moivre's Theorem provides a straightforward method for raising a complex number in polar form to a power.

• Using De Moivre's Theorem: $(\sqrt{3} + i)^{100} = \left[2\left(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}\right)\right]^{100} = 2^{100}\left(\cos \frac{100\pi}{6} + i\sin \frac{100\pi}{6}\right)$ $= 2^{100}\left(\cos \frac{50\pi}{3} + i\sin \frac{50\pi}{3}\right)$

Step 3: Simplify the argument $\frac{50\pi}{3}$.

Why? Simplifying the argument allows us to easily evaluate the trigonometric functions.

• $\frac{50\pi}{3} = \frac{48\pi + 2\pi}{3} = 16\pi + \frac{2\pi}{3}$ Since $16\pi$ represents 8 full rotations, we have $\cos \frac{50\pi}{3} = \cos \frac{2\pi}{3} = -\frac{1}{2}$ $\sin \frac{50\pi}{3} = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$

Step 4: Substitute the simplified trigonometric values and compare with the given expression.

Why? This allows us to determine the values of $p$ and $q$.

• $(\sqrt{3} + i)^{100} = 2^{100}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2^{99}(-1 + i\sqrt{3})$ We are given $(\sqrt{3} + i)^{100} = 2^{99}(p + iq)$. Comparing the two expressions, we get $p = -1$ $q = \sqrt{3}$

Step 5: Form the quadratic equation with roots $p = -1$ and $q = \sqrt{3}$.

Why? This fulfills the final requirement of the problem.

• The quadratic equation is given by $x^2 - (p+q)x + pq = 0$. $p + q = -1 + \sqrt{3} = \sqrt{3} - 1$ $pq = (-1)(\sqrt{3}) = -\sqrt{3}$ Substituting these values into the equation: $x^2 - (\sqrt{3} - 1)x - \sqrt{3} = 0$

Common Mistakes & Tips

• Be careful with the signs when determining the argument of the complex number.
• Ensure the argument is simplified to its principal value before evaluating trigonometric functions.
• Remember the correct formula for constructing a quadratic equation from its roots.

Summary

We converted the complex number to polar form, applied De Moivre's Theorem, simplified the argument, and compared the result with the given expression to find $p = -1$ and $q = \sqrt{3}$. Then we constructed the quadratic equation with these roots, which is $x^2 - (\sqrt{3} - 1)x - \sqrt{3} = 0$.

Final Answer

The final answer is \boxed{{x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0}, which corresponds to option (A).