Key Concepts and Formulas

• Polar Form of Complex Numbers: A complex number $z = x + iy$ can be written as $z = r(\cos \theta + i \sin \theta)$, where $r = |z| = \sqrt{x^2 + y^2}$ is the modulus and $\theta = \arg(z)$ is the argument.
• De Moivre's Theorem: For any complex number $z = r(\cos \theta + i \sin \theta)$ and any integer $n$, $z^n = r^n(\cos(n\theta) + i \sin(n\theta))$.
• Quadratic Equation Formation: A quadratic equation with roots $\alpha$ and $\beta$ can be written as $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.

Step-by-Step Solution

Step 1: Convert the Complex Number to Polar Form

We are given the complex number $z = 1 - \sqrt{3}i$. We need to express this in the polar form $r(\cos \theta + i \sin \theta)$.

• Calculate the Modulus ($r$): The modulus is the distance from the origin. $r = |1 - \sqrt{3}i| = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

• Calculate the Argument ($\theta$): The argument is the angle with the positive real axis. Since the real part is positive and the imaginary part is negative, the number lies in the fourth quadrant. We find the reference angle $\alpha$ such that $\tan \alpha = \left| \frac{-\sqrt{3}}{1} \right| = \sqrt{3}$. This gives $\alpha = \frac{\pi}{3}$. Since it is in the fourth quadrant, $\theta = -\frac{\pi}{3}$. Thus, the polar form is $1 - \sqrt{3}i = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$.

Step 2: Apply De Moivre's Theorem to Find the Power

We need to raise the complex number in polar form to the power of $200$. ${\left( {1 - \sqrt 3 i} \right)^{200}} = {\left[ {2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)} \right]^{200}}$ Applying De Moivre's Theorem: $= {2^{200}}\left( {\cos\left( {200 \times \left(-\frac{\pi}{3}\right)} \right) + i\sin\left( {200 \times \left(-\frac{\pi}{3}\right)} \right)} \right)$ $= {2^{200}}\left( {\cos\left( {-\frac{200\pi}{3}} \right) + i\sin\left( {-\frac{200\pi}{3}} \right)} \right)$ Simplify the argument: $-\frac{200\pi}{3} = -\left(66\pi + \frac{2\pi}{3}\right)$. Using the properties of cosine and sine, and their periodicity: $\cos\left( -\frac{200\pi}{3} \right) = \cos\left( -\frac{2\pi}{3} \right) = \cos\left( \frac{2\pi}{3} \right) = -\frac{1}{2}$ $\sin\left( -\frac{200\pi}{3} \right) = \sin\left( -\frac{2\pi}{3} \right) = -\sin\left( \frac{2\pi}{3} \right) = -\frac{\sqrt{3}}{2}$ Therefore: $= {2^{200}}\left( {-\frac{1}{2} - i\frac{\sqrt{3}}{2}} \right)$ $= {2^{199}}\left( {-1 - i\sqrt{3}} \right)$

Step 3: Determine the Values of p and q

We are given that ${\left( {1 - \sqrt 3 i} \right)^{200}} = {2^{199}}(p + iq)$. From Step 2, we found that ${\left( {1 - \sqrt 3 i} \right)^{200}} = {2^{199}}\left( {-1 - i\sqrt{3}} \right)$. Comparing the two expressions: $p = -1$ $q = -\sqrt{3}$

Step 4: Calculate the Roots of the Quadratic Equation

The problem states that $p+q+q^2$ and $p-q+q^2$ are the roots of the equation. First, calculate $q^2$: $q^2 = (-\sqrt{3})^2 = 3$

Now, the first root, $\alpha$: $\alpha = p + q + q^2 = -1 + (-\sqrt{3}) + 3 = 2 - \sqrt{3}$

The second root, $\beta$: $\beta = p - q + q^2 = -1 - (-\sqrt{3}) + 3 = 2 + \sqrt{3}$

Step 5: Formulate the Quadratic Equation

A quadratic equation with roots $\alpha$ and $\beta$ is given by: $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$ $x^2 - (\alpha + \beta)x + \alpha\beta = 0$

• Calculate the Sum of the Roots ($\alpha + \beta$): $\alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$

• Calculate the Product of the Roots ($\alpha\beta$): $\alpha\beta = (2 - \sqrt{3})(2 + \sqrt{3}) = (2)^2 - (\sqrt{3})^2 = 4 - 3 = 1$

• Construct the Quadratic Equation: $x^2 - (4)x + (1) = 0$ $x^2 - 4x + 1 = 0$

Common Mistakes & Tips

• Quadrant Awareness: Be very careful about the quadrant when finding the argument of a complex number. Use the signs of the real and imaginary parts.
• De Moivre's Theorem Application: Ensure the angle is simplified after multiplying by the power. Use trigonometric identities and periodicity to simplify the angle.
• Sign Errors: Double-check all signs, especially with the quadratic formula and complex number arithmetic.

Summary

We converted the complex number to polar form, applied De Moivre's theorem, and then converted back to rectangular form to find $p$ and $q$. Using these values, we calculated the roots of the quadratic equation and then formed the equation itself. The final quadratic equation is $x^2 - 4x + 1 = 0$.

The final answer is \boxed{x^2 - 4x + 1 = 0}, which corresponds to option (B).