Question:

Step 1: Convert the complex number to polar form. First, we express the complex number $1-\sqrt{3}i$ in its polar form $r(\cos\theta + i\sin\theta)$. The modulus $r$ is given by: $$r = |1-\sqrt{3}i| = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$$ The argument $\theta$ is found using $\tan\theta = \frac{-\sqrt{3}}{1} = -\sqrt{3}$. Since the real part is positive and the imaginary part is negative, the angle lies in the fourth quadrant. Thus, $\theta = -\frac{\pi}{3}$. So, $1-\sqrt{3}i = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$. Step 2: Apply De Moivre's Theorem. Next, we raise $1-\sqrt{3}i$ to the power of 200 using De Moivre's Theorem, which states $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$. $$(1-\sqrt{3}i)^{200} = \left[2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)\right]^{200}$$ $$= 2^{200}\left(\cos\left(200 \times -\frac{\pi}{3}\right) + i\sin\left(200 \times -\frac{\pi}{3}\right)\right)$$ $$= 2^{200}\left(\cos\left(-\frac{200\pi}{3}\right) + i\sin\left(-\frac{200\pi}{3}\right)\right)$$ Now, we simplify the argument $-\frac{200\pi}{3}$. $$-\frac{200\pi}{3} = -\left(66\pi + \frac{2\pi}{3}\right)$$ Since cosine is an even function and sine is an odd function, and $\cos(n\pi + \alpha) = (-1)^n \cos\alpha$ and $\sin(n\pi + \alpha) = (-1)^n \sin\alpha$: $$\cos\left(-\frac{200\pi}{3}\right) = \cos\left(66\pi + \frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$ $$\sin\left(-\frac{200\pi}{3}\right) = -\sin\left(66\pi + \frac{2\pi}{3}\right) = -\sin\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$ So, $$(1-\sqrt{3}i)^{200} = 2^{200}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 2^{199} \cdot 2 \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 2^{199}(-1 - i\sqrt{3})$$ Step 3: Equate with the given expression and find $p$ and $q$. We are given that $(1-\sqrt{3}i)^{200}=2^{199}(p+iq)$. From Step 2, we have: $$2^{199}(-1 - i\sqrt{3}) = 2^{199}(p+iq)$$ Dividing both sides by $2^{199}$: $$-1 - i\sqrt{3} = p+iq$$ By comparing the real and imaginary parts, we find: $$p = -1 \quad \text{and} \quad q = -\sqrt{3}$$ Step 4: Calculate the two roots for the quadratic equation. The problem states that $p+q+q^2$ and $p-q+q^2$ are the roots of the equation. Let these roots be $r_1$ and $r_2$. $$r_1 = p+q+q^2 = -1 + (-\sqrt{3}) + (-\sqrt{3})^2 = -1 - \sqrt{3} + 3 = 2 - \sqrt{3}$$ $$r_2 = p-q+q^2 = -1 - (-\sqrt{3}) + (-\sqrt{3})^2 = -1 + \sqrt{3} + 3 = 2 + \sqrt{3}$$ Step 5: Form the quadratic equation. A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1+r_2)x + r_1r_2 = 0$. Calculate the sum of the roots: $$r_1+r_2 = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$$ Calculate the product of the roots: $$r_1r_2 = (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$$ Substitute these values into the quadratic equation formula: $$x^2 - (4)x + (1) = 0$$ $$x^2 - 4x + 1 = 0$$ Step 6: Match the equation with the given options. The equation found is $x^2 - 4x + 1 = 0$. This matches Option 2. The final answer is $\boxed{\text{Option 2}}$.